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Electricity and MagnetismA.S. Ramsey (1867-1954) was a distinguished Cambridge mathematician and President of Magdalene College. He wrote several textbooks ‘for the use of higher divisions in schools and for first-year students at university’. This book on electricity and magnetism, first published in 1937, and based upon his lectures over many years, was ‘adapted more particularly to the needs of candidates for Part I of the Mathematical Tripos’. It covers electrostatics, conductors and condensers, dielectrics, electrical images, currents, magnetism and electromagnetism, and magnetic induction. The book is interspersed with examples for solution, for some of which answers are provided.

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Electricity and Magnetism

An Introduction to the Mathematical Theory

Arthur Stanley R amsey

CAMbRID gE UnIvERSIT y PRESS

Cambridge new york Melbourne Madrid Cape Town Singapore São Paolo Delhi

Published in the United States of America by Cambridge University Press, new york

www.cambridge.orgInformation on this title: www.cambridge.org/9781108002592

© in this compilation Cambridge University Press 2009

This edition first published 1937This digitally printed version 2009

ISbn 978-1-108-00259-2

This book reproduces the text of the original edition. The content and language reflect the beliefs, practices and terminology of their time, and have not been updated.

ELECTRICITY AND MAGNETISM

LONDONCambridge University Press

FETTER LANE

NEW YORK • TORONTO

BOMBAY • CALCUTTA • MADRAS

Macmillan

TOKYO

Maruzen Company Ltd

All rights reserved

ELECTRICITY ANDMAGNETISM

An Introductionto the Mathematical Theory

by

A. S. RAMSEY, M.A.President of Magdalene College, Cambridge

formerly University Lecturer in Mathematics

CAMBRIDGEAT THE UNIVERSITY PRESS

1937

PBINTED IN GBEAT BRITAIN

PREFACE

This book has been written in response to suggestions fromfriends who have asked for a text-book on the subject adaptedmore particularly to the needs of candidates for Part I of theMathematical Tripos.

A complete study of the theory of electricity and magnetism,as a logical mathematical development from experimentaldata, requires a knowledge of the methods of mathematicalanalysis far beyond what can reasonably be expected frommost readers of an elementary text-book. The knowledge ofpure mathematics assumed in the present volume amounts tolittle more than some elementary calculus and a few propertiesof vectors. The ground is restricted by this limitation. Itcovers the schedule for Part I of the Tripos, including thefundamental principles of electrostatics, Gauss's theorem,Laplace's equation, systems of conductors, homogeneousdielectrics and the theory of images; steady currents in wires;elementary theory of the magnetic field and the elementaryfacts about the magnetic fields of steady currents. There arealso short chapters on induced magnetism and induction ofcurrents.

From one standpoint it would be preferable that a bookon a branch of Natural Philosophy should consist of a con-tinuous logical development uninterrupted by 'examples'.But experience seems to indicate that mathematical prin-ciples are best understood by making attempts to applythem; and, as the purpose of this book is didactic, I have hadno hesitation in interspersing examples through the chaptersand giving the solutions of some of them. The text is basedupon lectures given at intervals over a period of many years,and the examples are part of a collection which I began tomake for the use of my pupils about forty years ago, drawnfrom Tripos and College Examination papers.

As regards notation, I felt much hesitation about abandoningthe use of V for the potential of an electrostatic field; but

Vi PREFACE

the custom of using a Greek letter to denote the scalarpotential of a vector field has become general, and the matterwas decided for me when I found ' £ = —grad <f>' in theCambridge syllabus.

I am greatly indebted to Mr E. Cunningham of St John'sCollege for reading a large part of the text and making manyappropriate criticisms and useful suggestions; and also toDr S. Verblunsky of the University of Manchester for readingand correcting the proofs, and to the printers and readers ofthe University Press for careful composition and correction.

A. S. R.

Cambridge

November 1936

CONTENTS

Chapter I : PRELIMINARY MATHEMATICSARTICLE PAGE

1-2. Surface and volume integrals 11-3. Solid angles 21-4. Scalar functions of position and their gradients . . . 41-5, 1-51. A vector field. Flux of a vector 61-52. Divergence of a vector field 6

Chapter I I : INTRODUCTION TO ELECTROSTATICS

2-1. The electric field 102-11. The nature of electricity 112-12. Conductors and non-conductors 122-13. Electricity produced by induction 122-14. The electrophorus of Volta 132-15. The electroscope 142-16. Comparison of electric charges 142-2. The electric vector 162-21. Experimental basis 172-22. Lines and tubes of force . . . . . . . 172-23. Coulomb's law of force 182-3. Gauss's theorem and its consequences . . . . 19i2-4. The potential function 282-41. Poisson's and Laplace's equations 292-42. Equipotential surfaces 322-43. Lines and tubes of force 322-44. Theorems on the potential 332-6. Special fields—point charges . . . . . . 342-53. Two-dimensional field 392-6. Cavendish's experiment . . . . . . . 41

Examples 43

Chapter I I I : CONDUCTORS AND CONDENSERS

3-1. Mechanical force on a charged surface . . . . 483-2. Uniformly charged sphere . . . . . . . 493-3. Condensers 493-31. Parallel plate condenser 503-4. Spherical condenser 513-5. Use of Laplace's equation 553-6. Sets of condensers 573-7. Energy of a charged conductor 59

VU1 CONTENTS

ARTICLE PAGE

3-8. Approximate expression for capacity 603-9. Electrostatic units 613-91. Worked examples 62

Examples 66

Chapter IV: SYSTEMS OF CONDUCTORS

4-2. Principle of superposition 734-21. Uniqueness theorems 744-3. Coefficients of potential, capacity and induction . . . 754-4. Energy of a system of charged conductors . . . . 784-41. Green's reciprocal theorem 794-5. Energy as a quadratic function of charges or potentials . 824-6. Mechanical forces 834-7. Electric screens 864-8. Quadrant electrometer 874-9. Worked examples 88

Examples 90

Chapter V: DIELECTEICS

5-1. Specific inductive capacity 975-2. Electric displacement 985-21. Generalized form of Gauss's theorem 1005-3. The potential 1035-4. Comparisons 1045-41. Worked examples 105

Examples 110

Chapter VI: ELECTRICAL IMAGES

6-1. Theorem of the equivalent layer 1146-2. Images—definition 1156-21. Point charge and infinite plane 1166-3. Sphere and point charge 1176-4. Evaluation of induced charge 1236-5. Orthogonal spheres 1286-51. Series of images 1296-6. Field due to an electric doublet 1316-61. Conducting sphere in a uniform field 1326-7. Dielectric problems 1346-8. Two-dimensional images 1386-82. Capacity of a telegraph wire 140

Examples 141

CONTENTS IX

Chapter VII: ELECTRIC CURRENTS

ABTIOLB PAGE

7-1. Current strength 1487-11. The voltaic cell 1497-12. Electromotive force 1497-2. Electrolysis 1507-3. Ohm's Law 1527-31. Resistance of a set of conductors 1537-33. Units 1557-4. Joule's Law 1567-5. Kirchhofi's Laws 1567-6. Wheatstone's bridge 1607-7. Currents in a network. Reciprocal theorems . . . 1637*71. Minimum rate of heat production 1647-8. Telegraph wire with a ' faul t ' 1667-9. Worked examples 168

Examples 171

Chapter VIII: MAGNETISM

8-1. Earth's magnetic field 1808-12. The fundamental vectors 1818-2. Field of a magnetic bipole 1828-24. Oscillations of a small magnet 1858-3. Potential energy of a magnetic bipole 1868-31. Mutual potential energy of two small magnets . . . 1878-34. Coplanar magnets 1918-4. Gauss's verification of the law of inverse squares . . 1938-41. Experimental determination of magnetic moment . . 1958-6. Finite magnets. Intensity of magnetization . . . 1978-61. Uniformly magnetized sphere 1988-7. Magnetic shells 200

Examples 203

Chapter IX: ELECTROMAGNETISM

9-11. Magnetic field of a straight current 2099-12. Potential. Ampere's circuital relation 2109-2. Comparison with magnetio shell 2139-3. Parallel straight currents 2159-4. Field due to a circular current 2179-43. Solenoids 2209-5. Potential energy of uniform shell 2229-6. Mechanical force on a circuit 2239-61. Plane circuit in a uniform field 2259-8. Tangent galvanometer 228

X CONTENTS

ARTICLE PAGE

9-83. Moving coil galvanometer . 2309-84. Ballistic galvanometer 231

Examples 233

Chapter X: MAGNETIC INDUCTION ANDINDUCED MAGNETISM

10-1. Magnetic induction 23910-11. Flux of induction out of any closed surface zero . . . 24010-2. Continuity of normal induction 24210-3. Induced magnetism 24310-4. Equations for the potential 24410-5. Examples of induced magnetism 246

Examples 250

Chapter XI: ELECTEOMAGNETIC INDUCTION

11-1. Faraday's experiments 25211-2. Law of electromagnetic induction 25311-3. Self-induction and mutual induction 25411-4. Single circuit with self-induction 25611-5. Periodic electromotive force 25711-51. Plane circuit rotating 25911-6. Circuit containing a condenser 26011-7. Examples of circuits with inductance 262

Examples 265

Table of Units

C.G.S. absolute unit of force = 1 dynec.G.s. absolute unit of work or energy = 1 erg

ELECTRICAL UNITS

Practical units

Charge 1 coulombPotential or electro- i , 1+

motive force M v o l t

Current 1 ampereResistance 1 ohmCapacity 1 faradInductance 1 henryRate of working 1 watt

Equivalent absoluteo.G.s. units

Electrostatic

3x10"

3-1 x 10-23x10°

3-a x 10-ii

32 x 10"3-2 x 10-"

10'

Electro-magnetic

IO-1

108

io-1

10"io-9

10»10'

One microfarad is one-millionth of a farad.An electromotive force of 1 volt drives a current of 1 ampere through

a resistance of 1 ohm and work is then being done at the rate of 1 wattor 10' ergs per second.

Chapter I

P R E L I M I N A R Y MATHEMATICS

i ' l . We propose in this chapter to give a brief account ofsome mathematical ideas with which the reader must befamiliar in order to be able to understand what follows in thisvolume.

1*2. Surface and volume integrals. Though the processof evaluating surface and volume integrals in general involvesdouble or triple integration and must be learnt from books onAnalysis, yet in theoretical work in Applied Mathematicsconsiderable use is made of surface and volume integralswithout evaluation, and we propose here merely to explainwhat is implied when such symbols as

\f(x,y,z)dS and \f(x,y,z)dv

are used to denote integration over a surface and throughouta volume.

f6Adefiniteintegralofafunctionofonevariable, say f(x)dx,

J amay be denned thus: let the interval from a to b on the #-axisbe divided into any number of sub-intervals 8l9 82, ... 8n, andlet/ r denote the value of f(x) at some point on 8r; let the sum»2 fr8r be formed and let the number n be increased without

nlimit. Then, provided that the limit as n-+oo of 2 fr8r exists

and is independent of the method of division into sub-intervals and of the choice of the point on 8r at which the valueof f(x) is taken, this limit is the definite integral of f(x) froma to b.

In the same way we may define \f(x,y,z)d8 over a given

surface; let the given surface be divided into any number ofsmall parts 8l9 82, ... 8n and let/ r denote the value off(x,y,z)

REM I

2 SOLID ANGLES [1*2-

nat some point on 8r, then the limit as n -> oo of 2 fr 8r, provided

r=l

the limit exists under the same conditions as aforesaid, is

defined to be the integral f(x, y, z) dS over the given surface.Any difficulty as to the precise meaning to be attributed to

' area of a curved surface' may be avoided thus: after choosingthe point on each sub-division Sr of the surface at which thevalue of f(x,y,z) is taken, project this element of surface onto the tangent plane at the chosen point, and take the planeprojection of the element as the measure of Sr in forming thesum.

The integral / (x, y, z) dv through a given volume may be

defined in the same way.

1*3. Solid angles. The solid angle of a cone is measuredby the area intercepted by the cone on the surface of a sphere ofunit radius having its centre at the vertex of the cone.

The solid angle subtended at a point by a surface of any formis measured by the solid angle of the cone whose vertex is atthe given point and whose base is the given surface.

Let PP' be a small element of area dS which subtends asolid angle da> at 0.

Let the normal to dS make an acute angle y with OP and letOP—r. Then the cross-section at P of the cone which PP'subtends at 0 is dS cos y, and this cross-section and the small

1*32] SOLID ANGLES 3

area du> intercepted on the unit sphere are similar figures, sot h a t dS cos y: da, = r 2 : l .Whence da, = (dS cos y)/r2) .or d8 = r28ecydu> j

I t follows that the area of a finite surface can be representedas an integral over a spherical surface, thus

8= Waecydw (2)

with suitable limits of integration.

1*31. dto in polar co-ordinates, dw is an element of thesurface of a unit sphere. Let theelement be PQB8 bounded bymeridians and small circles, wherethe angular co-ordinates of P are8, </>. Then since the arc P8 sub-tends an angle d<f> at the centre ofa circle of radius sin#, therefore

= sin 8 d(f>; and PQ=dd, so that

1*32. Solid angle of a right circular cone. A narrow zone ofa sphere of radius a cut off between parallel planes may beregarded as a circular band of breadth add and radius a sin. 8,so that its area = 2TT«2 sin Odd

= — 2nadx, where x = a cos 8.B

Hence the area of a zone of finite breadth

= 2TT<Z (XX — x2)

= circumference of sphere x axial breadth of zone.A right circular cone BOC of vertical angle 2a intercepts on

4 SOLID ANGLES [1*32-

a unit sphere of centre 0 a cap B'A'C of height MA' = 1 — cos aand area „ ,, . .,.

27r(l-cosa) (1),so that this is the measure of the sohd angle of the cone.

1#33. The idea of the sohd angle may easily be extended, ifwe observe that any bounded area on a unit sphere may beregarded as measuring a sohd angle.Thus a lune bounded by semi-circlesABD, A CD may be taken as measur-ing the sohd angle between thediametral planes ABD, ACD.

Let a be the angle between theseplanes. Because of the symmetryabout AD, it is evident that

area of lune: area of sphere = a.: 2TT.

But the area of the sphere is 4TT, SOthat the area of the lune is 2a; or the sohd angle between twoplanes is twice their inclination to one another.

1-4. Scalar functions of position and their gradients. Let<f> (x, y, z) be a continuous single-valued function of the positionof a point in some region of space. Suppose that the functionj> is not constant throughout any region, so that the equation

<f> (x, y, z) = const.

represents a surface. We assume that through each point ofthe region in which <j> is defined, there passes a surface^ = const. We also assumethat at every point P onthis surface there is adefinite normal PN andthat the tangent plane at Pvaries continuously withthe position of P on the surface.

From the definition of (f> two surfaces

<f>(x,y,z) = a and <f>(x,y,z) = b

cannot intersect; for if they had a common point it would be

1*4] GRADIENT 5

a point at which <f> had more than one value, in contradictionto the hypothesis that </> is a single-valued function.

Consider two neighbouring surfaces

(f>=a a n d <f> = a + 8a.

Let P, P' be points on each and let the normal at P to <f>=ameet <f> = a + 8a in N. For small values of 8a PN will also benormal to <f> = a + 8a.

Then using <j>P to denote the value of (f> at P, we have

<f>P—<f>P = 8a =<f>N-(/>P = <f>N-<l>p PNPP' PP' PP' PN ' PP'

_<!>N-<I>P a- pN cost/,

where 6 is the angle NPP'.Now if PP' = 8s and PN = 8n, and we make 8a and therefore

8s and 8n tend to zero, the limit of (<f>P> — <f>P)/PP' is the rate of

increase of </> in the direction 8s and is denoted by ^-; and

similarly the limit of (<f>N — (j>p)jPN is the rate of increase of </>

in the normal direction 8n and is denoted by -^-, and we haveon

(1).

Thus we have proved that the space rate of increase of <f> inany direction 8s is the component in that direction of its spacerate of increase in the direction normal to the surface <f> = const.;or that if we construct a vector of magnitude d<f>/dn in directionPN, then the component of this vector in any direction is thespace rate of increase of <f> in that direction.

The vector d<f>jdn with its proper direction is called thegradient of <j> and written grad <j>.

To recapitulate: ^ is a continuous scalar function of positionhaving a definite single value at each point of a certain regionof space, and the gradient of <f> is defined in this way: throughany point P in the region there passes a surface <j> = const., thena vector normal to this surface at P whose magnitude is thespace rate of increase of <j> in this normal direction is defined to

6 PLTJX OF A VECTOR [1-4-

be the gradient of <f> at P, and it has the property that its com-ponent in any direction gives the space rate of increase of <f>at P in that direction. It is clear that the gradient measuresthe greatest rate of increase of ^ at a point.

1*5. A vector field. If to every point of a given regionthere corresponds a definite vector A, generally varying itsmagnitude and direction from point to point, then the regionis called a vector field, or the field of the vector A; e.g. electricfield, magnetic field.

1*51. Flux of a vector. If a surface S be drawn in the fieldof a vector A and An denotes the component of A normal to

an element dS of the surface, then the integral AndS is called

the flux of A through S. Since a surface has two sides the senseof the normal must be taken into account, and the sign of theflux is changed when the sense of the normal is changed. Theflux of a vector through a surface is clearly a scalar magnitude.

1-52. Divergence of a vector field. Let A denote avector field which has no discontinuities throughout a givenregion of space. Let 8v denote any small element of volume

containing a point P in the region and let L4md/8I denote the

outward flux of A throughthe boundary of 8v, then the

limit as of8v

O.is defined to be ths diver-gence of A at the point Pand denoted by div A.

I t can be shewn that,subject to certain conditions, this limit is independent ofthe shape of the element of volume Sv, but for our presentpurpose, which is to obtain a Cartesian form for div A, itwill suffice to calculate the limit for a rectangular elementof volume. Using rectangular axes let P be the centre (x, y, z)

1*52] DIVERGENCE OF A VECTOR FIELD 7

of a small rectangular parallelepiped with edges parallel tothe axes of lengths 8x, 8y, 8z.

Let the vector A have components Ax, Ay, Az parallel tothe axes at P .

Consider the contributions of the faces of the parallelepipedto the flux of the vector out of the element of volume. The twofaces parallel to the xy plane are of area 8a; 8y, the componentof A normal to these at the centre (x, y, z) of the parallelepipedis Az. The co-ordinates of the centres M, N of these faces arex,y,z — \8z and x,y,z + \8z; so that if the magnitude of As

at P is f(x,y,z), its magnitude at M is / (x, y, z — \8z) or

f(x, y, z) — -~8z, to the first power of 8z, i.e. Az — --~8z,Z oz Z oz

1 dAand similarly the magnitude at N is Ae + — - ~ 8z, and both

A ozthese components are in the direction Oz. Then assumingwhat can easily be proved, that, subject to certain conditions,the average value of the component over each small rectangleis the value at its centre, the contributions of these two facesto the total outward flux are

and

dAgiving a sum

vzFinding similarly the contributions of the other two pairs

of faces, we have for the total outward flux

\ 3a; dy c

to this order of small quantities.But the volume 8v of the small element is 8x8y8z, so that in

accordance with our definition, dividing the flux by the volumeand proceeding to the limit in which the terms of higher orderin the numerator disappear, we have

dAx dAy dAz~~X r* ~ T N " "T* ~7C • • • • • • • • • • • • • • • ( X l *

ox oy oz

8 DIVERGENCE IN POLAR CO-ORDINATES [1"53-

1*53. Divergence in polar co-ordinates. Let P be thepoint (r, 8, to) and suppose it to be the centre of an element ofvolume ABGDA'B'O'D', whose faces ABCD, A'B'C'D' areportions of spheres of radii r + \8r, ADD'A', BGG'B' areportions of cones of angles 8 +|80, andABB'A', DGC'D' are planes a> + |8w. Thelengths of the edges of the element ofvolume are Sr, r86 and rsin08a> and itsvolume is r2sin08r808a>.

Let Ar, AB, Am denote the componentsof the vector A at P in the directions inwhich r, 8, u> increase, i.e. perpendicular tothe faces of the element. The cross-sectionof the element through P at right angles to Ar is of arear2sin08#8to, so that the flux of A through this cross-sectionis Arr

28in.8888(o. Hence the outward flux across the parallelsection ABCD which only differs from that through P byhaving r — J8r instead of r is

- \ Arr* 8in 8 88 8o>-^r.^iArr* sin 8 88 8co)\+ei,

and the flux across A'B'C'D' is in like manner

where ex, e2 are small quantities of the fourth order in Sr, 88,8<o.Hence this pair of opposite faces contribute an amount

j£ (rMr) sin 88rS88(o + ex + e2

to the total outward flux.It may be shewn in the same way that the faces ADD'A'

and BCC'B' contribute

and

1*54] CYLINDRICAL CO-ORDINATES

and that the faces ABB'A', DCC'D' contribute

and i ^ \

where e3, e4, e5, e6 have like meanings.Hence the total outward flux from the six faces is

\-21- (r*Ar) + —i-g A (sin 0Ae) + —l—Q % M r* sin 68r 80 8w + e,(r2dr r rsinddd ° raw.8 dm)

where e is of the fourth order in Sr, $6, Sw.Therefore if we divide the flux by the volume and then make

8r, 88, 8a> tend to zero, we get for the divergence at P

i | 1 4 ^ ^ ...(1).

1*54. Divergence in cylindrical co-ordinates. Usingcylindrical co-ordinates r, 6, z and taking an element of volumeof edges Sr, rSO, 8z with its centre at (r, 9, z), it can be shewnin the same way that

1 3 . , . 19^0 dA.(A) + ^ + ^

where Ar, Ag, Az are the components of A in the directions ofthe increments in r, 6 and z.

Chapter II

INTRODUCTION TO ELECTROSTATICS

2*1. The electric field. I t was known to the Greeks andRomans that when pieces of amber are rubbed they acquirethe power of attracting to themselves light bodies. There areother substances which possess the same property; thus, if astick of sealing-wax is rubbed on a piece of dry cloth it willattract bran or small scraps of paper sufficiently near to it.The same result is obtained if a glass rod is rubbed with a drypiece of silk. It is also found that the cloth and the silk acquirethe same property as the sealing-wax and the glass rod.Further, if the sealing-wax is suspended so that it is free tomove, it is found that the cloth attracts the sealing-wax, buttwo pieces of sealing-wax similarly treated repel one another.In the same way the glass rod and the silk attract one another,but two such glass rods repel one another.

We describe bodies in such a state as electrified or chargedwith electricity. The word was derived by William Gilbert* fromrjXeKTpov or amber, the first substance upon which such experi-ments were performed.

If we experiment further we find that an electrified stick ofsealing-wax is attracted by an electrified glass rod, but repelledby the piece of silk with which the rod has been rubbed.

These and kindred phenomena are explained by the state-ment that electricity is of two kinds, and that charges of the samekind repel while charges of opposite kinds attract one another.

It is convenient to describe the two kinds of electricity aspositive and negative, that produced as above on the sealing-wax is called negative and that on the glass rod positive; butit must be pointed out that this is merely a convenient arbi-trary nomenclature and that the opposite would have answeredall purposes equally well.

* William Gilbert (1540-1603), a native of Colchester, Fellow ofSt John's College, Cambridge.

2'11] THE NATURE OP ELECTRICITY 11

If we try the same experiment using a metal rod held in thehand instead of the glass rod or stick of sealing-wax, no resultis obtained.

If a small pith ball coated with gold leaf is suspended by asilk fibre and allowed to touch an electrified rod, it appears toacquire a charge of the same kind as the rod, for after contactit is repelled from the rod.

If such a charged pith ball is removed to a distance from allother bodies, it does not appear to be acted upon by any forcessave its weight and the tension of the supporting fibre; butwhen brought into the neighbourhood of other charged bodiesit appears to be subject to an additional force, which at everypoint has a definite direction.

It follows that the properties of space in the neighbourhoodof charged bodies appear to differ from those of the rest ofspace. The space in the neighbourhood of charged bodies iscalled an electric field. The field in general extends throughall space, but its intensity (2-2) diminishes as distance fromcharged bodies increases. There are natural electric fields ofgreat intensity, such as those due to the presence of highlycharged regions in the atmosphere; e.g. thunder-clouds.

I t is our object to formulate a working hypothesis and buildupon it a mathematical theory which will describe correctlythe phenomena which take place in an electric field. But beforewe attempt to do this we must make a further appeal toexperiment.

2-11. The nature of electricity. The modern answer tothe question 'what is electricity?' is that it is 'a fundamentalentity of nature', an answer which gives the minimum ofinformation. For a long time electricity was considered to beof continuous fluid form, though it only resembles a fluid inthat it moves in 'currents'. I t is now known to be of 'atomicstructure'. The existence of the electron or ultimate indivisiblenegative charge was demonstrated by J. J. Thomson in 1897.The corresponding positive charge is called a proton. Inaccordance with the electron theory of matter, an atom of anelement consists of a central nucleus composed of protons and

12 CONDUCTORS AND NON-CONDUCTORS [2*11-

electrons with a total positive charge, round which electronsmove in orbits like planets round the sun, the number ofelectrons in an uncharged atom being just sufficient to balancethe positive charge of the nucleus. The nucleus is small in sizecompared to the dimensions of the orbits of the revolvingelectrons; but the mass of the atom is almost entirely con-centrated in the nucleus. Recent experimental work in theCavendish Laboratory has demonstrated the existence of theproton and also of an uncharged particle of the same masscalled a neutron.

A charged body means one which contains either protonsor electrons, or both protons and electrons, apart from thosewhich compose the atoms of the body.

2*12. Conductors and non-conductors. An electrifiedmetal-coated pith ball looses its charge if touched by a metalrod held in the hand, but not if touched by a dry glass orebonite rod.

The explanation of the disappearance of the charge is thata metal rod acts as a ' conductor' and a passage of electricitytakes place along it. Thus a positive charge disappears eitherbecause it escapes to earth along the rod, or because an equalquantity of negative electricity passes from the earth along therod to neutralize the charge on the metal-coated ball, orpartly from the one cause and partly from the other. Dryglass and ebonite do not permit of a like passage of electricity.

This and kindred experiments enable us to classify substanceseither as 'conductors' of electricity, or as 'non-conductors','insulators', or 'dielectrics'. Among conductors are the earth,metals, water, carbon and the human body; and among in-sulators are dry air, glass, sulphur and ebonite. Some sub-stances are better conductors or better insulators than others,and there are some substances such as wood which may beclassed either among bad conductors or among bad insulators.

2-13. Electricity produced by induction. Unchargedconducting bodies contain both positive and negative elec-tricity, i.e. protons and electrons, so distributed as to neu-tralize one another's effects. But when such a body is brought

©c

2'14] INDUCTION 13

into an electric field the distribution of the electricity isaltered and becomes positive and negative charges on differentparts of the surface of the body.

Thus let A represent a metal body supported on an insulatingstand. When a positively charged body B is brought into theneighbourhood of A, negative electricity in A is attracted to-wards B and positive is repelled from B to the remoter partsof A. The presence of thelatter can be demonstrated byallowing a metal-coated pithball C to come into contactwith A; part of the positivecharge on A then passes toC and it is repelled from A.

If, further, the conductorA is touched by the finger, thepositive charge on A is repelled by the charge on B through thehuman body to the earth, and when contact is broken theconductor A is left with a negative charge which it retainsafter A is moved out of the range of the influence of thecharged body B. The body A is then said to have been chargedby electrostatic induction.

2*14. The electrophorus of Volta.* This is the simplestmachine for producing a succession of electric charges. Itconsists of a circular metal plate A to one side of which aninsulating handle B is attached, and aslightly larger circular slab of shellac orresin C. The diagram shews the apparatusin section. The upper surface of the slabof shellac is electrified by rubbing withcat's skin, and the metal plate is thenplaced upon it. Owing to the roughness I ^3of the surface, actual contact onlyexists at a few points. The electrified shellac plays the partof the body B of 2*13 and the metal plate the part of A.The negative electricity on the shellac attracts positive

* Alessandro Volta (1745-1827), Italian physicist.

14 THE ELECTROSCOPE [2*14-

electricity to the under side of the metal plate A and repelsnegative electricity to the upper side. The plate is touched bythe finger so that the negative electricity on its upper sideescapes to earth. The plate can then be lifted off the shellacby the insulating handle and it now possesses a positive charge.

2#15. The electroscope. An electroscope is an apparatusfor detecting the presence of a charge of electricity. A simpleform consists of a pair of small sheets of gold leaf a, a, so sus-pended that when uncharged they hang in a vertical plane.They are connected by a brass rod b to a brass plate or aknob 0. The rod passes through a cork in theneck of a glass bell-jar which protects the goldleaves from air currents. To avoid irregulareffects which might result from a possibleuneven electrification of the glass, a cylinder ofmetal gauze slightly smaller than the cylindricalpart of the bell-jar is placed inside it and con-nected to earth. This effectively screens the leavesfrom any electrification of the glass. When acharged body is brought near to the plate C,electricity is repelled into the gold leaves andcauses them to repel one another and separate.In fact the parts a, b, C of the apparatus can be chargedby induction in the manner described in 2*13. If whenthe gold leaves are positively charged a negatively chargedbody such as a stick of sealing-wax is made to approachthe plate C, the divergence of the leaves will diminish, and theapproach of a sufficiently strong negative charge will cause theleaves to collapse and then re-separate charged negatively.

When the gold leaves become charged, electricity is inducedon the metal gauze cylinder and this tends to increase thedivergence of the leaves and so renders the instrument moresensitive.

2'16. Comparison of electric charges. Experiments maybe performed by standing on the plate of an electroscope ametal vessel which may be closed by a metal lid in which area few small holes through which silk threads can be passed.

2'16] COMPARISON OF ELECTRIC CHARGES 15

By this means a charged body can be suspended inside theclosed vessel and raised or lowered at will without contactwith the vessel.

The effect of suspending a body charged with electricityinside the vessel is to attract electricity of the opposite kindto the inner surface of the vessel, and repel electricity of thesame kind as the given charge to the outer surface of the vesseland so to the gold leaves. And it is found that the amount ofdeflection of the gold leaves does not depend on the positionof the charged body inside the vessel so long as the body andthe vessel are not in contact. It follows that the externalelectrification of the vessel only depends on the total chargeof the body inside it and not on the position of this charge.If two charged bodies suspended in turn inside the vesselproduce the same divergence of the gold leaves, we say thattheir charges are equal; and if two charged bodies when sus-pended simultaneously inside the vessel produce no divergenceof the gold leaves, we say that their charges are equal andopposite.

By experiments of this kind it can be shewn that whenelectricity is produced by friction the amounts of the twokinds of electricity produced are equal and opposite.

We can also shew that the total charge on two bodies isunaltered by allowing them to touch or by connecting themwith one another by a conductor.

Again, if the outer surface of the vessel is connected to theearth (e.g. by touching it with a finger) its external chargedisappears and at the same time the gold leaves collapse, sothat the total charge inside the vessel is now zero, whichimplies that the charge on the inner surface of the vessel isequal and opposite to the charge on the body suspendedwithin it.

Such experiments therefore give us a standard of com-parison, in that, if a charged body G suspended inside thevessel produces the same divergence of the gold leaves as isproduced by the joint effect of two bodies A and B similarlysuspended, we say that the charge on G is equal to thesum of the charges on A and B, and hence that electricity is

16 THE ELECTRIC VECTOR [2'16-

meaaurable as regards quantity. We shall have occasion later todefine a unit of electricity; for the present it is sufficient torecognize the fact that charges are measurable in terms ofsome unit of charge.

2*2. The electric vector. When an electric field is exploredby placing a small charged body at different points in turn, itis found that the body is acted upon by a force having adefinite magnitude and direction at every point of the field,and that the magnitude F of this force is the product of twoquantities, one of which e denotes the charge on the smallbody and the other E depends upon the field; so that, writtenas a vector equation,

F = eE.

E therefore denotes the force per unit charge at any point ofthe field; it is called the electric intensity, or the electricvector, or simply the field.

The foregoing exploration of the field involves the carryingout of a number of experiments, in some of which the chargeon the small body is constant while its position in the fieldis altered; and in others the charge on the body is alteredbut its position remains fixed.

There may be points or regions of equilibrium where E iszero, but in general E is variable from point to point and hasa definite magnitude and direction at each point of the field.

It must be understood that when we speak of exploring agiven field as above with a small charged body, we assume thatthe charge carried by the body does not disturb the field by itspresence. We also assume the existence of a mechanical sup-port for the small charged body capable of balancing the forceF due to the electric field. If there were no such mechanicalsupport, the small charged body would move under theinfluence of the electric field.

We may now define a conductor explicitly as a substancein which, if there is an electric field, then there is a flow ofelectricity; or alternatively, as a substance in which, if thereis no flow, then there is no field. A non-conductor is asubstance in which there is no flow, though there may be a field.

2-22] EXPEBIMENTAL BASIS 17

Later we shall have to consider the motion of electricity, butour present object is to establish a theory of electrostatics;i.e. of electric fields in which there is no motion of electricity.

2-21. Experimental basis. We shall assume that thefollowing are demonstrable facts, though in some cases thestrongest evidence for the truth of such statements lies not indirect experiment but in the accord of the general theorydeduced therefrom with experiments of a more general kind:

(a) When electricity is produced by friction or by inductionthe quantities of positive and negative electricity producedare equal.

In a state of equilibrium(6) there is no electric field inside a hollow conductor which

contains no charge;(c) there is no free electric charge within the substance of

a conductor;(d) at the surface of a conductor the electric intensity is

normal to the surface.

I t will appear later that (d) is contained in the definitionof a conductor. For it will be shewn that, in crossing aboundary surface, the tangential component of the electricintensity is continuous, and, as there is no tangential com-ponent inside the conductor, neither can there be a tangentialcomponent on the outside of the surface.

I t follows that in equilibrium electric charges reside on thesurfaces of conductors. Also that there are abrupt changes inthe electric field at the surfaces of conductors, there being afield outside a charged conductor and no field in its substance,and these abrupt changes are due to the presence of chargeson the surfaces of the conductors. We may therefore say thatan electric field is bounded by electric charges.

2*22. Lines and tubes of force. A line in an electric fieldsuch that its direction at every point is the direction of theelectric vector at that point is called a line of force. I t followsthat lines of force cannot intersect one another, for there isonly one direction for the electric vector at any point in the

18 COULOMB'S LAW OP FORCE [2*22-

field. If lines of force are drawn through every point of a smallclosed curve, they lie on a tubular surface called a tube of force.

In order to be explicit we shall assume that the smallcharged body which we use for exploring an electric field hasa positive charge, so that Unes of force are considered as directedin the sense in which a small positive charge would move if itwere free to do so.

It appears from 2*21 (d) that if a conducting surface ischarged positively lines of force start away from the surfacealong the normals, and since these lines cannot intersect andare always in the sense in which a positive charge would movethey must continue through space until they arrive at anegatively charged surface.

Consider the field of a conductor with a charge + e, andsuppose that we divide its surface into areas each of whichcarries a unit of charge and draw the tubes of force throughthe boundary curves of these areas; we may call these unittubes since each starts from a unit of charge. These tubes willcontinue through space until they fall on some other con-ducting surface or surfaces.

2*23. Coulomb's Law of Force. It is possible and has beencustomary to build up a theory of electrostatics starting fromthe fact that the force between two small charges is proportionalto the product of the charges and inversely proportional to thesquare of the distance between them, without inquiry at theoutset as to how this action at a distance is produced. This lawis generally known as Coulomb's Law of Force, since Coulomb*was the first to publish an experimental verification.

The graphic representation of an electric field obtained bydrawing Unes of force was introduced by Faraday,f who wasthe first to endeavour to explain electrical effects as the resultof interactions in the medium in which charges are placedrather than as the result of charges acting upon one anotherat a distance.

* Charles Augustin Coulomb (1736-1806), French physicist.t Michael Faraday (1791-1867), Director of the Laboratory of the

Royal Institution. Author of Experimental Researches in Electricityand Magnetism.

2*3] GATJSS'S THEOREM 19

2-231. The ether. According to modern views actions inan electric field require a medium for their transmission, andcharges at a distance can only affect one another by means ofstresses in this intervening medium. I t is usual to speak ofthis medium as the ether. The ether is regarded as filling allspace including the space occupied by material bodies. It trans-mits radiation (heat, light, electromagnetic waves), but doesnot emit or absorb it. On this hypothesis it is necessary to con-sider electric effects as effects localized in a medium and expressthe mathematical formulae of the subject in a localized form.

We have seen how by means of Faraday's lines of force wecan make a picture of an electric field which shews the directionof the field at every point, and, in order to proceed further onthis basis, we must make a hypothesis concerning the magni-tude of the electric intensity, and it must be a hypothesiswhich will accord with the facts of observation so far as wecan state them. In the following article we shall state thisfundamental hypothesis.

2*3. Gauss's Theorem. Theoutwardfluxof electric intensitythrough any closed surface is proportional to the total chargewithin the surface.

Thus if E denotes the electric intensity, the outward fluxof E through a closed surface 8 means the surface integral over

8 of the normal component of E, i.e. EndS, where En is the

component of E normal to dS (1*51); and this fundamentaltheorem, known as Gauss's Theorem,* states that this integralis proportional to the sum of the charges inside 8, or

End8=ftZe,

where h is a constant and 2e denotes the sum of the charges.In electrostatic units, which we shall presently define, theconstant h is 4TT, and the theorem is

P (1).

* Karl Friedrich Gauss (1777-1855), German mathematician andphysicist.

2-3

20 CONSEQUENCES OP GAUSS'S THEOREM [2"3-

If the electric charge be considered to exist as a volumedistribution of density p inside 8, including the possibility ofp = 0 through the whole or any part of the region, the theoremmay be written

( (2),

where the integration on the right extends to all parts of theregion inside S at which there is electric charge.

2'31. Consequences of Gauss's Theorem. The reader willnotice that we do not offer any proof of Gauss's Theorem butstate it as a fundamental hypothesis. The justification fordoing so is that the theory which we are able to build up onthis hypothesis accords with experiment. It will appear laterthat Coulomb's Law of Force can be deduced from Gauss'sTheorem, or vice versa, and on the whole it is more satisfactoryto take the latter rather than the former as our fundamentalhypothesis and so avoid basing our theory on the idea ofaction at a distance.

We shall now indicate how Gauss's Theorem accords withsome results already obtained.

(i) Let the surface S of the theorem be drawn in the sub-stance of a conductor not containing a cavity; then, bythe definition of a conductor, E is zero at every point of 8;hence by Gauss's Theorem the total charge inside 8 is zero, andthis is true however small the space enclosed by 8. It followsthat there can be no charge at any point in the substance ofa conductor in equilibrium, thus confirming 2-21 (c).

(ii) If the region bounded by S contains no charge, then thetotal flux of E through S iszero, or the flux out of theregion is balanced by an equalflux into the region.

In particular if the region bea portion of a narrow tube offorce between two cross-sections of areas w, co' and E, E'denote the magnitudes of the electric intensity at these cross-sections, then since E has no component at right angles to the

2*31] CONSEQUENCES OF GAUSS'S THEOREM 21

sides of the tube and the tube contains no charge, Gauss'sTheorem gives

orso that the electric intensity along a tube of force variesinversely as the cross-section of the tube.

It follows that the mapping of an electric field by unit tubesof force gives an indication not only of the direction but alsoof the magnitude of the intensity at any point.

Again consider the whole length of a tube of force startingfrom one charged conductor and ending on another, and sup-pose it to be prolonged at each end a short distance into eachconductor and then closed. The electric intensity E has nocomponent at right angles to the sides of the tube in air andE is zero inside the conductors, hence for the complete boun-dary of this prolonged tube \End8 = 0, and therefore the total

charge contained is zero. Rut the only charges are those on theelements of the conductors which form the ends of the tube,so that these charges must be equal and opposite.

(iii) Let a body A with a charge + e be surrounded by anuncharged closed conductor whose inner and outer surfaces are8X and 82. Let the surface 8 of Gauss's Theorem be drawn in theconductor between the surfaces St and Sz so as to surround 8X.Then at every point of S there is no electric intensity, so that

End8 = 0, and therefore 8 contains no total charge. Hence

there must be on 8X a charge equal and opposite to the givencharge e on the inner conductor A. This is also a consequenceof (ii) above, since every tube which starts from A must endon S1, and the charges on the ends of a tube are equal andopposite.

Since the closed conductor has no total charge, its outersurface 82 must have a charge + c, and the tubes of force whichproceed from this charge must continue until they reach someother conductor, possibly the walls of a room.

In this case there are two electric fields, one inside the con-

22 A UNIFORMLY CHARGED SPHERE [2 '3 i -

ductor and the other outside, and though they both owe theirorigin to the charge on A, they have an independent existence;for the surface 8Z can be discharged by touching it with thefinger. The field outside S2 will then disappear, but the field

inside S1 will remain-unaltered; or, A may be allowed to touch8X, when the charges on A and 8t will neutralize one anotherand the field inside St will disappear, but the charge on S2

and the external field will remain unaltered.

2-32. A uniformly charged sphere. Let a charge e beuniformly distributed over the surface of a sphere of radius a.To find the electric intensity E ,. -^at a point P at a distance rfrom the centre of the givensphere, draw a concentric sphereof radius r. By symmetry theelectric intensity has the samevalue at all points of this sphereand is directed radially, so thatthe total flux of intensity out ~~ -"of this sphere of radius r is 47rr2E, and, by Gauss's Theorem,if r > a,

4 i r r 2 ^ = 47re (1)

or E=ejr2 (2).

2*33] UNIT OF CHARGE 23

But, if r < a, the sphere of radius r contains no charge so that

or E = 0 (3);a result in accordance with 2*21 (6).

There is therefore no field inside the charged sphere and theexternal field does not depend upon the radius a of the sphere,and would have the same intensity e/r2 no matter how smallthe radius of the given sphere. We infer that 'a point charge'e produces a radial field of intensity e/r2 and that it would repela like charge e' at a distance r with a force ee'/r2; so that by thisreasoning Coulomb's Law of Force is a consequence of Gauss'sTheorem.

We shall shortly prove the converse and thereafter forspecial problems in electrostatics we shall use whichever ismore convenient for the purpose in hand, but for the logicaldevelopment of the general theory we shall regard Gauss'sTheorem as the basis.

2-321. The electrostatic unit of charge may now be definedin the language of action at a distance, as one which willproduce a field of unit intensity at unit distance, i.e. the chargewhich will repel an equal charge at unit distance with unit force.

It is now apparent why in Gauss's Theorem (2-3) the con-stant h was taken to be 4TT. For, if we retained h in 2*32, then(1) would read

and, if units are so chosen that E = 1 when e = 1 and r = 1, wemust have h = 4TT.

2*33. The local form of Gauss's Theorem. Let us applyGauss's Theorem to a small element of volume 8v enclosing apoint P. If E denotes the electric vector at P, we have from1-52 / (• \ I

divE= Urn I \End8\ 8v,

where the integral is the outward flux of E through theboundary of Sv. Therefore

>8v (1),

24 LOCAL FORM OP GATTSS'S THEOREM [2'33-

where e is a small quantity which vanishes with Sv. But if prepresents the average density of electricity inside Sv, the left-hand side of (1) is, by Gauss's Theorem, equal to kirphv.

Therefore (divE + e)Sv = 4i7p8«;.

Whence, by dividing by 8v and then making 8«->-0, we get

divE =-krp .(2),

where the left side is the divergence of the intensity and theright side is the density at any point P in the field at which Eis continuous.

This is a fundamental equation of electrostatics. It in-cludes the fact that at all points at which there is no charge,i.e. at which p = 0,

divE = 0 (3).

We must comment on the fact that equation (2) contains a volumedensity p of electricity; and we have stated that charges reside uponsurfaces and we should therefore expect that they would enter intocalculations as surface density or charge per unit area. But a chargeper unit area is really a charge occupying a volume though concen-trated by the fact that one dimension of the volume is indefinitelydiminished, and it is often convenient in theoretical work to assumethe existence of volume densities and treat surface densities as limitingcases.

We must also comment on the fact that when we use such integral

forms as I pdv we imply that p represents something mathematically

continuous and having a definite value at each point of a region ofspace. But electricity exists in the form of electrons and protons andis not continuous physically or mathematically, so that we are com-pelled to put a special interpretation upon such an integral as I pdv;

and we take it to mean what the value of such an integral would be ifwe imagined a continuous p to exist and to have at every point of theregion of integration a value which is the average density of the actualcharges in a small but finite element of volume surrounding the point.*

2*34. Gauss's Theorem for a surface distribution. Wehave seen that discontinuities in the electric field, i.e. in thevector E, arise at charged surfaces. We assume in 2*33 that

* On this subject see J. G. Leathem's Tract, Volume and SurfaceIntegrals used in Physics.

2'34] SURFACE DISTRIBUTION 25

there is no surface of discontinuity of E within the regionconsidered. We shall now consider the field in the neighbour-hood of a surface of discontinuity.

Let a charge of electricity be spread over a surface. To makethe argument as general as possible we shall not limit thesurface to be that of a conductor but include the case of thesurface of a dielectric body such as a piece of glass, so that theremay be an electric field on both sides of the surface.

About a point P on the surface draw a short narrow cylinderat right angles to the surface with its ends parallel to thesurface and its length small compared to the linear dimensionsof its cross-section.

Let o) be the cross-section of the cylinder and a the surfacedensity of the charge at P, i.e. thecharge per unit area, so that wo- is thetotal charge within the cylinder.

The charge on the surface causesa discontinuity in the electric field.Let Ex, E2 denote the electric vectorat points close to P on opposite sides of the surface. Nowapply Gauss's Theorem to the region bounded by the cylinderand take the cylinder so short that the flux of E through itssides is negligible. Hence we get

where the suffix n indicates a normal component. Therefore

(I)-

This is the form which Gauss's Theorem takes for a surfacedistribution.

I t will be noticed that the normal components are directedaway from the surface, and that if we change the sign of (E2)n

so that both are directed in the same sense, the theorem shewsa jump in the value of the normal component of the electricvector of amount 4ira in crossing the surface.

In the special case in which the surface is that of a con-ductor, if we take Ex and E2 as referring to the regions outsideand inside the conductor, we have E2 = 0, since there is no

26 GAUSS'S THEOREM [2*34r-

field in the substance of a conductor, and in this casesuppressing the suffix, the result may be •written

En = ±™ (2).

This is Coulomb's expression for the normal electric intensityjust outside the surface of a charged conductor.

2*35. Examples, (i) Deduce Oauss's Theorem from Coulomb's Lawof Force; i.e. to prove that for any closed surface 8

Consider the field due to a single point charge e at a point O. Drawa cone of small solid angle dtu and vertex 0 and let it cut S in smallelements of area dSlt dS2, dS3, ... at the points Plt P2 , P3 , . . .; and

let the outward drawn normals to these elements make angles 8X, 8%,d3,... with the line OPX P2.... The 0's are obtuse or acute angles accord-ing as the line OPX P2... is entering or leaving the region bounded by S.

The point charge e at 0 produces at a point P a field of intensitye/OP* directed along OP (Coulomb's Law), so that the contributionsto the outward flux of E which arise from the elements of area in whichthe cone cuts the surface amount to

ecos9t ecos6>2 ecosgd S + d S +

But from 1-3 (1), for each of these elements

•UF-2dS=+da>,

+ or — according as 8 is acute or obtuse, i.e. according as the conedrawn from 0 is leaving or entering the closed region. Hence expression(1) is of the form

— edw + edco — edco -t~...;

2*36] PLANE DISTRIBUTION 27

and if O is outside /S the number of entrances of the cone is equal to thenumber of its exits from the region and the sum is zero; but if 0 isinside <S the cone makes a final exit from the region and thesum is edw.

Then taking cones in all directions round 0, since I dw = 4TT, it follows

that, for a point charge e,

EndS=O or iire (2),

according as the point charge is outside or inside S.Proceeding in this way and adding together the effects of all the

point charges in the field, we get

/ '(3),

where Se means the sum of the charges inside S.

(ii) Prove that the electric intensity at all points on either side of auniformly charged plane is 2n<r, where a is the charge per unit area.

Shew that, of the total intensity Ina at a point A at a distance of half aninch from the plane, one-half is due to the charge at points within an inchof A. [M. T. 1918]

Here it is assumed that the medium on both sides of the plane sheetof electricity is air and that the plane is of infinite extent. By sym-metry the tubes of force are at right angles to the plane everywhereand of constant cross-section, so that by 2*31 (ii) the electric intensityhas the same value everywhere, and, taking a to be positive, it isdirected away from the plane on both sides. Then if as in 2'34 we applyGauss's Theorem to a short cylinder of cross-section o> at right angles tothe plane with plane ends parallel to the plane on opposite sides of it,denoting by E the constant value of the electric intensity, we get

Ew + Eo) = 4:7T<Ta>,o r E=2iro.

This is in fact a special case of 2-34 (1), in which (EJn = (E2)n = E; andtaking note of the sense of the intensityon opposite sides of the plane it accordswith the general rule that there is a jumpof 4TT(T in the value in crossing the plane.

For the second part of the question,the charge on an element of area dSat P is adS, and if r is its distance fromA it produces at A an intensity odS/r*along PA, and resolving this along thenormal MA gives adS cos 0/r2, where 8is the angle MAP, or, from 1'3 (1), adw,where dw is the solid angle subtended at A by the element dS.

But the points of the plane which lie within an inch of A, lie withina circle of centre M and radius MQ, where AQ = 1 inch. The charge on

28 THE POTENTIAL FUNCTION [2#35-

this circular area therefore contributes to the total intensity at A anamount oca, where <a is the solid angle which the circle subtends at A.But AM = \ inch, so that the solid angle is that of a right circular coneof semi-vertical angle 60°, and therefore

<o = 27r(l-cos60°) (1-22)

Therefore an intensity rra at A, or half the total intensity, is due to thecharges which lie within an inch of A.

2*4. The potential function. In order to complete thebasis of the theory of electrostatic fields in air we must nowintroduce another function—a scalar function of position likethat discussed in 1*4—called the potential function. Weassert as our second fundamental hypothesis for the buildingup of the theory of the electrostatic field that there is, forevery electrostatic field, a single-valued potential function suchthat the electric intensity is its negative gradient; i.e. we assumethat there exists a single-valued function </> such that, atevery point of the field,

E = — grad <j>.

We proceed to shew that this function <f> bears an inter-pretation in terms of work.

Suppose that a unit charge could be displaced in an electro-static field without causing a disturbance of the other charges;then the work done by the field on the unit charge as it

CQ

moves from P to Q by any path would be given by Esds,JP

where 8s denotes an element of the path and Es the componentof E in the direction 8s. But, by hypothesis, Es= —so that the work done

or the work done is equal to the excess of the potential at Pover the potential at Q. If we assume that the field is offinite extent, we may suppose the potential at a great dis-tance to be zero, and the potential of the field at a pointP is then the work that would be done by the forces of thefield on a unit charge as it moved from P to an infinite distance,

2'41] EQUATIONS OF POISSON AND LAPLACE 29

always supposing the field to be undisturbed by the presenceof the unit charge.

Physically therefore potential is of dimensions 'work perunit charge', or 'energy per unit charge'.

Since E is zero in the substance of a conductor in electro-static equilibrium, therefore the potential has a constantvalue throughout such a conductor.

For laboratory purposes the potential of the earth may beregarded as invariable, for it is obvious that any chargewhich might in the course of an experiment be added to thatof the earth could not affect its potential appreciably.

Further, it is clear that the addition of a constant to thefunction $ will not affect E, neither will it affect the differenceof potential between any two points. It is therefore permissibleand convenient to take the potential of the earth to be zero.

2'41. Equations for the potential. The relation

E= -g rad^ (1)

implies that for rectangular components

T? w w ty d<f> tyEx, Ey, Ez = -Tx, - ^ , -Vz (2).

Hence by substituting in the fundamental relation 2*33 (2),

div E = 4T7-/>,

or (see 1-52) » . + S + ».

av ^ ^weget dS+w+wThis is known as Poisson's Equation. It is satisfied by the

potential at every point of the field at which there is a volumedensity of electricity. At points at which there is no electricitythe equation becomes

known as Laplace's Equation.

30 SURFACE DISTRIBUTION [2*41-92 32 32

The operator «-s + ~--s+7r-oi known as Laplace's operator,3a;2 dy2 dzz r

is usually abbreviated into the form V2, so that (3) and (4)are written _„ . . , _„ ,

V2£ 4 a n d V2<f> = 0.To obtain Laplace's equation in polar co-ordinates r, 6, 00,we take polar components of the electric vector, viz.

and substitute in the equation

divE = 0,

using the expression for divergence given in 1*53 (1); this gives

= 0 ...(5).( r \ \r*dr\ 9rj r2sin080

Similarly, in cylindrical co-ordinates r, 6, z, we take

and substitute in the equation

div E = 0,

using the expression for divergence given in 1*54; this gives

2*411. For a surface distribution. As explained in 2-34,there are different fields on opposite sides of an electrifiedsurface, and the potential function has therefore differentforms (£x, <f>2 on opposite sides of the surface, so that

Ex = — grad fa and E2 = — grad <f>2.

Therefore 2'34 (1) may be written

3% 3w2

where 9% and dn2 are elements of the normal directed awayfrom the surface on each side.

2*411] SURFACE DISTRIBUTION 31

As we have already seen therefore, at every electrified sur-face there is a discontinuity in the electric vector or, what is thesame thing, a discontinuity in the gradient of the potentialfunction, and the potential functions on opposite sides of thesurface must satisfy (1).

At the surface of a charged conductor we have 2*34 (2),which may now be written

where the differentiation is along the outward normal and <f> isthe potential outside the conductor.

Further, the potential function is assumed to be physicallycontinuous, i.e. between any two assigned values it assumes allintermediate values, save, as we shall see later, possibly at acommon boundary of two different substances where there isa constant difference. At all points of such a surface we have

<t>r-h=C (3),where G is a constant, which is in most cases negligible; and<f>lf <f>2 denote the potential functions on opposite sides of thecommon interface.

Let Ss be a small arc drawn in any direction on the surface ofseparation, then by differentiating (3) we get

%-%-* <*>

or E^E^ (5);

i.e. the tangential component of the electric vector is con-tinuous in crossing the common surface of two media, whetherthis surface be electrified or not.

To summarize: the potential function must satisfy 2*41 (3) atall points where there is a volume density and (4) at all points inemptyspace. At a charged surface the normal components of itsgradient must satisfy (1) above but the tangential componentsof its gradient are continuous. The function itself must be con-tinuous everywhere save in exceptional cases where there maybe a constant difference at an interface between two media.

32 BQTJIPOTBNTIAL SXTBFAOBS [2*411-

COROLLARIES. (i) If the potential is constant through anyregion, e.g. in the substance of a conductor in equilibrium, thenfrom 2-41 (3), since <j> = const., therefore V2<£ = 0 and p = 0; sothat there can be no charge in the substance of a conductor inequilibrium.

(ii) If the normal component of the gradient of the potentialis continuous across any surface, then from (1) there can beno charge on the surface.

2-42. Equipotential surfaces. A surface at every pointof which the potential <f> has the same constant value is anequipotential surface.

In any case in which the potential <f> has been determined asa function of x, y, z, the equation

<f>(z,y,z) = a. (1),

for different values of the constant a, represents the family ofequipotential surfaces. It is clear that two such surfaces ofdifferent potentials cannot intersect.

The electric vector at every point is normal to the equi-potential surface through the point. This follows from the factthat the vector is the negative gradient of the potential, orsimply from the fact that in any direction along the surface<j> is constant, so that d<f> = 0, and there is therefore no componentof the vector in that direction.

Conversely a surface to which the electric vector is every-where normal must be an equipotential surface, for dfyjds = 0in every direction at right angles to the vector, and therefore<j> = const, over the surface.

It follows that if space is mapped out by tubes of force andequipotential surfaces, the former cut the latter at right angleswherever they intersect.

2-43. On lines and tubes of force. Since the electricintensity is the negative gradient of the potential, therefore aline of force, which is at every point in the direction of theelectric intensity, may be regarded as having a positive sensein the direction in which potential decreases. Hence we regardlines and tubes of force as drawn from places of higher to

2-44] LINES AND TtTBES OF FORCE 33

places of lower potential; and as in 2*22 lines and tubes of forcestart from positive charges and end on negative charges.

(i) Since the potential of a conductor is constant it followsthat no line of force can begin and end on the same conductor,nor can it begin and end on conductors at the same potential.

(ii) In a field containing several conductors the conductorof highest potential must be positively charged; for a negativecharge would imply tubes of force arriving there havingstarted from a place of still higher potential.

(iii) The conductor of lowest potential must be negativelycharged; for a positive charge would imply tubes of forcestarting thence and proceeding to a place of still lowerpotential.

(iv) Apart from the earth there must be at least one con-ductor whose charge is wholly positive or wholly negative, forif there is an electric field its potential must have either ahighest or lowest value or both a highest and a lowest.

(v) On a conductor insulated and uncharged the distributionof electricity, if any, is partly positive and partly negative andas many unit tubes of force fall on the conductor as proceedfrom it. These are separated by a line of no electrification on thesurface, and along this line, since a=0 , therefore by 2*411 (2)d<f>/dn = 0, or <j> is constant in the direction normal to the con-ductor, and therefore at all points of this line of no electrifica-tion the conductor cuts an equipotential surface at right angles.

(vi) When all the conductors but one are connected toearth (i.e. at zero potential) and that one has a positive charge,the charges induced on the others are all negative and theirnumerical sum cannot exceed the positive charge on theinsulated charged conductor. This follows because the unittubes of force which start from the positively charged con-ductor have at their further ends to account for all the negativecharges in the field.

2*44. Theorems on the potential, (i) / / a closed equi-potential surface contains no charge, the potential is constantthroughout the region bounded by the surface.

If there are no charges inside a closed surface 8, no lines

34 THEOBEMS ON THE POTENTIAL [2-44-

of force can begin or end inside S; so that, if there is any fieldinside S, the lines of force must be continuous lines traversingthe region bounded by S. But this is impossible since allpoints on S have the same potential, and no two points on aline of force can have the same potential. Therefore there isno field inside S, or the potential is constant.

(ii) The potential cannot have a maximum or minimum valueat a point at which there is no charge.

Let the potential have a maximum at a point P; then thepotential decreases in every direction round P, so that if wedraw a small sphere round P, the electric vector is directedoutwards at every point of the sphere, so the outward flux ofthe electric vector is a positive number, and by Gauss'sTheorem there must be a positive charge inside the sphere.Hence the potential cannot have a maximum at P unlessthere is a positive charge there. Similarly at a point wherethe potential has a minimum there must be a negativecharge.

2-5. Special fields. Point charges. It is possible to give aformal proof that if the potential function in any field satisfiesthe relations 2-41 (3), (4) and 2-411 (1), (3) at different points ofthe field, then the potential at any point is equal to the sum ofeach element of charge divided by its distance from this point.But the proof requires too much analysis to be introduced here,and we shall content ourselveswith shewing that the definitionof potential in terms of work in2-4 leads to the result statedabove in the case of a field dueto point charges repelling in ac-cordance with Coulomb's Law.

Thus let there be a charge e at0. To find its contribution tothe potential at a point P, letQQ' be an element ds of a path from P to an infinite distancein any direction. Let OP —a and OQ=r. Then taking theelectric intensity at Q to be e/r2 along OQ, the work done in

2-51] COLLINEAR POINT CHARGES 35

repelling a unit charge from P to infinity along this path isrepresented by

f " e dr . rm e , eJPr zds ~]a r2 ~ a

Similarly if there are other point charges e', e", ... at dis-tances a', a", ... from P, they give like contributions to thepotential at P, so that the potential is

- + -, + — + ... or S - .a a a a

2-51. The field due to a set of collinear point charges.Let a number of charges ex, e2, e3, ... be situated at collinearpoints^!, A2, A3,...; let P be a point at distances r_, r2, r3,...from J.1; A2,A3, . . . ,andletAXP, A2P, A3P, ... make ______£_angles 91,92,93,... with the

The point charges makeradial contributions ejr^2,e_/r2

2, es/r32, ... to the in-

tensity at P, and if weresolve these along the normal to the line of force through P,the sum of the components is zero, since the resultant intensityis tangential to the line of force. Hence

e i ddx e2 ^ 2 e3 d93

where ds is an element of the line of force.

Therefore — d9x+—d92 + — d93 +... = 0.r i r2 r3

so thatex sin 9xd9x + e2 sin 92d92+e3 sin 93dd3 + . . . = 0,

and by integration

ex cos 9X + e2 cos 92 + e3 cos 93 +... = const (1),

and for difEerent values of the constant this equation representsall the lines of force.

3-2

36 COLLINBAR POINT CHARGES [2-52

2*52. Examples, (i) Two point charges e±,e%of the same sign.Since the charges have the same sign, no line of force can pass from

one to the other and all the lines of force must go to infinity.Let the charges ex, e2 be at the points A, B and let P be a point on the

line of force which starts from A at an inclination a to BA produced.The equation of the lines of force is

e1 cos 9± + e2 cos 02 = const.,where Blt 0a are the inclinations to the line BA of the radii from A, Bto any point on such a line. But a sP ->- ion the line considered wehave 0!-»-a and 02->O; therefore its equation is

ex cos B1 + ea cos 0a — el cos a + ea.

The direction of the line at infinity is got by putting 61 = 8i = 6, so that(ei + e2>cos 0 = «i cos a + e2

gives the slope of the asymptote.Again if the tangent at P cuts AB in T, since there is no electric

intensity at P at right angles to TP, therefore

But

so that

sinAPT _shxAPT sinBTPAT BPsin.BPT~sin.ATP ' sinBPT~AP ' B~T"

AP- .AT = BP3 . BT.

Now let P move along the curveto infinity, then AP/BP-+1 andT moves up to a point G on ABsuch that

And as this is independent of a, itfollows that the asymptotes to all Bthe lines of force pass through this fixed point G.

2-52] COLLINEAR POINT CHARGES 37

Further there is a point of equilibrium O on the line AB, where therepulsions e1/AO2 and eijBO% are equal and opposite.

The equipotential surfaces are surfaces of revolution about the lineAB given by the equation

A = -1 + — = const.,

where rx, rt denote distances from A and B.For small values of rx and of rt, i.e. for large values of the constant,

these are disconnected surfaces surrounding the points A, B respec-tively. The potential decreases as we proceed outwards from either

charge. For a particular value of the constant the two surfaces becomea single surface passing through 0, the point of equilibrium; the vanish-ing of the components of intensity -$-,-$•, •$• at this point being the

occ oy uZconditions that the surface <f>(x, y, z) = const, should have there aconical or nodal point and therefore no definite normal. For smallervalues of the potential the equipotential surfaces are single sheetssurrounding both charges.

(ii) Charges 4e and — e.Let the charges 4e and — e be at A and B. There is a point of equili-

brium O on AB produced so that BG=AB, for this makes

Of the unit tubes of force which start from A only one-quarter willend on the negative charge at B and the rest will go to infinity, and theequilibrium point O will separate those which go to B from thosewhich go to infinity.

If P be any point on a line of force and PAO=8 and PBC— 8', theequation of the lines of force is given by

4e cos 8 — e cos 8' = const (1).Consider the possibility of a line of force passing through C. At Owe

have 6 = 8' = 0, so that its equation would be4cos0-cos0' = 3 (2);

38 OOLLINEAE POINT CHARGES [2-52-

and as the point P approaches O, 0, 0' are small and (2) may be written

which gives 0' = 26. From this we deduce that BP = AB—BC; andtherefore as P approaches C the curve cuts BO at right angles. In theplane of the paper a similar line of force would approach BO from below.

The angle a which these lines make with AB when they start from Ais got from (2), by putting6 = a and 8' = n, giving

4oosa+l = 3,so that a = 60°.

The equipotential surfaces B

are surfaces of revolution about the line AB, given by the equation

? = const., where r, r' denote distances from A and B.

For small values of r' the potential is negative and the surfacessurround the point B. The section of the surfaces by a plane throughAB consists of loops round the point B increasing in size, until we

reach the value of the potential at C, viz. -r-~ — =^ or - -=, for which

2-531] TWO-DIMENSIONAL FIELD 39

the equipotential surface is a surface which crosses itself with a conicalpoint at O; the section consisting of two loops the larger of whichencloses loops which surround, the point A with potentials which con-tinually increase as the loops shrink round the point A.

Outside the surface through O there are equipotential surfaceswhich surround both A and B with potentials which decrease steadilyto zero as the surfaces expand to infinity.

2*53. A two-dimensional field. When a field is producedby a number of long uniformly charged cylinders, with parallelaxes and ends so far distant from the region of space consideredthat the field may be deemed to be the same in all planes atright angles to the cylinders, we may regard the field as two-dimensional.

Consider a single circular cylinder of radius a carrying acharge e per unit length. To find the electric intensity at apoint P at a distance r from the axis of the cylinder, describea coaxial cylinder of radius r and consider the flux out of a unitlength of this cylinder. By symmetry the field E is radial sothat the total flux is 2irrE, and if r > a the charge enclosed inunit length of the cylinder is e, therefore by Gauss's Theorem

or E = 2e/r (1).

But if r < a, no charge is contained in the cylinder of radiusr and therefore F — 0 CZ\

If <j> denotes the potential of which E is the negative gradient,we have j i a.

oi> _ 2e . .—£=E=— (r>a)dr r

or <j> = const. — 2e log r (3)

and, for r<a, <£ = const (4).

We notice that the formulae (1) and (3) are independent ofthe radius a of the given cylinder.

2-531. Field produced by a set of long charged parallelfine wires. Let long parallel fine wires carrying charges e1,e2,e3, ... per unit length cut the plane of the paper at right anglesat the points Alf A%, A3

40 FIELD DUE TO LINE CHARGES [2-531-

We shall speak of the flux across a line on the paper, meaningthereby the flux across a unit length of a cylindrical surfaceof which the line is a cross-section.

With this understanding the total flux of intensity acrossa curve which surrounds the point A1 is, by Gauss's Theorem,

and the amount of flux which lies between two radii

inclined at an angle ddx is -^. 4:ne1 or 2e1dd1.AT*

Let an element PP' of a line of force subtend angles dd1,dd2, dd3, ... at Ax, A2, A3, ..., then since the total flux ofintensity across PP' is zero, therefore

2exddx + 2e2d62 + 2e3d63 +... = 0,

or, by integration,

ex 01 + e202 + e303 + ... = const (1),

where the d's are the angles made by the lines AXP, A2P,A3P, ... with any fixed direction.

For different values of the constant, equation (1) representsthe lines of force in any cross-section of the field.

I t should be noticed that if any point A lies on the oppositeside of the line of force to the other points the sign of thecorresponding term in the equation must be changed.

In like manner by using 2-53(3) and assuming that thepotential due to a number of charged wires is got by adding

B o

2-6] FIELD DTTE TO LINE CHARGES 41

the potentials which they produce separately, we get for thepotential of the field

<£ = const. — 2ex log r1 — 2e2 log r2 — 2e3 log r3 — (2),

where the r's are the distances of a point from Alt A2, A3, ....

2*532. Examples, (i) Two wires with equal and opposite charges.Let e, — e denote the charges per unit length on wires which cut the

paper at right angles at A, B.Let radii from A, B to a point yP make angles 8, 8' with BAproduced. Then the lines offorce are given by

e9—e8' = const.,

or the angle APB is constant.So the lines of force are coaxialcircles passing through Aand B.

The potential is given by

^ = const. - 2e log AP + 2e log BP,

so that the equipotential curves are given by AP/BP = const., i.e. thefamily of coaxial circles which have A, B as limiting points and cut theformer family at right angles.

(ii) Two wires with equal charges of the same sign.

Here the lines of force are seen to be

8+8' = const.

Taking rectangular axes so that A, B are the points ( ± a, 0), this is

tan"1 ———I- tan"1 —-— = tan"17 ,

x—a x+a k

say, or x2 — 21cxy — y 2 = a2,

a family of rectangular hyperbolas passing through A and B.As above, the equipotential curves are the ovals

rr' = const.

and we thus have a physical proof that the orthogonal trajectories ofthe family of ovals is a family of rectangular hyperbolas.

2*6. Experimental verification of the Law of InverseSquares. Cavendish's experiment. A metal globe rests on anebonite ring inside another metal globe formed of two tightly fittinghemispheres supported on an insulating stand. There is a small holein the outer globe which can be closed by a metal lid to which is affixeda short wire and this makes contact with the inner globe when the lid isclosed. When the lid is lifted by a silk thread, a wire can be passed

42 CAVENDISH'S E X P E R I M E N T [2-6

through the hole to connect the inner globe with an electrometer andso test whether it has a charge. The figure shows the essential part ofthe apparatus in section.

A charge is given to the outer sphere with the lid closed so that theinner and outer are in communication. The lid is then lifted, the outersphere discharged and the inner sphere is tested and found to bewithout charge, thus demonstrating experimentally that there is noelectric charge inside a uniformly charged spherical conductor.

Now let A be any point inside a sphere which is charged with elec-tricity of uniform surface density a. Let a cone of small solid angle du>and vertex A cut the sphere in elements of area dS, dS' at P, P'. LetAP = r, AP' = r', and let O be the centre of the sphere. The charges on

the elements are adS and odS'. But dS=r3da> sec OPA, anddS' = r'*da>aec OP'A, and the angles OPA, OP'A are equal. So if thelaw of force were the inverse square of the distance, the forces due tothe elements, odS/r* and crdS'/r'*, would be equal and opposite, andby taking cones in all directions round A the whole surface of the sphereis divided into like pairs of elements exerting equal and opposite forcesat A so that the resultant would be zero and there would be nothing tocause a transfer of electricity between the outer and inner sphereswhen they are in communication during the experiment.

Now consider the effect of taking a modified law of force r-*-*instead of r~2. A plane through A at right angles to AO divides thesurface of the sphere in such a way that, for all such cones as we con-sidered above, AP' <AP, where P' is above this plane and P is below it.

The repulsions of the corresponding elements are —£fp and ,i+p,

where as above —j- = —^ - , but r' < r. Hence if p is positive, the force

along PA is less than the force along P'A, and vice versa if p isnegative: and therefore with this law of force the plane through Adivides the surface into portions the charges on which exert unbalancingforces at A, and there would be a resultant force at any point A insidethe sphere, other than the centre. Thus at all points on the inner

2-6] EXAMPLES 43

sphere in the experiment there would be a force in the same sense,inwards if p is positive and outwards if p is negative, which mustresult in the passage of electricity when the spheres are in com-munication and the presence of a charge on the inner sphere. Sinceexperiment indicates that there is no such charge, we conclude that ifthe law of force is some power of the distance it must be the inversesquare.

EXAMPLES

1. Prove that, if there be only one charged conductor in an electricfield, the amount of electricity of either kind induced on any otherconductor cannot exceed the charge on the first. [I. 1891]

2. An insulated conductor is under the influence of a point charge— e; if the total charge on the conductor is positive and greater than e,prove that the distribution of electrification on the conductor is every-where positive.

3. Shew that, if a conductor A encloses a conductor B, and the twoconductors are both charged, there will be no change in the field outsideif B is connected to A by a wire. [St John's Coll. 1913]

4. The total flux of intensity through a closed surface of one sheetin an electric field is zero, also the surface encloses all points of the fieldat which there is any distribution of electricity and is not a surface ofzero potential. Prove that it encloses points of positive and points ofnegative and points of zero potential. [M. T. 1909]

5. Prove that, if there are two positively charged conductors in thefield, neither enclosed within the other, there is at least one point inthe field where the intensity is zero. [I. 1912]

6. Two conductors carrying respectively a positive charge ex and anegative charge — e2 are introduced into the interior of a larger con-ductor maintained at potential V. Shew that if ex > e2 the potential ofthe first conductor is greater than V. [M. T. 1909]

7. A field of force is due to three quantities of electricity ex, e2, egat A, B, G. Shew that the direction of the line of force at a point P isgiven by joining P to the centre of gravity of masses

APa>

at A, B, O respectively, and find the magnitude of the force at P.[I. 1895]

8. Two particles each of mass m and charged with e units of elec-tricity of the same sign are suspended from the same point by strings

44 EXAMPLES

each of length a. Prove that the inclination of the strings to the verticalis given by the equation

4msrassin30=eacos0. [I. 1894]

9. I t is required to hold four equal point charges e in equilibrium atthe corners of a square. Find the point charge which will do this ifplaced at the centre of the square. [M. T. 1922]

10. Three equal small spheres each of mass m grams and each carry-ing a charge e electrostatic units are suspended by strings of lengthI cm. from the same point. If d cm. is the distance between a pair of thespheres when the system is in equilibrium, shew that

Why, for a similar system consisting of charges Ae and strings oflength XI, do the masses have to be m (and not Am) for the charges torest at distance \d apart? [M. T. 1930]

11. Assuming only that the potential due to a small quantity e ofelectricity is ejr, prove that the potential due to a uniform sphericaldistribution at any internal point is equal to the potential at the centre,and at any external point is the same as if all the electricity were atthe centre. [I. 1894]

12. The potential is given at four points near each other and not allin one plane; obtain an approximate construction for the direction ofthe field in their neighbourhood. [M. T. 1894]

13. If any closed surface be drawn not enclosing a charged body orany part of one, shew that at every point of a certain closed line on thesurface it intersects the equipotential surface through the point atright angles. [M. T. 1897]

14. Sketch the form of the equipotential surfaces and lines of forceof two equal and opposite point charges. [M. T. 1927]

15. Positive and negative unit charges are situated at two pointsA,B. Find at what distance from AB the plane normal to and bisectingAB is cut by the lines of force which issue from A in a direction parallelto this plane. [M. T. 1919]

16. Draw a diagram of the lines of force and the equipotential sur-faces in a field in which the only charges are 2e at a point A, and — e ata point B. [I. 1905]

17. Sketch the lines of force and level surfaces of two positive pointcharges of 1 and 2 units respectively. [I. 1914]

18. Sketch roughly the field of force due to two point-charges 4e and— eat A and B respectively. Sketch also the equipotential surfaces ona separate diagram.

What is the locus of the points at which the lines of force are parallel[I. 1914]

EXAMPLES 45

19. Two point charges e and — e' (e>e') are situated at A and Brespectively. Prove that the extreme lines of force which pass from Ato iSmake, on leaving A, an angle 9 with AB, where cos0 = (e—2e')/e;and indicate by a figure the general form of the lines of force.

[M. T. 1907]

20. Draw a rough diagram to show the distribution of lines of forceand equipotential surfaces due to two point charges, one of them being4e and the other - 3e. [I. 1909]

21. Charges +1 , — 4, +1 are placed at coUinear points A, B and O,where AB = B0. Sketch the lines of force; and shew that any line offorce leaving C will reach B at an inclination to BO less than \n.

[I. 1923]

22. Point-charges e, — e' and — e' are placed at O, A and B respec-tively, which are in a straight line, and OA — OB. If e is > 2e', shewthat the greatest angle a line of force leaving O and entering A can makewith OA is a, where

e sin2 = e'.

Draw a diagram, giving the lines of force approximately, when4e'=e. [I. 1900]

23. Charges of electricity 1, —2, 8 units respectively are placed ina straight line at points A, B, C at unit distances apart. Prove thatthere is a point of equilibrium between A and B and make a roughdrawing of the lines of force. [I. 1906]

24. Indicate by a sketch the forms of the equipotential surfaces andlines of force for three point charges +e at (a, 0), +e at ( — a, 0) and- 2 e a t ( 0 , 0). [M. T. 1931]

25. Give careful sketches indicating the most significant features ofthe lines of force and equipotential lines in the following cases:

(i) Two equal point charges of the same sign.(ii) Two equal and opposite point charges,

(iii) Two charges 4e and — e.(iv) A point charge placed in a uniform field of force. [M. T. 1933]

26. Charges 1,1, — 2 are placed at three points A, B, O forming anequilateral triangle. Shew by a diagram the arrangement of the linesof force within the triangle and just outside it in its plane. [I. 1906]

27. If a total charge e were uniformly distributed throughout thevolume of a sphere of radius a, what would be the electric intensity ata distance r from the centre of the sphere?

46 EXAMPLES

28. One side of a circular glass plate of radius a is uniformly chargedwith electricity, the charge per unit area being a. Prove that the electricintensity at a point on the axis of the plate at a distance c from the

29. Three infinite straight cylinders of negligible section, electrifiedto line densities e, e' and — (e + e'), cut a plane perpendicular to themin the points B, C, A respectively, and P is a variable point. Prove thatthe equations of the lines of force in the plane are

e. BRA ± e'. CPA = const. [I. 1892]

30. If three infinitely long thin wires whose charges per unit lengthare 1, — 2, 1 respectively cut a plane at right angles in collinear pointsA, B, C, where AB = BO=a, prove that the equation of the lines offorce in this plane is

r2 = a2 cos (20 + a) sec a,B being the origin, BO the axis from which 6 is measured, and a avariable parameter. [I. 1904]

31. Three infinite parallel straight lines, with charges e, — \e,, — \e, perunit length, meet a plane perpendicular to them in A, B, C respectively,where AB = A0. Prove that the lines of force reaching B, after havingpassed indefinitely close to the equilibrium point, make with BO anangle = BAG. Draw a rough sketch of the lines of force and the equi-potential surfaces. [I. 1909]

32. Three infinitely long parallel straight wires carrying charges2e, —e,—e per unit length cut a plane at right angles at the cornersA, B, O of a triangle right-angled at A. Prove that no line of forcewhich starts from A inside the triangle emerges from the triangle.

33. Three infinite parallel wires cut a plane perpendicular to them inthe angular points A, B, O of an equilateral triangle and have chargese, e, — e' per unit length respectively. Prove that the extreme lines offorce which pass from A to O make at starting angles

2e-5e' , 2e + e'_ _ „ and - ^ - w

with AC, provided that e' > 2e. [M. T. 1904]

34. Three long thin straight wires, equally electrified, are placedparallel to each other so that they are cut by a plane perpendicular tothem in the angular points of an equilateral triangle of side s/3 c; shewthat the polar equation of an equipotential curve drawn on the plane is

r6 + c6 — 2r3c3 cos 30=const.,the pole being at the centre of the triangle and the initial line passingthrough one of the wires. [I. 1894]

EXAMPLES 47

35. The electric intensity in certain regions of a discharge tubeunder certain conditions is a linear function of x, the distance from oneend of the tube. Shew that the volume density of electricity is constant.

[M. T. 1927]

36. If the potential at a point outside a conducting sphere of radius

- F cos 6 (r-aP/r*),

where r is the distance measured from the centre, 8 is the angulardistance measured from a fixed diameter and F is constant, find thesurface density at any point of the sphere and shew that the totalcharge on the sphere is zero. [M. T. 1935]

37. Give a physical proof that the families of curves

and r^i raet r3

ei ... = const.

cut one another at right angles at all their intersections.

ANSWERS

( e e e l7. 1-3-Bs + ^ p + Tjpat £*&> where G is the centre of gravity of the

masses.

9. - e ( l + 2V2)/4. 15. ABV3/2.

18. The sphere for which AP- 2%BP.

27. er/a3, r<a; and e/r2, r>a.

Chapter III

CONDUCTORS AND CONDENSERS

3*1. Mechanical force on a charged surface. Thoughelectric charges can move freely through the substance andalong the surface of a conductor, yet they cannot movenormally outwards from the surface, although the field exertson them a normal electric force. The mechanical effect in anelectrostatic field is the same as if the conductor exerted aforce on each element of charge on its surface preventing itfrom moving normally, and, since action and reaction areequal and opposite, the conductor is therefore subject tonormal forces equivalent to the forces exerted by the fieldupon the charges on its surface.

Consider a surface distribution of density a. The charge onan element of area dS is adS, and we want to find the resultantforce on this charge.

It is clear that the charge adS will not contribute to the fieldwhich produces the force upon it; for if it existed alone itwould have no tendency to move, hence the required force ondS is due to the charges in thefield other than the charge ondS. Now let Ex, E2 denote theelectric vector at points P, Qclose to dS near its centre andon opposite sides of it. Thecharge on dS may be regarded as an infinite plane sheet inrelation to the points close to it near its centre, and thereforeproducing at these points a normal field of intensity 2-na(2*35 (ii)) away from itself on both sides. The rest of the chargesin the field will make practically the same contribution R to theintensity at both P and Q, so that we have vector equations

and E2 = R — ,where n is a unit normal vector in the positive sense from 2 to 1.

3-3] UNIFORMLY CHARGED SPHERE 49

Therefore R = £ (Ex + E2), and the force on the charge adS is

RodS, or KE^ + EJodS.Hence the mechanical force per unit area on the surface ofthe body is

If the body is a conductor, and the sides 2, 1 are inside andoutside, E2 = 0 and Ex is along the normal and equal to AITV

(2#34). Hence there is an outward normal force 2TTCT2 per unitof area over the surface of a charged conductor.

3*2. A uniformly charged conducting sphere. Revertingto 2#32, if (f> denotes the potential of the sphere at a distance rfrom its centre, then, for r > a,

so that (f> = - + const.

But, when the sphere is alone in the field, <f>->0 as r->oo, sothat the constant is zero and, for r > a,

(D-

Similarly, for r <a, —^- = E = 0,

so that <f> = const.

inside the sphere, the constant value being the value at thecentre, viz. e/a, or the value on the surface obtained by puttingr=a in (1), since the potential is continuous.

3*21. The capacity of a conductor is denned to be the chargeneoessary to raise it to unit potential when itis alone in the field,or when all other conductors are at zero potential. It followsthat the capacity of a sphere is equal to its radius, and that thephysical dimensions of capacity are one dimension in length.

3*3. Condensers. A condenser consists of a pair of con-ductors insulated from one another and generally so placedthat a surface of one is parallel to a surface of the other; e.g.

50 PARALLEL PLATE CONDENSER [3-3-

parallel planes, concentric spheres, coaxial cylinders; butparallelism of the surfaces is not an essential condition.

When the two conductors are so arranged that they haveequal and opposite charges, e.g. by earth-connecting one ofthem, the capacity of the condenser is denned to be the positivecharge on one of the conductors which will make the differenceof the potentials of the conductors equal to unity.

3*31. A parallel plate condenser. Let two parallel con-ducting plane plates be at a distance t apart and let us considerthe field between them, neglecting any external field andthe irregularities in the field caused by their edges and assume

—cr

that the tubes of force are everywhere straight. Then if a is thesurface density of the charge on one plate, — a is that on theother, and the electric vector E = 4ncr (2-34 (2)) is at right anglesto the plates and constant in value.

Then if <j>^, <f>2 denote the potentials of the plates <f>i—<f>2 = Et(the work done on a unit charge in moving from one plate tothe other) _ . .

Hence the capacity per unit area of the condenser, or thecharge when j>x — <f>2 = 1, is 1 j^nt, and the capacity for area A isA\^ttt, when the irregularities at the edges of the field areneglected. From 3*1, there is a force of attraction between the

p l a t e s w h i c h = 2ITO2A o r ——z {<f>x — <f>2)2.

3*32. The guard-ring. The effects of the edges may verylargely be eliminated by the following device: a circular plateA is surrounded by a concentric annular plate B in the sameplane, the inner radius of B being only slightly larger than the

3-4] GUARD-RING 51

radius of A, a metallic connection between them is made by apiece of wire so that they will have the same potential, andthey are placed parallel to a third conducting plate C. When thecondenser is charged the distribution of the unit tubes offorce which fall on C will be practically uniform save at itsedges, so that so far as the plate A is concerned the 'edgeeffects' have been eliminated; but we must take into accountthe tubes which fall upon the part of C opposite to the gapbetween A and B and this we can do by assuming that half ofthem start from A and half from B, so that if A' denotes thearea of A plus half the area of the gap between A and B thecapacity of the condenser formed by A and C is A'/4irt.

Again the force of attraction on the plate A towards G is

2TTAOZ or ——- (<}>! — <f>2)2, where A denotes the area.

O77t

If the apparatus is so arranged that the plate A is sus-pended from one arm of a balance, this force can be measured,and we have thus a practical method of determining a potentialdifference. Such an apparatus is called the Guard-Ring orAttracted Disc Electrometer.

3*4. Spherical condenser. Let a conducting sphere ofradius a be surrounded by a concentric conducting sphericalshell of internal and external radii b, b'.

Let the sphere of radius a have a uniformly distributedcharge e. The tubes of force which start from this charge allend on the sphere of radius b, so that its charge is — e. If theshell is insulated and has a total charge e' the charge on itsouter surface must be e + e', but by 3*33 the field between the

4-3

52 SPHERICAL CONDENSER [3-4-

conductors is independent of the charges on the surfaces of theshell and depends only on the charge on the inner sphere,having the value e/r2 at a distance r from the centre. Hence,if <f>a, <j>b denote the potentials ofthe sphere and the shell, the differ-ence <f>a—<f>bia the work done on aunit charge as it moves from thesphere to the shell, i.e.

e+e'

We notice that this is inde-pendent of the external radius ofthe shell and of the charge on itsouter surface, and that the argument is not affected bysupposing the shell to be earth connected. The only differencein that case being that <f>b is zero.

Therefore the capacity of the condenser, or the charge when<f>a-<f>b is u n i t y , is ab/(b-a) (2).

If on the other hand the sphere of radius a is connected toearth by a fine wire passing through a small hole in the shelland insulated from it, so that the inner sphere is at zeropotential, the capacity of the con-denser will be the total chargeon the shell when its potential isunity.

Let e, e' be the charges on theinner and outer surfaces of the shell.Since the tubes of force which startfrom the charge e all end on thesphere of radius a, its charge is — e.

Then, by 2-32, the field in thespace between the conductors is— e/r2 in the direction of r in-creasing; and the field outside theshell, being due to the sum of all the charges, is e'/r2.

Hence calculating the potential differences as the work done

'Earth,

3-42] CONCENTRIC SPHERES 53

on a unit charge as it passes firstly from the sphere to the shelland secondly from the shell to an infinite distance, we get

e + e' = <f>b Uf— + b');

and

Therefore

and the charge, when the potential difference is unity, is

— b'b — a

3-41. We have seen in 3*21 that the capacity of a sphere alone inthe field is equal to its radius, and in 3*4 that if a sphere of radius a issurrounded by a shell of internal radius 6 the capacity becomes 06/(6 — a)and this can be increased by diminishing the difference between 6 and o.To illustrate the utility of this increase of capacity, suppose that thewire in the second diagram in 3*4 is connected with the terminal of avoltaic cell whose potential <j> is too low to cause a deflection of theleaves of an electroscope, and that the shell is connected to earth. Thecharge e on the sphere is then given by

oh

Now let the wire be detached and removed, leaving this charge on thesphere; and let the shell, which might consist of two separable hemi-spheres, be also removed.

The potential of the sphere now alone in the field is e/a or b<f>/(b — a);so that we have now got a conductor whose potential is the originalpotential <j> multiplied by the factor 6/(6 — a).

3*42. Concentric spheres. Consider a number of concentric con-ducting spheres. Let the radius of the innermost be a and let the innerand outer radii of the spheres next in order be 6, V; c, c'; etc.

If the charges on the spheres are given we can find the potentials,and conversely. Let the charges be e, e', e", ...; then as in 3*4 theremust be charges — e and e + e' on the spheres of radii 6, b' respectively,and similarly — e — e' and e + e' + e" on the spheres of radii c and c' andso on if there are more spheres.

Again the spheres divide space into regions numbered 1, 2, 3, ... inthe diagram, in which the potentials may be denoted by fa, tj>s, <f>a

54 CONCENTRIC SPHERES [3-42-

Then if r denotes distance from the centre, the field in region 2 is e/r2,so that as in 3*4

*-*=/.*£*=•(5-5) (1)-Similarly, the field in region 4 is (e + e')/r2, so that

) (2);

and the field in region 6 is (e + e' + e")\r%, so thatie + 6' + e"dr=

e±lp^ (3).7* 0

By adding (2) and (3) we find fa, and by adding (1), (2) and (3) wefind fa, and these are the potentials of the conductors.

Alternatively, by using the fact that a uniformly distributed chargee on a sphere of radius a produces a field of potential e/a inside thesphere and ejr outside the sphere, we may take the sum of the effects ofeach spherical distribution in each of the regions in turn and writedown

* -

ce + e' + e"

oe + e' + e"

The formulae exhibit the continuity of the potential function, thusthe variable function <f>t becomes fa or fa according as we put r=a or b

3-5] USE OF LAPLACE'S EQUATION 55

<f>t becomes (f>3 or ^5 according as we put r = b' or c, and <f>t becomes <f>f

or zero according as we put r = c or oo.

3-43. Condenser formed of coaxial cylinders. Let a bethe radius of the inner cylinder and b the internal radius of theouter, then when a charge e per uni t length is uniformly dis-tr ibuted on the inner and the cylinders are long enough for usto neglect the effects of their ends there is a field of intensity2ejr between the cylinders (2*53), irrespective of whether theouter cylinder be charged or not . Hence for the potentials <f>a

and <f>b we have a relation(*62e b

T d

and therefore the capacity of the condenser is 1 /12 log- I per

unit length.

3*5. Use of Laplace's equation. In 2*41 we saw that inempty space the potential function <f> is a solution of Laplace'sequation V2<f> = 0. There are many simple applications of thisfact, especially where symmetry enables us to reduce thenumber of independent variables in the equation. We willtake as illustrations the three types of condensers consideredin 3-31, 3-4 and 3-43.

(i) The parallel plate condenser, neglecting edge effects. Let <j>lt <f>2

be the potentials of the plates P, Q(& > i)- Take an axis Ox at rightangles to the plates, with the origin JO on the plate P . Then if the plates "are large enough for us to neglectthe edge effects, the potential ^ at apoint between the plates will dependon its x co-ordinate alone, so that<f> is independent of y and z, and OLaplace's equation takes the simpleform tax

.(1).ex*-" -

By integration we get0)

a n d <j,

where F and O are arbitrary constants.•(3),

56 U S E O F L A P L A C E ' S E Q U A T I O N [3-5-

But, if t is the distance between the plates, when x=0, <f>=<f>i and,when x=t, <f> = (f>2. Substituting these values in turn in (3) we find that

©=& and *-=-(&-*•)/* W,

so that * = 4*-jfa-'M (6)

gives the potential at any point between the plates.Again the surface density a on the positively charged plate P is

given by Coulomb's Theorem, which is in this case

and, from (2), this is 4cira= — F

or, from (4) or (5), a = ^lJ% <6);

in agreement with the result obtained in 3*31 from the considerationof tubes of force.

(ii) Spherical condenser. In the air space between two spheres ofradii a, b uniformly charged, by symmetrythe potential function <j> depends uponr alone. Laplace's equation in polar co-ordinates (2*41) then takes the form

which gives, on integration,

~8r~7* ( 2 )

and t=-j+G (3),

where F and O are arbitrary constants.Hence, if <j>ly ^a are the potentials of the conductors, we have

when r = a, and <j> = a when r = 6, so that

b—a b—a

and ab(^-^) 6^-a^r r(b — a) b — a v '

gives the potential anywhere between the conductors.Again, if a be the surface density on the sphere r=a, we have, by

Coulomb's Theorem,

or fa^r-^^b — a

3-8] U S E O F L A P L A C B ' S E Q U A T I O N 57

But lna2a is the total charge e on the inner sphere, so that

e~ b^, '

and the capacity -; = =-— ,<l>i-<l>2 b-a

as in 3*4.The reader will find it an instructive exercise to apply this method

to a succession of concentric spheres as in 3*42, obtaining the potentialsin the successive regions as solutions of Laplace's equation.

(iii) Coaxial cylinders. Neglecting the effects of the ends of thecylinders, the problem is a two-dimensional one and there beingsymmetry about the axis the potential <j> is a function of r alone, wherewe must now use Laplace's equation in cylindrical co-ordinates (2*41),

!('§*)=» <"^ = 7 (2)

and (£ = Flogr+O (3).

Then, with the notation of 3*43, <j> = <j>ll when r = a and <f> = <j>i, whenr = b, so that

and <£=(V0log--'Mog"jy'log- (5)

gives the potential anywhere between the cylinders.Then Coulomb's Theorem gives, for the surface density a on the

inner cylinder, /8(A\ F4:irar= —(£)

so that the charge per unit length of cylinder is

) f ilogl (6).

6 1This makes the capacity -; 7-= jn—TTT^> as in 3*43.

<l>a-4>* 21og(6/a)'

3*6. Sets of condensers. A set of n condensers is said to bearranged in series when the plate of lower potential of the firstis connected with the plate of higher potential of the second,the plate of lower potential of the second to the plate of higherpotential of the third and so on as in the figure.

Let Clt C2, ... Cn be the capacities of the condensers, andlet a charge e be given to the outer plate of the first. There will

58 SETS OF CONDENSERS [3-6-

then be a charge — e induced on the opposing plate of the firstcondenser and therefore a charge e on the plate of the secondconnected with it, and so on, so that the charges on all theplates are e and - e alternately. Then since every pair of plates

e

—e

c,

* *e

—e

0,

* *—e

Ca

t t e-e

Cn

1+1

connected together have the same potential, we may denotethe successive potentials by j>x, <£2, • • • <f>n+ias m the figure; andthen we have

so that, by addition,I 1

+

But if C denotes the capacity of the compound condenser

c~cx+c2

+• - + cn ( '•

It follows that if the n condensers have the same capacity C,the capacity of the set arranged in series is C/n.

i

-«1

0,

iThe n condensers may also be arranged in parallel by

connecting all their plates of higher potential and alsoconnecting all their plates of lower potential as in the figure.

3-71] ENERGY OF A CHARGED CONDENSER 59

Then if <j>, <f>' denote the higher and lower potentials, thecharges on the successive condensers are given by

e^G^-n e2=C2(^-f), ... eB = Cntf-f)so that

e1 + e, + ...+eB = (C1 + Ca + ... + CB)fo-f).But if C denotes the capacity of the compound condenser

therefore C= C1 + Ct + ... + Cn (2).

It follows that if n condensers of the same capacity G arejoined in parallel the capacity of the compound condenseris nC.

3*7. Energy of a charged conductor. The energy of acharged conductor is measured by the work done in chargingit, and we assume that the work done in bringing a small charge8e to a place of potential 0 in an electric field is <j> he.

Consider a conductor alone in the field so that the potentialis due only to the charge on the conductor. If G is the capacity,the charge e and potential <f> are connected by the relatione=Cf

At any stage in charging the conductor let e' denote thecharge and </>' the potential, so that e' = C<f>'. Then the workdone in increasing the charge by a small amount de' is </>'de' or

Yje'de'. Hence the whole work done in placing a charge e on

the conductor is

jV<fe' = | g , or & or iC<p (1),

and this represents the energy.

3*71. Energy of a charged condenser. Let a condenserhave one of its plates earth-connected. Let G denote thecapacity of the condenser, and at any stage in charging thecondenser let ± e' be the charges on the opposing surfaces and</>' their potential difference so that e' = Cc/>'.

60 ENERGY OS A CHARGED CONDENSER [3-71-

The work done in simultaneously increasing the positivecharge by de' and the negative charge by —de'ia

fide' or ^e'de'.

Hence the whole work done in placing charges ± e on theopposing surfaces is

de' or ^ or \e$ or \C? (1),

where <f> is the final potential difference.

3*72. Application to the parallel plate condenser. Referringto the parallel plate condenser of 3*31, taking A to be the area of eitherplate and neglecting edge effects, we see that the capacity C—Al^nt;also from 3'1 each plate is acted on by a normal force 2irAo2 tendingto draw the plates together.

Let us consider the mechanical work that would be required and thechange in the energy that would result if the distance between theplates were increased from * to t+St; according as the charges or thepotentials are kept constant during the change.

Firstly, when the charges are constant, the work necessary to over-come the force of attraction is ZTTAO* St; and, from 3*71 the initial energyis ea/2(7, where e = Aa and O=Aj4mt, so that the initial energy is2ITAOH and this undergoes an increment 2nAa*U, or the increase inenergy is equal to the mechanical work done.

Secondly, when the potentials are kept constant; since, from 3*31,cr=(^1 — <f>2)/4*rt, the force of attraction 2nAa2 is A (^x — S)

2/87T«2, andthe mechanical work required is A (fa — fa) Uji-nt2. But in this casethe energy, expressed in terms of the potentials, is %G (fa — fay orA (fa — <j>2)

2/8Trt, and the increment in this consequent on an incrementStintis —A (fa — fa)2 8^8*^,80 that there isaloss of energy numericallyequal to the mechanical work done. The explanation of this lies in thefact that to maintain the plates at a constant difference of potentialthey would have to be connected to the terminals of a battery, i.e. asource of electric energy, and in order, in this case, to get an increasein energy equal to the mechanical work done the battery would have tosupply an amount of energy equal to twice the mechanical work done.

A general proof of the theorem of which this is a special case will begiven in the next chapter.

3-8. Approximate expression for the capacity of a con-denser. Let a condenser be formed of two nearly parallelconducting surfaces. Let a be the surface density on an elementof area d8 of the positively charged surface, and t the distance

3-9] ELECTROSTATIC UNITS 61

between the surfaces at a point on d8. Then if t is small andfa, <f>2 are the potentials of the conductors, we have from 2*34

= (<f>i — <j>2)lt approximately.

Hence the total positive charge

C ,„ <£i-<£2 CdS . ^ .= \adS= —— approxnnately;

J w J 'and therefore an approximate expression for the capacity ofthe condenser is — — .

4TTJ t

3*9. Electrostatic units for fields in air. The absoluteelectrostatic unit of charge in the O.G.S. system of units is, inaccordance with 2-321, the charge which will repel an equalcharge at a distance of one centimetre with a force of one dyne.

If the fundamental units of mass, space and time are denotedby M, L, T, since Coulomb's Law may be expressed by aformula ee'/rz = force, we have for the dimensions* [e] of electric

charge [e]2 = MLT~2 x L*, or [e] = MlL$T~\

Then since dectric intensity E is force per unit charge, wehave for the dimensions of electric intensity

[E] = M LT~* -s. M*I*T-X = M*L-lT-i;

and potential difference, or, as we shall call it later, electro-motive force, being the line integral of E, we have for itsdimensions

Then capacity being the ratio of charge to potential, itsdimensions are given by

[C] = [e]-[<£] = £,

in agreement with 3*21.The practical units are multiples or sub-multiples of the

absolute units. For present purposes it is sufficient to state thatThe practical unit of charge

1 Coulomb = 3 x 109 absolute electrostatic c.G.s. units.* See Dynamics, Pt. I, 4-8.

62 PRACTICAL UNITS [3-9-

The practical unit of potential difference or electromotiveforce

1 Volt= 1/300 absolute electrostatic C.G.S. unit.The practical unit of capacity

1 Farad = 9 x 1011 absolute electrostatic C.G.s. units;and the Microfarad is one-millionth of a Farad.

We remark that these units are consistent, in thatCoulomb = Volt x Farad,

or charge=potential difference x capacity.The reader should notice that the arguments by which we

deduce the dimensions of these physical quantities in terms ofmass, space and time will need qualification when we con-sider fields in other media than air; also, that there is anothersystem of units—the electromagnetic—which starts from adifferent basis and differs from the electrostatic system.

3*91. Examples, (i) Two condensers of capacities 0-1 microfaradand 0-01 microfarad respectively are charged to voltages 1 and 10 respec-tively. Shew that if they are then connected in parallel there is a loss ofenergy amounting to 3-7 ergs. [M. T. 1933]

Using the result of 3*71 for the energy of a charged condenser, anddenoting the capacities of the condensers by Ct, C2 and their potentialdifferences by <j>lt <f>it the initial energy is

where <71 = 0-l microfarad = 9 x 10* abs. units ,<7a = 0-01 microfarad = 9 x 10s abs . units ,<j>x=l volt = 1/300 abs. unit , and ^2 = 10 volts = 1/30 abs. un i t ;

so t h a t t he energy = 5-5 ,ergs; and the charges on the z_J ZJS-upper plates are Oj = 300 c t C2

and O2^s = 300 abs. units. g 0Now let the condensers be / j)

connectedinparallel.i.e.theupper plates are connected and the lower plates remain connectedthrough the earth. Let <j> be the common potential of the upper plates,then their charges become Ct $ = 90000^ and Ct<f> = 9000^.

But their total charge is 600 abs. units, so that 99000^ = 600 or^=1/165.

And the energy of the compound condenser is half the chargemultiplied by the potential difference, i.e.

i x 600 XjJ-5 = g = 1-8 ergs,

so that there is a loss of energy of 3-7 ergs.

3*91] EXAMPLES 63

The energy might also be obtained from the formula £Ctya, whereC the capacity of the compound condenser =C 1 + C2 (3"6(2)).

(ii) The two plates of a parallel plate condenser are at a distance d apartand are charged to a difference of potential of 300 volts (1 electrostaticunit). If one of the plates can turn about an axis through its centroid,prove that it will be in equilibrium when parallel to the other plate, but thatthe equilibrium will be unstable, the couple required to hold it at a smallangle 0 with the other plate being J0/4«d8, where I is the moment of inertiaof its area about the axis. It is assumed that to the approximation required(he lines of force between the plates are nearly straight and the field outsideis negligible. [M. T. 1932]

When the plates are parallel the surface density is uniform and themechanical force 2no% per unit area is uniform and gives a resultantthrough the centroid of the plate, sothat there is no force having a momentabout the axis and the plate is in equi-librium. We shall prove that theequilibrium is unstable by shewingthat when the movable plate is turnedthrough a small angle 6 it is acted onby a couple tending to increase the displacement.

At a distance x from the axis the distance between the plates isd + Bx, so that the surface density a in this position is given by

and the force per unit area is

Let y be the breadth of the plate parallel to the axis at a distance xfrom the axis, and let the limiting values of 2 be ± a; then the momentabout the axis of the forces tending to decrease 8

yxdx _ _1_ fa yx<

to the first power of 8.

/a [a

yxdx=0; also I yx*dx = I.-a J -a

Therefore there is a couple of magnitude I8/4ird* tending to increase 6.(iii) A conducting sphere of radius a carrying a charge Q is divided into

any two parts which remain in electrical contac with one another. Shewthat the 'repulsion between them reduces to a single force through thecentre of the sphere and that its component in any direction is equal toQi8l(8nal), where S is the area of the curve formed by the projection onthe plane normal to that direction of the curve dividing the two parts.

64 EXAMPLES [3-91

What meaning must be attached to S if its boundary has a doublepoint? [M. T. 1934]

The mechanical force per unit area on the surface of the sphere is27TO-2 outwards (3*1), where in this case a= Q/iira2.

Hence on an element of area du> there is a force 2no2da> passing throughthe centre 0 of the sphere; and since the force on every element passesthrough 0 therefore the resultant force on any portion of the sphericalsurface passes through 0. To find the component in an assigned direc-tion Ox of the resultant force on a specified area of the spherical surface,weobservethat the element of areadco gives a contribution 2ira2da> cos 6,where 8 is the angle between Ox and the normal to the element. Butda> cos 6 is the projection dS of the element dw on a plane perpendicularto Ox. Hence the element makes a contribution 2ttald8 to the requiredforce. And by summing for all such elements the required force is seen

(Ill)to be 2TT<72<SI, where S has the meaning assigned in the question, or

Let the curve dividing the surface of the sphere be the curveQHKLM, as in the second figure, where OHK is on the front of thesphere and KLMO is on the back. It is easy to see that if we take thedirection of the required component force to be at right angles to theplane of the paper, the projection of the boundary will have a node.I t will in fact be a curve roughly resembling the curve QHKLMcrossing itself at a point X. But the portion ONKH of the sphere willcontribute a force acting upwards through the paper, and the portionKNOML will contribute a force acting downwards through the paper,so that the required component of the resultant force perpendicularto the plane of the paper will correspond to the difference of the pro-jections of these areas, which will be the difference of the two loops intowhich S is divided through having a double point, and it is in this waythat we must interpret S when its boundary has a double point.

(iv) Two conducting concentric spherical surfaces, radii a, b (a>b),insulated from each other and from earth, are given charges Qlt Q2.After the spheres have been charged, the inner sphere is connected to a

3-91] EXAMPLES 65

distant uncharged gold leaf electroscope whose capacity is equal to thatof a sphere of radius c. Find the charge x which the electroscope receives.

The two spheres are now connected to each other without affecting thedivergence of the leaves of the electroscope, and the final charge on theelectroscope is thus either x or —x. Shew that

«i = ? or «i= - ° ( a + 6 +9

2 l [M. T. 1924]Qt c Q3 bc + ca + 2ab L J

After the inner sphere has been connected to the electroscope let<f>t be their common potential, and fa that of the outer sphere. Sincethe inner sphere has given up a part x of its charge to the electroscope,it retains Qa — x, and the charges on the outer sphere are therefore— Q% + x on its inner and Qx + Q2—x on its outer surface.

If we assume that the charge on the electroscope is too far distantto affect directly the potential at the centre of the spheres, we have

fa—bT + T (1)-

But, since c is the capacity of the electroscope,x = c<j,2 (2),

so that by eliminating <f>2 from (1) and (2) we get

* - a(b + c) Wm

In the second case the two spheres are connected and the divergenceof the leaves of the electroscope remains unaltered, so that the chargeon the electroscope must be either a; or — x.

The electricity on the inner sphere now passes to the outer sphere andits total charge becomes Qt + Qt + x according as the charge on theelectroscope is ± x. By equating the potentials of the spheres and theelectroscope, for they are all now in contact, we get

Q1 + Q,Tx=±x c(Q1 + Q2)a 0 a+c v '

Then by equating the values of a; found in (3) and (4), we get either

9l = a or Ql=

Q3 c Qi bc + ca + 2ab'REM 5

66 EXAMPLES [3-91

(v) A circular disc of non-conducting material of radius a is free torotate in its plane about its centre O. At a point in its plane at distanceb (b>a) from C one terminal of an electrical machine is placed and this ischarged up until sparks pass across to the edge of the disc. Explain whythe disc commences to rotate.

After a time the rate of rotation of the disc steadies down to a constantvalue, the moments of the electrical and frictional forces balancing. If theterminal at A be regarded as having a point charge E and if the charge onthe disc be limited to its rim and at P is E'e~^ per unit length of rim,where A OP = 6, shew that the moment of the electric forces round the axisof the disc is [** anEWe-faneM m T 8 „

The sparking across from A to the disc means that electric chargepasses from A to the edge of the disc, and begins to accumulate theresince the disc is a non-con-ductor. There is then a re-pulsion between the chargeat A and the charge at theopposing point of the disc,and as sparking is an irregularprocess this latter charge willnot be uniformly distributedand the equilibrium will beunstable so that the discbegins to rotate.

Taking the charge on an element add of the rim at P to be aE'e~^$dd,the repulsion between this charge and the charge E at A is

and its moment about O is got by multiplying by a sin GPA orab sin 8/AP. But AP2 — a2 + b2 — 2ab cos 8, therefore the total moment is

EXAMPLES

1. Calculate the capacity of a condenser formed by two circularplates of tinfoil mounted on glass, taking the diameter of eaoh plateto be 40 cm. and the thickness of the glass to be equivalent to an airthickness of 1-5 mm. Express the result in terms of a microfarad.

[M. T. 1914]

2. One plate of a parallel plate condenser is a circle of radius 10 cm.and it is surrounded by a guard ring. The distance between the platesis 5 mm. Prove that when the force of attraction between the plates is5 grams weight the difference of potential between the plates is approxi-mately 3000 volts. [I. 1910]

EXAMPLES 673. The trap-door of an electrometer is a circle of 6 cm. diameter, and

is at a distance of 2 mm. from the lower plate; it is found that a weighti gram will balance the attraction between the plates. Calculatethe potential difference in volts. [M. T. 1913]

4. The trap-door of a guard ring electrometer is circular, 6 cm. indiameter, and the difference.of potential between the plates is 4 electro-static O.G.s. units; if the force of attraction is 50 dynes, what is thedistance between the plates? [St John's Coll. 1907]

5. What difference of potential (in electrostatic units) between theplates of a guard ring electrometer is required to produce an attractionon the trap-door of 5 dynes per square centimetre, when the distancebetween the plates is 0-3 centimetre? What is the potential differencein volts? [I. 1910]

6. Three parallel plates A, B,G, each 10 cm. square, are maintainedat potentials 50,100 and 10 volts respectively. B is between A and C,A and B being separated by 1 mm. and B and C by 3 mm. Find themagnitude and direction of the force required to prevent B moving.

[M. T. 1925]

7. Shew that if the distance between the plates of a parallel platecondenser be halved the electric energy is halved or doubled accordingas the charges or the potentials are kept constant during the change.Explain this on physical grounds. [I. 1915]

8. Neglecting edge effects, find the capacity of a condenser in airconsisting of two sets of parallel square plates of side a each at distanceh from its neighbours, and connected together alternately as in thediagram.

If the outer set are fixed in position and the inner set are free to moveall together to the right or to the left, determine, from a considerationof the mechanical strains on the plates, whether the equilibrium of thecentral position is stable or unstable. [M. T. 1930]

9. A conductor is charged from an electrophorus by repeatedcontacts with a plate which after each contact is re-charged with aquantity E of electricity from the electrophorus. Prove that, if e isthe charge of the conductor after the first operation, the ultimatecharge is Eej(E-e). [M. T. 1893]

5-a

68 EXAMPLES

AB

10. Two condensers AB, OD have capacities Gt, Gt respectively.They are initially uncharged, and A is permanently connected to earth(potential zero). The two con-densers are connected in series(Fig. a) and D is charged to apotential <j>. CD is then discon-nected, and joined in parallel withAB (Fig. 6). Shew that there is aloss of energy equal to

ABCD

What becomes of this energy? la)Draw two diagrams similar toFigs, a and b, and mark against each plate (1) its potential, (2) itscharge. [M. T. 1925]

11. Two condensers A and B both consist of a pair of parallelcircular plates of radius r. Initially the plates of A are separated by adistance a and are charged to a potential difference cf>, the plates of Bare separated by 6 and are uncharged. The plates of A are now separatedto a distance na and connected one to each of the plates of B. Shew that

there is a gain of energy to the system of -~-. —T units.

[M. T. 1926]

12. A circular disc of radius a surrounded by a wide coplanarguard ring is placed with its plane inclined at a small angle e to anotherplane. The centre of the disc is at a perpendicular distance d from theplane. The plane and the disc with its guard ring are maintained at apotential difference <£. Shew that the mechanical forces acting on thedisc reduce to a force (j>2aa/(8a!2) perpendicular to its plane and a couple

[M. T. 1920]

13. Four equal large conducting plates A, B, C,D are fixed parallelto one another. A and D are connected with the earth, B has a chargeE per unit area of its surface, and G a charge W. The distance betweenA and B is a, between B and G it is 6, and between G and D it is c.Find the potentials of B and G. [I. 1905]

14. A spherical conductor of radius 10 cm. has initially a charge of6284 units. Electrical connection is made, for an instant, with adistant insulated non-charged spherical conductor of radius 15 cm.What are the final charges of the two spheres? [M. T. 1917]

15. The radii of two concentricconductingspheresare5-4and5-6cm.,the outer is earthed and the inner raised to a potential of 200 volts.Calculate the mechanical stress per sq. cm. on the inner sphere.

[I. 1914]

EXAMPLES 69

16. Two concentric spherical conductors of radii a and 6 (a<b)carry charges e and e'; find the fields of force in the region between thespheres and in outside space. How would these fields be altered if thespheres were connected for an instant by means of a wire? Whatwould be the loss of electrostatic energy? [St John's Coll. 1907]

17. On the outer of two concentric spherical conductors of radiia, b (a < b) is a charge Q; what charge must be given to the inner tobring its potential to the value U ? Obtain expressions for the potentialfunction in all space. [St John's Coll. 1909]

18. Two insulated conducting spheres, of radii 10 and 20 cm.respectively, carry equal charges of 6 electrostatic units. They are sofar apart that their influence on one another may be neglected. Shewthat if they are joined by a wire the energy lost is equal to the workrequired to raise a weight of one milligram through a height of about3-06 mm. [M. T. 1932]

19. A condenser is formed of a conducting sphere of radius r, anda conducting concentric spherical shell of internal radius B, insulatedfrom it, the dielectric being air. Shew that, if there be placed betweenthese two spherical conductors n conducting concentric sphericalshells, the capacity becomes

1

where alt at,... are the internal radii and blt bit... the external radiiof the shells. [I. 1891]

20. A circular gold leaf of radius 6 is laid on the surface of a chargedconducting sphere of radius a, a being large compared to 6. Provethat the loss of electrical energy in removing the leaf from the con-ductor—assuming that it carries away its whole charge—is approxi-mately J&2.E2/a8, where E is the charge of the conductor and thecapacity of the leaf is comparable to 6. [M. T. 1896]

21. Three concentric thin spherical shells are of radii a,b,c(a<b<c);the first and third are connected by a fine wire through a small hole inthe second, and the second is connected to earth through a small holein the third. Shew that the capacity of the condenser so formed is

i^-+-^-r. [M. T. 1909]0 — d C — 0

22. Three insulated concentric spherical conductors, whose radii(in ascending order of magnitude) are a, b, c, have charges elt ea, e8respectively. Find their potentials, and shew that, if the innermostsphere be connected to earth, the potential of the outermost is diminished

[M. T. 1901]

70 EXAMPLES

23. A condenser is formed by a sphere B of radius b inside a con-centric sphere A of radius a. A second condenser is formed by a sphereJD of radius d inside a concentric sphere G of radius c. Connection canbe made with B, D through small holes in A, G. Initially A, G areearthed and B, D are at potential T. The sphere G is now insulated andafterwards B is joined to C by a fine wire and D is earthed. The spheresA and C are so far apart that the inductive effect of either on the othermay be neglected. Shew that, if O is the final potential of B and G,

(a-6) T_(a — b)

24. If themiddle conductor of acondenser, formed by a long cylinder(radius a) inside a coaxial shell (internal radius 2o and external 3a)with another coaxial cylinder (internal radius d) outside, be chargedand the other two put to earth, determine the minimum value of d inorder that the parts of the middle shell may not separate, when it iscut into two parts by a plane through its axis. [I. 1912]

26. A condenser is formed of three concentric cylinders of which theinner and outer are connected together. Obtain a formula for thecapacity, neglecting end effects, and shew that, if the middle plate is10 cm. long and the radii of the cylinders are 3-9, 4-0, 4-1 cm., thecapacity is equal approximately to that of a sphere of 4 metres radius.

[M. T. 1920]

26. On a certain day the vertical electric force in the atmosphereat the earth's surface was 100 volts per metre and at a height of 1-5kilometres it was 25 volts per metre. Prove that the total charge in theatmosphere per square kilometre of the earth's surface up to thisheight was 1-99 x 108 electrostatic units, and that the effect of thischarge is to increase the barometric pressure by about 4 x 10~' dyneper sq. cm., assuming the charge uniformly distributed. [M. T. 1921]

27. Sketch the lines of force of the system of charges e at (2a, 0,0)and — \e at ($a, 0,0). Prove that the sphere r = a is the equipotential(£ = 0 .

Prove that, if the sphere r = a is a conductor at zero potential, thesurface density at (a, 0,0) is — 3e/47ra2, and the mechanical stress onthe conductor 9e2/87i-a* dynes per (cm.)2. [M. T. 1920]

28. Prove that the electric capacity of a conductor is less than thatof any other conductor which can completely surround it.

[M. T. 1904]

29. A condenser formed of two concentric spheres of radii a and bis divided into two halves by a diametral plane, the inner and outersurfaces being rigidly connected. Prove that the force required to keep

the two halves together is Je2 ( p ^) > where e and — e are the total

charges on the spheres. [M. T. 1922]

EXAMPLES 71

30. An electrified spherical soap bubble of radius a is surroundedby an insulated uncharged concentric spherical conducting shell whoseinternal and external radii are b and c; shew that when the bubble burststhe loss of electric energy is independent of c.

If the spherical shell consists of two hemispheres in contact, provelea

that the increase in the mechanical force between them is = j - a , where

e is the original charge of the bubble. [I. 1899]

31. A conducting sphere of radius r is electrified to potential V. Ifthe sphere consists of two separate hemispheres, shew that the forcebetween them is $V2; and if the whole be surrounded by an uninsulatedconcentric spherical conductor of internal radius R and the potentialof the solid sphere is still V, prove that the force between the hemi-spheres is iV^B^KR-r)". [I. 1895]

32. An insulated spherical conductor, formed of two hemisphericalshells in contact, whose inner and outer radii are b and b', has within ita concentric spherical conductor of radius a and without it anotherconcentric spherical conductor of which the internal radius is c. Thesetwo conductors are earth connected and the middle one receives acharge. Shew that the two shells will not separate if 2ac >bc + b'a.

[M. T. 1896]

33. Outside a spherical charged conductor there is a concentricinsulated but uncharged conducting spherical shell, which consists oftwo segments; prove that the two segments will not separate if thedistance of the separating plane from the centre is <a6/(aa + 62)*,where a, b are the internal and external radii of the shell. [I. 1897]

34. An infinite circular conducting cylinder has charge q per unitlength, and is surrounded by a thin coaxial cylindrical shell which is aconductor connected to earth. If the shell is divided along two gener-ators subtending an angle 28 at the axis of the cylinders, determine theresultant force between the portions. [Trinity Coll. 1898]

35. Calculate the capacity in fractions of a microfarad of a condenserformed by coating the inside and outside of a cylindrical j ar with tinfoil.The radius of the jar is 12 cm., the thickness of the glass is equivalentto an air thickness of £ mm. and the tinfoil reaches to a height of30 cm. as well as covering the base of the jar.

[The microfarad is equal to the capacity of a sphere whose radius is9 kilometres.] [St John's Coll. 1915]

36. Shew that if a conductor is slightly deformed so that the elementdS is displaced normally outwards a distance £ the consequent changeSO in the capacity is given by

72 EXAMPLES

E being the total charge. For example if a spherical conductor isslightly deformed the change in capacity is proportional to the changein volume. [St John's Coll. 1911]

ANSWERS

1. 0-00074. 3. 1260 approx.4. 6 mm. 5. 3-36= 1008 volts.6. 7 dynes towards C. 8. 3a2/2wfe. Unstable.

{ }a+b+c ' a+b+c

14. 2513-6 and 3770-4 units. 15. 0-47 dyne per sq. cm.

16. ie*Q-I). 17. a{U-(Q/b)}. r<a,<f,=, . aU aQ Q , . aU aQ Qa<r<b, * = — -£ + %; 6 < r # = — - £ + *

24.34. {q*/nb) sin 6 per unit length, where 6 is the internal radius of theshell.35. 00048.

Chapter IV

SYSTEMS OF CONDUCTORS

4-i. When an electric field is produced by charges on anumber of conductors, it is clear that the distribution of thecharge on each conductor is affected by inductive effects ofthe charges on the other conductors. The surface density isgiven at every point by Coulomb's Law (2-411 (2))

4mr=En= -d<f>ldn (1),

so that a knowledge of the potential function <f> would enableus to calculate the distribution of charge everywhere; and thetotal charge on any conductor may be represented by

— — ^- dS integrated over the surface.4TTJ on °

Generally either the total charges on the separate con-ductors are given, or else the constant values which thepotential function takes on the conductors are given, andfurther, we know that the potential function <j> must satisfyLaplace's equation V2<£ = 0 at all points at which there is nocharge. While there is no general method for finding a valueof <j> which satisfies Laplace's equation and takes given valuesover a given set of surfaces, or provides for given total chargesover a given set of conductors, yet it is possible to establish agood many general theorems about the kind of electric fieldwhich we are discussing, as will be seen in what follows; but,because the proof requires too much analysis, we shall assumethat the potential at a point can always be representedby the sum of each element of charge divided by its distancefrom the point. (Cf. 2*5.)

4*2. Principle of superposition. Let A1} A2, As, ... bea set of conductors and let a distribution of variable surfacedensity a give the conductors total charges e1} ea, e3, ... andproduce a field of potential (f> which takes constant values0i> 0a > 03 > ••• o n *n e surfaces of the conductors. Let there be

74 UNIQUENESS THEOREMS [4-2-

an alternative distribution of surface density a' giving totalcharges e/, e2', e3, ... and producing a field of potential <j>'which takes constant values <f>^, <j>2', <f>3', ... on the conductors.

Then <f> = integrated over the conductors

and (f>' = integrated over the conductors.

Hence a distribution in which the surface density is a + a'

will produce a field of potential dS = cj) + <}>', and th is

will take the values (j>i + <f>i, 02 + 02', 03 + 03'> " . o n the con-ductors, and their total charges are now ex + e-l, e2 + e2',e3 + e3', ...; and these charges are still in equilibrium becausethe surfaces are at constant potentials; so that we have provedt h a t if charges ex, e2, e3, ... give rise to potentials (f>x, (f>2, <j>3, ...and charges ex', e2 ', e3, ... give rise to potentials <j>x, i^2', <f>3, ...,then charges ex + e-{, e2 + e2', e3 + e3,... will give rise to potentials4>\ + $i'> 02 + 02) 03 + 03> This is the principle of super-position of electric fields.

4-21. Uniqueness theorems, (i) There is only one vxty inwhich given charges can be distributed in equilibrium over agiven set of conductors.

Let Ax, A2, A3, ... be the conductors and e1, e2, e3, ... thecharges placed on them. If possible let there be two differentways of distributing the charges on the conductors in equili-brium; and let a, a' denote the surface densities at the samepoint in the two distributions. Change the sign of the electri-fication at every point of the second distribution and thenimagine the two to be superposed. This joint distributionwould still be in equilibrium and it would be a distribution inwhich each conductor has a total charge zero. Hence, unlessa' = a at every point, each conductor has a charge partlypositive and partly negative. Let <£x, <f>z, <£3,... now denote thepotentials of^41,^L2,^43, .... Then since there is no maximumof 0 in empty space, one of these potentials, say <f>t, must be thehighest in the field. Therefore tubes of force start from A± and

4-3] COEFFICIENTS OF POTENTIAL 75

none end on it; but this contradicts the hypothesis that it hasboth positive and negative electricity on its surface; thereforea = a at all points on A±, and no tubes of force start from it orend on it. By similar reasoning we prove that a' = <J at all pointson the rest of the conductors, or the equilibrium distributionof given charges is unique.

(ii) There is only one distribution of electricity on a given setof conductors which will cause those conductors to have givenpotentials.

If possible let a, a' denote surface densities at the same pointin two different distributions of electricity which both causethe conductors Ax, A2, As,... to have potentials <f>±, <f>2, cf>3,....Then the potentials <j>, $' of the two distributions may berepresented by

/ *? and f -

integrated over the surfaces of the conductors, where

o n ^ 4 1 ; (f> = (f>' = (f)1; o n . 4 2 , <f> = <f>' = <f>2; a n d s o o n .

It follows that a distribution in which the surface density isa—a' would produce a field in which the potential of eachconductor would be zero. But the potential has no maximumor minimum in empty space, hence the potential of the chargesof density a — a' is zero everywhere. Therefore there can be noelectricity in the field, i.e. a = a everywhere, or the distributionwhich produces the given potentials is unique.

4*3. Coefficients of potential, capacity and induction.Let an electric field be produced by charges residing on a setof n conductors.41; A2, ... An of prescribed shapes occupyingfixed relative positions in space.

In all calculations of potential made hitherto charge hasentered linearly into the expression for potential, so we assumethat if a charge ex is given to the conductor Ax, the other con-ductors being uncharged, the potentials of all the conductorswill be proportional to ex and that they may be denoted by

76 COEFFICIENTS OF POTENTIAL [4-3-

Similarly when the conductor A2 has a charge e2 and all theother conductors are uncharged, the potentials may be denoted

> 2>22e2> 2>23e2> •

and so on. Finally when the conductor An has a charge en andall the other conductors are uncharged, the potentials may bedenoted by

Pnlen> Pn2en> Pn3en> ••• Pnn^n-

Now imagine these n fields to be superposed and we have afield in which the charges on the conductors A1; A2, ... An aree1; e2, ... en and the potentials of the conductors may bedenoted by <£1; <f>2, ... <f>n, where, by the principle of super-position,

Them's are called the coefficients of potential of the system ofconductors; they depend only on the shapes and sizes andrelative positions of the conductors.

We see from the definition that prs denotes the potential ofthe conductor As when the conductor AT has a unit charge andall the other conductors are uncharged. We shall prove later

Since the equations (1) express the <£'s as linear functions ofthe e's, they can be solved for the e's, so that the e's are expres-sible as linear functions of the c£'s in the form

where the q'B are functions of the p ' s .The coefficients with double suffixes qn,q22>... qnn are called

coefficients of capacity, and the other q's are called coefficientsof induction. Thus q^. is the charge on Ar when it is at unitpotential and all the other conductors are at zero potential(earth connected). This agrees with the definition of the

4-32] COEFFICIENTS OF POTENTIAL 77

capacity of a conductor given in 3-21. Also qra is the chargeinduced on the conductor As when the conductor Ar is raisedto unit potential and all the conductors except Ar are earthconnected.

4*31. Coefficients of capacity are positive, those of inductionare negative and the sum of the latter belonging to any given con-ductor is generally numerically less than the coefficient of capacityof that conductor.

Let Ax be insulated and raised to potential <£x and let allthe other conductors be earth connected, so that

Then from 4-3 (2)

Now if ex is a positive charge, <j>x is a positive potential, forthe tubes of force all start from A1 and proceed to places oflower potential (in this case zero). Therefore qn is positive.But the charges induced onA2,A3t... are clearly all negative,being at the other ends of the tubes, therefore q12, q13, ... qln

are all negative. Also the sum of the charges e2, e3, ... en

cannot numerically exceed ex, therefore

Sii + fca + • • • + Jm * fci numerically.

4'32. Coefficients of potential are all positive and prs cannotexceed Prr or p8S.

Let Ar have a positive unit charge and let all the otherconductors be uncharged, then, as in 2-43 (ii), pn is the highestpotential in the field and is positive. If As lies inside Ar, then,since it has no charge, its potential is the constant potentialinside Ar, i.e. hi this case prs=prr. But when As lies outside Ar,then, since A3 has no total charge, as many unit tubes of forceleave it as fall upon it, and those falling on it come from placesof higher potential and those leaving go to places of lowerpotential, so that pra lies between the highest and lowestpotential in the field, i.e. between p^ and zero. Similarly pra

cannot exceed

78 ENBBGY OF A SYSTEM OF CONDUCTORS [4-4-

4-4. The energy of a system of charged conductors.If it were possible to create a permanent electrostatic field inwhich there were no changes or movement of electricity owingto imperfect insulation or other cause, the maintenance ofsuch a state would not require any expenditure of energy; but,in the creation of the field, work would have to be performedin the separation of the electric charges by some mechanicalprocess. This work is to be regarded as conserved in the fieldin the form of electrical energy and is available for trans-formation into heat or mechanical work when the field ceasesto be static.

The energy of the field of the n charged conductors of 4-3may be calculated as the work required to place the chargeson the conductors. In accordance with the definition ofpotential of 2*4 the work which an external agent would haveto perform in order to bring a small charge de to a place ofpotential <f> is <f>de. Now suppose the n conductors to be chargedsimultaneously in such a way that at any instant in the processeach conductor has received the same fraction x of its finalcharge, so that the charges are

xe^, xe2) ... %&n.

Then, at this instant, the potentials being linear functionsof the charges are

where <f>lt <f>2, ... <j>n denote the final potentials; and the worknecessary to increase the charges by further small amounts

e^dx, e2dx, ... endx

is j d +

or

And the whole process of charging is completed by summingfor values of x between 0 and 1, so that the total work done orthe energy of the field is

W = ~Z(e<f>) | xdxJo

or PF = £Z(e<£) (1).

4-41] GREEN'S RECIPROCAL THEOREM 79

The theorems of 3'7 and 3*71 are particular cases of thisresult.

4'41. Green's Reciprocal Theorem.* / / the charges andpotentials of a set of n fixed conductors are altered from

e1,(f>1; e a ,<£ 2; ... en,<f>n

to e i ' ,& ' ; e . ' > & ' ; • • • «»' .*»' .

then e ^ ' + « ,&' + . . . +en<f,n' = e / ^ + e2'<£2 + . . . +en'<j>n

or S(ef ) = S(e'<^).

Let the alterations take place in such a way that at anyinstant each conductor has received the same fraction x of itsfinal increment of charge, so that the charges are

H + x(ei'-ei)> e2 + x(e2'-e2), ... en + x(en'-eJ;their potentials must then be

and the work necessary to make further small increments ofC a r g e (e/-ex)dx, (e2'-e2)dx, ... (en'-ejdx

is

and the total work necessary to alter the state of the con-ductors as prescribed is found by summing for values of xbetween 0 and 1, i.e.

fJoor £ S (^ + (£')(e'_e) (1),

i.e. the sum of the increment in charge of each conductor multipliedby the mean of its initial and final potentials.

But this work must measure the difference between theinitial and final energy of the field, so that

which reduces to2(e f ) = 2(e'<£) (2).

* George Green (1793-1841), mathematician.

80 RECIPROCAL THEOREMS [4-41-

COR. By adding 2 (e<£') — 2 (e'^)to expression (1), it becomesi S ( e + e') (<£' — <f>), expressing the work necessary to changethe state as the sum of the increment in potential of each con-ductor multiplied by the mean of its initial and final charges.

4*42. To prove that qrs=qv and prs=par. (i) To prove thefirst result, consider first the field in which Ar is insulated andgiven a charge which makes its potential unity while all theother conductors are earth connected, i.e. <£r= 1 and all other</>'& are zero.

Then, from 4-3 (2),

6 1 = 3 , 1 . C 2 = 9'r2> ••• ^r = 9rr> ••• ^ i r s * ••• e B = ? r n -

Consider next the field in which Aa is insulated and given acharge which makes its potential unity while all the otherconductors are earth connected, i.e. <f>s' = 1 and all other <f>"aare zero. Then, as above,

« i ' = fci. ««' = ? * . • " « / = &•» ••• « . ' = ? « . ••• en' = isn-

But by Green's Reciprocal Theorem

S(ef) = S(e'fl,therefore qra = q^.

(ii) To prove the second result, consider first the field inwhich Ar has a unit charge and all the other conductors areinsulated and uncharged, i.e. er= 1 and all other e's are zero.

Then, from 4-3(1)

< £ l = 2 > r l > 4>t=Pri> ••• 4>r=Prr> ••• <f>8=Prs> ••• K = P r n -

Consider next the field in which Aa has a unit charge and allthe other conductors are insulated and uncharged, i.e. ea' = 1and all other e"s are zero. Then, as above,

$ 1 = P s l > 4>a =P*a> ••• <f>r'=Psr> ••• <t>8=Pss, ••• <f>n'=Psn-

But 2(ef) = 2(e'<£),SO that Psr^Prs-

It would be sufficient to prove one of these theorems becausethen the other would follow from the relations which expressthe £>'s in terms of the q'a or vice versa.

4-43] GREEN'S RECIPROCAL THEOREM 81

4'43. Further applications of the Reciprocal Theorem, (i)Shew that the locus of the positions in which a unit charge will induce agiven charge on a given uninsulated conductor S is an equipotentialsurface of 8 when 8 is freely electrified and alone in the field.

Here we have two conductors, viz. S and an infinitesimal conductorat a point P .

In the first state (Fig. (i)) a given charge ex is to be induced on S atzero potential by a unit charge at P, so that

ex is given, e2 = 1, fa = 0 and fa is unknown.

In the second state (Fig. (ii)) 8 is freely electrified and alone in thefield, i.e. there is no charge at P and the conductor at P is regarded astoo small to disturb by its presence the electrification on S; in thisstate therefore ex' and fa' are definite constants, ea' is zero and fa' isunknown. But by the Reciprocal Theorem

«i fa' + ea fa' = ex'fa + et'fa,so that, by substituting the particular values above, we get

e1fa' + fa' = O; or fa^-e^fa'.But ex is a given charge and fa' is the potential of S when freely

electrified and therefore a constant, so that fa' is constant.But fa' is the potential at P in the second state; hence the required

locus of P is an equipotential surface of 8 when 8 is freely electrifiedand alone in the field.

(ii) Two closed equipotential surfaces fa and fa are such that fa,encloses fa, and <f>p is the potential at a point P between them. If now acharge e be placed at P and the equipotential surfaces are replaced byconducting surfaces at zero potential, prove that the charges induced onthese surfaces are given by

ei = e» = 6

<f>p — fa fa — <f>p fa—faThe distribution of potential in a field in air is not affected by placing

a thin conducting sheet in the position of any equipotential surface,for the lines of force will still continue to cross the surface at right

REM 6

82 ENERGY [4-43-

angles. Hence we may suppose the thin conducting surfaces to existin both the fields described in the problem.

In order to complete the specification of the first field (Fig. (i)) wemust suppose it to be produced by a charged conductor, having acharge e0 and potential <f>0 say, inside the given equipotential surfaces,and let this be the only charge in the field. We can suppose also thatthere is a small conductor at P which carries the charge e in the secondfield (Fig. (ii)).

Then in the first field we have

andcharges e0, 0, 0, 0potentials <j>0, <j>1, <f>p, <£a in order.

And in the second field, when there is no charge on the innermostconductor, and the other conductors save that at P are at zero potential,we have

charges 0, ex, e, e2

and potentials 0, 0, <f>P', 0 in order,

where </>?' denotes the potential at P in the second field, and theinnermost conductor is at zero potential because it is uncharged andlies inside a conductor at zero potential.

Applying the Reciprocal Theorem we get

(1).

Also since in the second field all the tubes of force which start fromthe charge e end on one or other of the conducting surfaces, therefore

^ + 6 + 6,5=0 (2);

and from (1) and (2) we obtain the required result.

4-5. Energy as a quadratic function of charges orpotentials. Taking the expression for the energy from 4-4, viz.

and substituting from 4-3 (1) for the potentials in terms of the

4-6] MECHANICAL FORCES 83

charges, we get a quadratic expression for W in terms of thecharges, which we may denote by We and write

We = %p11e1*+plie1e2 + ip!i!ie!*+p13exe3 + (1).

If instead we substitute from 4*3(2) for the charges interms of the potentials, we get a quadratic expression for W interms of the potentials which we may write

^ = k i i & a + ? i 2 ^ 2 + k 2 2 ^ 22 + 9 1 3 ^ 3 + (2);

where, of course, We = W$ = W.

We observe that, differentiating partially,

e

From the foregoing quadratic expressions various theoremsabout the p'a and q's may be proved. Thus taking the un-charged state of the conductors as the state of zero energy,We is essentially positive, therefore pn, p22, p33, ... must allbe positive. Then by taking all the e's to be zero except ex andc2, it follows that pup22—p12

2 must be positive, and by similararguments each of the successive discriminants

Pu, Pn Pn.Pl2 #22

Pn Pn. PaiPiz #22 #32P13 P23 Pas

must be positive.

4*6. To deduce the mechanical forces between the con-ductors from the energy. First suppose that the charges onthe conductors are given and remain constant. Wemaysup'posethat the relative positions of the conductors are determined bythe values of certain co-ordinates of position x1,xi,x3,.... Thecoefficients of potential, the p'a, then depend on the x'a. Achange in any co-ordinate x will in general alter some or allof the p'a and therefore alter the energy We. Let X be theforce which tends to produce a displacement 8a;. The work thatwould be done in such a displacement is X Sx and this could only

6-3

84 MECHANICAL FORCES [4-6-

be done at the expense of the store of electrical energy and istherefore equal to the corresponding decrease in We, i.e.

X8x=-8We

*--£ a);and in this expression the e's are constant and the differentia-tion is performed on the p'& which alone are functions of x.

Now we observe that although W$ = We it is not open to usto substitute Wx for We in (1), because of our hypothesis thatthe charges remain constant during the displacement.

If we take as an alternative hypothesis that the potentialsare kept constant while the conductors are displaced, thisintroduces another physical consideration, viz. that anyalteration of position of the conductors would alter the poten-tials and in that case the potentials can only be kept constantby connecting each conductor with a battery, i.e. with asource of energy on which it can draw, and we cannot in thiscase argue that the mechanical work done in a small displace-ment is the exact equivalent of the loss of energy, because weare assuming that there is a limitless store of energy availablefor doing work.

To deal with this case we proceed by an independent argu-

ment to prove that -5-5 = —5-^ , thus:r dx dx

We have We + Wx = 2W = £ (er^r).v r=l

Consider a small displacement 8x without any restrictionon charges and potentials, i.e. suppose that charges, potentialsand positions of conductors all undergo small changes.

Then

or id^Se^w ^ lt de r dx t i d(f> rr dx

r=l r=ldW

But from 4-5 (3) -^-£=<f>r, so that the terms in Ser cancel out,

4-61] MECHANICAL FORCES 85

and from 4*5 (4) - ^ = er, so that the terms in S<hr cancel out,

and there remains

^ + ^ - 0 (2).dx dx

By combining (1) and (2) we may therefore also express theforce X tending to produce a displacement 8a; by the formula

(3)

Hence the force is, by (1), the rate of decrease of the energywhen expressed in terms of constant charges, or, by (3), it isthe rate of increase of the energy when expressed in terms ofconstant potentials, the increase being supplied by thebatteries which are used to maintain the constant potentials.

In a finite displacement which only changes the co-ordinatex, in which the potentials are kept constant, the work done bythe mechanical forces is

i.e. = the gain in electrical energy.Thus we have mechanical work done and an equal gain in

electrical energy, so that when the potentials are kept constantduring a displacement, the batteries must supply an amountof energy equal to twice the mechanical work done.

4*61. Example. We shall again use the parallel plate condenser toillustrate the theory of the precedingarticle. <pi

Let there be two parallel plates ? 0~of area A with charges of surface 'density a, —a and potentials <$>x, ^2 at ia distance x apart, and neglect the i —<7"effect of the edges and consider only d>the field between the plates.

Then <j>1-^i = ^irax (3-31),and as in 3'71 or 4*4 the energy is

so that W. = 2irA<r»x and W^= ^

86 ELECTRIC SCREENS [4-61-

Then the force tending to increase x is either

So that either formula gives a force per unit area 2TT(72, as in 3*1, pullingthe plates together.

Now suppose that the plates approach one another so that theirdistance becomes x'.

(i) If the charges are kept constant, the work done by the attraction

[X -2irAa*dx=2TrAoi{x-x') = W,-We'J X= loss of electrical energy.

(ii) But if the potentials are kept constant, the work done by theattraction is

Jx 8irx2 °° Sir \*' */

= gain in electrical energy;and in this case the battery to whose poles the plates are attached inorder to maintain the constant potentials would have to supply boththe gain in the electrical energy and an equal amount of energy torepresent the mechanical work done by the force of attraction.

4-7. Electric screens. We saw in 2-31 (iii) tha t it is possibleto have two independent ^ ^ •—velectric fields inside and yS >v V joutside a conductor. We / ^.^ \ 3

shall now shew that a con- / f )ductor Ax if enclosed in a [ L-/A, jA2second conductor A2 isthereby screened from theeffects of an external con-ductor A3.

For any such system of conductors the charges are con-nected with the potentials by the relations

(1),

where the q'a are independent of the c's and <f>'a. In order tofind out what we can about the q'a take a special case: let A2

be at zero potential and A1 without charge. Then since A2

contains no charge the potential is constant (i.e. zero) inside it

4-8] QUADRANT ELECTROMETER 87

(2*44 (i)). Hence in equation (1) we have e1 = 0> ^ = ^ = 0 , sothat the equation reduces to

but <f>3 is unrestricted in value, therefore q31 — 0.Hence A3 may be raised to any potential without affecting

Ax and vice versa; so that the conductor A2 screens Ax fromthe external field.

4-8. The quadrant electrometer. This is an instrument inventedby Sir William Thomson (Lord Kelvin*) for measuring differences ofpotential. It consists of a metal short cylindrical box divided intofour quadrants, shown in plan in thefigure. Each quadrant is supported ona separate insulating stand but oppositequadrants are connected by wires.Inside the box a flat piece of aluminiumO is suspended by a silk fibre so thatit can turn in a horizontal positionabout the axis of the cylinder. Fromthe lower surface of G along its axishangs a fine metal wire which connectsG with the inner surface of a condenser,which is maintained at a constant highpotential. The opposing pairs of quad-rants A, A' and B, B' are connected to the two bodies the differenceof whose potentials is to be found, one of which might be the earth.

When A, A' and B, B' have the same potential, the needle G will besymmetrically placed with regard to them; but when the potentials ofA, A' and B, B' differ, C will rotate and take upaposition of equilibriumin which the couple on it produced by the electric field inside the boxis balanced by the torsion of the silk thread. We have now to determinethe relation between the potentials and the angle turned through bythe needle G.

It will be convenient to denote the potentials of the pairs of quadrantsA, A' and B, B' and the needle C by A, B and G respectively. Then theexpression for the energy in terms of the potentials is

W+ = J ? u i ! + Mti B a + iq33 Oa + <j-28 BG + qn GA + qxiAB,where the q'a depend in general on the position of the needle. Hence if0 denotes t h e angular displacement of the needle in t h e position ofequilibrium, the couple tending to increase 8 is

18qas 9g a 3 8q31 dql

+ ° + +dW*_ 1 8 g l l 18gaa 18qas 9g a 3 8q31 dqlt

88 ~ 2 89 + 2 88 + 2 88 + 89 ° + 89 + 89 A '

But by hypothesis there is no couple when A = B, no matter what

* William Thomson, Lord Kelvin (1824-1907).

88 QUADRANT ELECTROMETER [4-8-

value G may have; so that, when A = B, the coefficients of the differentpowers of 0 in the last expression must vanish, i.e.

Therefore, when A and B are unequal,

Also it is evident that if A, B and G are all increased by the sameamount so that the potential differences inside the box are unaltered,the couple will be unaltered. Therefore

9 f t

W~ d6 +* 06 ~ 'and lastly, by the symmetry of the arrangement, it is clear that if<7ii = <?o+a0> then qS2 = q()—a.8, because an increase in the area of theneedle opposite the surfaces of A, A' is accompanied by an equaldecrease opposite B, B'.

Hence the expression for the couple reduces to

But this couple is balanced by the torsion of the silk fibre, and that isproportional to the angle through which it is twisted from the equi-librium position, so that we have a relation

6 = h(A-B){G-i(A + B)},

where k is a constant.For measuring small potential differences, if the condenser is highly

charged, so that G is large compared to $ (A + B), then the deflection 6is proportional to the difference of potential A — B.

For measuring a large difference of potential it is usual to connectthe condenser and therefore the needle with one pair of quadrants—say A, A'. Then we have G=A, and

so that the deflection is proportional to the square of the difference ofpotential.

4*9. E x a m p l e s , (i) If qlx, qi2, q3i are the coefficients of capacity ofthree conductors A, B and G re-spectively, of which G encloses A and B,and ifq1%, qls, qis are the coefficients ofinduction, then the capacity of O whenaloneinspace is qst - (gn + 2qls + qn).

[M. T. 1907]The relations between charges and

potentials are

4-9] EXAMPLES 89Let A, B be without charge and let G be raised to unit potential by

a charge e8, then since A, B are uncharged their potentials are alsounity (2'44 (i)). Therefore the above relations become

whence we get e8 = g88 - (gu + 2q12 + q22),

but this is the charge which will raise C to unit potential when alonein space, i.e. its capacity.

(ii) Shew that, if a new insulated conductor be brought into the field ofany system of conductors, their coefficients of potential of the type p , , arein general diminished, and ifPrr>Pri>Pss oe three coefficients of potentialbefore the introduction of the new conductor, pn', prs', p,s' the same coeffi-cients afterwards (PrrPit—PrrPts) ^s not ^ess t^an (Prs~Prs')2-

[M. T. 1896]When the new conductor is introduced there will be a redistribution

of the charges in the field, though the charge on each conductorremains the same, and positive work will be done by the electric forcesso that there will be a diminution of the electric energy. Hence thedifference

is positive for all values of the charges.

Hence pn-pn'-&§ and p,,-p,a'>0

and (pr, -pra')2 <(Prr-Prr') (P,s ~P,,')

i.e. <PrrP,, + Prr'P,s'-PrrP,/ -Prr'Pss-

Therefore (pr, -pra'Y <PrrP,> +Prr'P.>'-Prr'P,,' -Prr'Pn'

i.e. <PrrPts-Prr'P»-

(iii) Three insulated spherical conductors of the same size are placedat the corners of an equilateraltriangle whose sides are large incomparison with the radii of thespheres. If the conductors are thentouched in turn by another smallcharged spherical conductor, shewthat the charges they receive arein geometrical progression.

[I. 1899]Let A, B, O be the fixed and

D the movable conductor. Let ebe the initial charge on D, and6i. e>i e3 the charges which A, B,C in turn receive from it.

o

90 EXAMPLES [4'9

When A and D are in contact their common potential fa satisfiesrelations ,

so that, by subtraction,O = (Pii-Pii)e1 + (pli-pn)(e-e1) (1).

Next, when D is in contact with B, the charges on B and D are ea ande — — ea and the coefficients of potential of B and D are by sym-metry the same as the coefficients of potential of A and D when D wasin contact with A, save that the potentials of B and D will both containa term due to the charge ^ o n i , which we may take to be ejr, wherer is a side of the triangle regarded as large compared to the dimensionsof the conductors. Hence we have for the common potential of B

0X1 D <!>

so that, by subtraction,0=(Pu-Pi2)«2 + (?'i2-Ps2)(6-ei-ea) (2).

Similarly, for the common potential of C and D when in contact,fa =Pix 63 +P12 (e - ex - ea - e3) + (ex + e2)/r,<£s =:Pi2 e3+P22 (<* - et - ea - e3) + (ex + ea)/r,

giving 0 = ( p u - p l a ) e 3 + (p i a - p 2 a ) ( e - e 1 - e 2 - e s ) (3).From (1), (2) and (3) we get

which give

e—e1 _ e — e t — ea _ e — e^ — ea — e3

«i el ~ e~3 '

EXAMPLES

1. If the radii of two concentric spheres be a, b (6 > a), and if eachsphere be electrified with a positive charge e, shew that the energy ofthe system will be e2 (6 + 3a)/2a&. [I. 1912]

2. If n conductors in space be connected together by wires, provethat the capacity of the compound conductor thus formed is, with theusual notation, SgVr + 2Sgr,, where the first summation is taken for allintegral values of r from 1 to n and the second for all unequal integralvalues of r and s from 1 to n. [I. 1899]

3. A system of insulated conductors having been charged in anymanner, charges are transferred from one conductor to another till allare brought to the same potential <f>. Shew that <f> = El(sy + 2sa), wherest , sa are the algebraic sums of the coefficients of capacity and inductionrespectively, and E that of the charges. Prove that the effect of theoperation is a decrease of the electrostatic energy equal to what wouldbe the energy of the system if each of the original potentials werediminished by the amount <f>. [M. T. 1901]

EXAMPLES 91

4. If there are two conductors, A, B, and when B is uncharged theoperation of charging A to unit potential raises B to potential ^, provethat when A is kept at zero potential a unit charge on B will induce onA a charge -$. [M. T. 1904]

5. If one of n conductors entirely surrounds the others, shew that2n — 1 coefficients of potential are equal; and that if they are each equalto p, shew that the loss of energy when the outside conductor is con-nected with the earth is ipe2, e being the quantity of electricity thatpasses to earth. [I. 1915]

6. If a conductor could be made larger without change of shape,would its capacity be increased or diminished? Give a reason for youranswer.

If a conductor carrying a given charge is completely surrounded byanother uncharged conductor, is the energy of the system increasedor diminished? [St John's Coll. 1906]

7. Two equal uncharged conducting spheres are placed so that thecoefficient of capacity for each is a and the coefficient of mutual induc-tion is 6. Each encloses a charged conductor, the charges being e.x, e,respectively. Prove that when the spheres are connected by a wire theloss of energy is {et - e2)

2/4 (a - 6). [I. 1914]

8. Shew that if a given charge of electricity be distributed betweentwo conductors of the same shape and size, and symmetrically placedwith respect to one another and neighbouring conductors, the maxi-mum energy obtainable is when the whole charge is given to one, andthe minimum when it is equally divided between them. [I. 1891]

9. Shew that if a given charge is distributed over a number of con-ductors so that the potential energy of the system when in electricalequilibrium is least, the conductors are at the same potential.

[M. T. 1897]

10. If conductors A, B, G, D, ..., where O completely surrounds Aand B only, are charged in any manner and A and G are connected by athin wire, shew that the charge remaining on A bears a constant ratioto the charge on B. [M. T. 1908]

11. Two equal and similar condensers, each consisting of a con-ducting sphere surrounded by a concentric spherical conducting shellof negligible thickness, are insulated and placed at a great distance rfrom one another, and charges e, e' are given to the inner surfaces ofthe condensers. Prove that, if the outer surfaces are connected by a

fine wire, Nthe loss of electric energy is approximately i(e — e')2U — j ,

where 6 is the radius of the outer surface of either condenser.[M. T. 1904]

92 EXAMPLES

12. Shew that if two conductors, for which

be connected by a fine wire, the new charges will be Ex + x, E% — x,where x is such as to make the expression

p x x { x ) p x l ( x ) ( t ) p n ( t )a minimum, and determine the apparent loss of energy.

What happens to this energy? [St John's Coll. 1910]

13. A system of conductors consists of three, Cx, <72, (7S; and Cxcompletely surrounds (7a. If G3 were annihilated, the coefficients ofpotential of Ot and C3 would be P n , Pls, P33; and if Ca were annihilated,the coefficients of potential of Ox and Ca would be w n , « l a , t»aa. Shewthat the actual coefficient of potential of <7a on itself is

and write down the remaining coefficients of potential of the system.[M. T. 1909]

14. Two insulated fixed conductors are at given potentials whenalone in the electric field and charged with quantities et, ea of electricity.Their coefficients of potential axepxi, p22, plt. But if they are surroundedby a large spherical conductor at potential zero with its centre nearthem, the two conductors require charges e/, e2' to produce the givenpotentials. Prove that, neglecting terms involving the inverse squareof the radius of the enclosing conductor,

e 1 ' - e 1 : e a ' - e a =p a a -^ 1 2 :p 1 1 -p 1 2 . [M. T. 1895]

16. Two conductors of capacities Gx, G2, when each is at an infinitedistance from any other body, are brought to a distance B apartwhere B is very great compared with their dimensions. Prove that the

O Gcoefficient of mutual induction is ^-2 and the coefficients of

( G G \ f G G \1H—W^J > ^2 (1H—jijr) • Verify the results by shewing

that if the charges be Qx, Q2 the increase of electrical energy on the

approach of the bodies is % r ? - [I. 1892]

16. There are two concentric conducting spheres of radii a and6,theyare insulated and the inner (a) has a charge e, but the outer, which is tobe regarded as a shell of negligible thickness, has no charge. Shewthatifthe outer is connected by a thin wire with an insulated and unchargedconducting sphere of radius c at a great distance d, the loss of electricenergy is 1 a^ ^ 26- \ „

I1 d(6+T)r [

EXAMPLES 93

17. The equipotential surface (j>lt due to the charges Alt Atf ....encloses all those charges, and the equipotential surface <j>t, due to thecharges Blt J5a, ..., encloses all those charges, the two systems beingtoo far apart to influence one another appreciably. If each of thesurfaces is now made a conducting surface, without resultant charge,and if the two surfaces are joined by a fine conducting wire, shew thattheir common potential will be

J t ? ' PL T. 1907]

18. Two conductors A, B of a system are connected by a wire ofnegligible capacity so as to form a single conductor A'. Shew that thecoefficient of potential of a third conductor C is diminished by

(Pis -Pss)al(Pii +Pzt ~ 2Pu)>where ptl, plt, ... denote as usual the coefficients of potential of theoriginal system, and the suffixes 1, 2, 3 refer to the conductors A, B,C.

[I. 1906]

19. If a system of charged conductors is surrounded by a very largespherical sheet at zero potential with its centre in the neighbour-hood of the conductors, obtain an approximate expression for theloss of energy due to the presence of the sphere in terms of its radiusand the given charges, neglecting squares and higher powers ofthe ratios of the linear dimensions of the system to the radius of thesphere.

How is the result modified if the spherical sheet is insulated withoutcharge instead of being at zero potential? [I. 1904]

20. A system consists of five conductors of which the second sur-rounds the first, the third surrounds the second and the fifth sur-rounds the fourth. Determine which of the coefficients of inductionare zero. Find, independently, the relations which exist between thecoefficients of potential and shew that only six of them are independent.

[I. 1908]

21. A condenser is formed of two spheres, one inside the other, withtheir centres at a distance x from one another. Assuming the capacityto be a known function/(x) of a;, write down a formula for the forcetending to increase x, when the charges on the spheres are e and — e.

[St John's Coll. 1906]

22. Three equal spheres are placed at the corners of an equilateraltriangle. When their potentials are (j>, 0, 0 their charges are E, E', E'respectively. Shew that, when each sphere is at a potential <f>', each hasa charge (2JB' + E) <f,'l<f,.

Find the potentials when the charges are E", 0, 0. [M. T. 1934]

94 EXAMPLES

23. Prove that in a small displacement accompanied by smallchanges (SE) in the charges, the identity

ZE&<f, - S08.E?=8pn Ej* + 2Spla ExEi+...

holds good. Deduce the equation

and explain the relation of each of these expressions to the mechanicalwork done by the forces of the system in the displacement.

[St John's Coll. 1911]

24. Four perfectly equal uncharged and insulated conductorsAx, A2, A3, At are such that when they are placed at the angularpoints of a square, they are symmetrically situated with respect toeach other and the centre of the square. Another charged conductor ismoved so as to touch each in succession in the same way, beginningwith Ax; and elfei,et, e4 are the charges on the conductors after theyhave been each touched once. Shew that

(e1-ei)(e1ea-e!i*) = e1{eie3-e1ei). [M. T. 1898]

25. An insulated sphere of radius 25 cm. is charged and afterwardsconnected to an electrometer by a long fine wire, the deflection being75 divisions. The system is then joined to a distant insulated sphere ofradius 12 cm. and the deflection falls to 53 divisions. Calculate thecapacity of the electrometer. [M. T. 1918]

26. Three concentric spherical conductors, radii a, b, c (a<b<c),have charges EX,E%, Es respectively. Shew that, if the inner conductoris now connected with the earth, the potentials of the conductors arediminished by amounts inversely proportional to their radii, and thatthe loss of energy is

Kf f )2 [M.T.1912]

27. Three concentric spherical conductors, radii a, b, c (a<b<c),have charges e1, ea, es respectively and potentials V1, F2 , F a . Writedown the potentials in terms of the charges and shew that the electro-static energy is

{ ( ) G £ M G*'1- 1916]28. A system consists of three concentric spheres (a, 6, c). The

middle sphere has a charge e and the other two are first put to earthand then insulated, find the charges induced on the first and thirdspheres.

Calculate what would be the loss of energy (1) if the two innermostspheres were joined by a wire, (2) if the two outermost spheres werejoined. [St John's Coll. 1911]

EXAMPLES 95

29. Use the theorem, that if a set of conductors suffer given smalldisplacements the loss of electrical energy when the charges are keptconstant is equal to the gain of electrical energy in the same displace-ments when the potentials are kept constant, to find the force X per

V -X

V, V2

unit length acting on the upper plate of a parallel plate condenserarranged as shewn, edge on, with the potentials of the plates, in thefigure. [M. T. 1929]

30. If three equal spherical conductors are placed at the corners ofan equilateral triangle and raised to potentials fa, fa,, fa, prove that,when they are put into electric communication with one another, theenergy of the system is reduced by

where W is the energy of the system, when the potentials are 1 , - 1 ,and 0 respectively. [I. 1912]

31. A spherical conductor of radius a has a charge E and is sur-rounded by three insulated spherical conducting shells, the internalradii of the shells being 2a, 4a and 6a, and the external radii 3a, 5a, 7a,and their charges are Ex, Et, Ez. Find the change in the energy of thesystem when the first and second shells are connected by a wire.

[I. 1909]

32. If one conductor of a system of charged conductors contains allthe others and there b e n + 1 in all, shew that there are n relationsbetween either the coefficients of potential or the coefficients of in-duction, and if the potential of the largest be fa and that of the othersfa, <f>%, •••<j>n> then the most general expression for the energy is \Cfa?increased by a quadratic function of fa — fa,, fa —fa, ••• <f>n — <f>o> whereO is a definite constant for all positions of the inner conductors.

[M. T. 1905]

33. If three condensers of capacities Olt (7a, Ct are joined in cascadeand charged with a charge Q, and if then this connection is broken andthe condensers are joined in parallel, shew that the loss of energy is

[M. T. 1907]34. Recent experiments lead to the inferences that in O.G.S. electro-

static units the charge on an electron is 3-4 x 10~10, and that the massof an electron is 6-1 x lO"28.

If electrons pass between two metal plates at a potential differenceof 2000 volts, and the whole of their lost potential energy is convertedinto kinetic energy, calculate the velocity with which they strike thesecond plate, on the above data. [M. T. 1913]

ANSWERS

12. i{E1(p1,-p11) + Et(pn-plt

19. (Ee)2/2r, where Se is the sum of the charges and r is the radius ofthe sphere.

20. qlt, qlt, qu, ga4, 22B, qu are zero. 21. e2/' (a;)/2 {/(a;)}2.

22. (E + E')E"/{E* + EE'-2E'*), -E'E"I(E* + EE'-2E'*).

25. 3-9 abs. units.

q(c-6) _cj6-o) le2a(6-a)(o-b)ii 1 e2c(6-a)'(o-6)^8* 6(c-o)6' 6(c-a)e ; 2 6s(c-a)2 ' 2 68(c-o)2

29. (2F-71-Fa)(F2-F1)/8w«. 31.

34. 1-92 x 109 cm. per sec.

Chapter V

DIELECTRICS

5-i. Thus far we have assumed that the electric fields underconsideration are produced by charged conductors in air.Faraday found that when some non-conducting substanceother than air fills the space between the plates of a condenserits capacity is altered, being always increased in a definiteratio when the same substance is used. The ratio, usuallydenoted either by K or by e, is called the specific inductivecapacity of the substance or its dielectric constant.

As compared with a vacuum the specific inductive capacityof air has been found to be 1-000585, so that it is immaterialwhether we take air or vacuum as the standard for com-parison provided that the air is at constant temperatureand pressure.

Taking the specific inductive capacity of air to be unity,those of other gases do not differ much from unity, but theymay vary with the pressure. The specific inductive capacitiesof solids and liquids vary considerably, e.g. solid paraffin 2-29,sulphur 3-97, glass 3-2 to 7-6, distilled water (a semi-conductor)76.

5*11. Inference drawn from parallel plate condensers.Consider two parallel plate condensers A and B which areexactly similar save that in A the medium between the platesis air while in B it is a dielectric of specific inductive capacity K.

t air A ^ilii-iS

Let 0, C be the capacities per unit area of A and B, then,by definition of K, O' = KG (V)

REM 7

98 FIELD IN A DIELECTRIC [5'i i-

Let the upper plates of the condensers have equal charges aper unit area, let their potentials be <f>, <f>'; and let the lowerplates be earth connected.

Then, by definition of capacity,

CW/<£ and C' = (7l<f>' (2).

Hence from (1) and (2)

4>' = <f>IK ( 3 ) ;or, for the same charges, the drop of potential in the dielectricis \jK of what it is in air under like conditions.

Neglecting edge effects, in both condensers the tubes offorce are straight, and if E, E' denote the electric intensitiesin the two fields and t the distance between the plates of thecondensers, we have

Et=<j> a n d E't = <j>',

so that E' = E\K (4);

or the gradient of potential in the dielectric is \jK of what it isin air under like conditions; and though we have only demon-strated this by a comparison of fields in which the gradientsof potential are constants, yet we shall base our theory onthe hypothesis that it is true for condensers of any shape.

We assume therefore that if a point charge e is embeddedin a medium of uniform specific inductive capacity K the in-

tensity of the field at distance r is -^-^, and that the force

between two small charged bodies with charges e, e' at adistance r apart is ee'lKr*.

5*2. Electric displacement. Maxwell's* method of explain-ing the relations of the electric field, including the phenomenonjust described, involves the assumption that when an electricfield is set up there is a displacement of electricity in thedielectric medium as well as in conductors; with the differencethat in a dielectric the displacement is controlled and checkedby some kind of elastic force which does not exist in conductors.

* James Clerk Maxwell (1831-1879), Cavendish Professor of Experi-mental Physics at Cambridge.

5-2] ELECTRIC DISPLACEMENT 99

Thus if the plates A, B of an uncharged condenser are con-nected by a wire and by the action of an electromotive force,a quantity Q of electricity is transferred along the wire fromthe plate B to the plate A; then, at the same time, the electri-fication of the plates produces a certain electromotive forcefrom A towards B in the dielectric between the plates and thiscauses a displacement of electricity in the dielectric. Thisdisplacement is not a continuing flow,but a displacement which takes placethroughout the dielectric while the electro-static field is in process of creation, theamount of electricity which is displacedacross any surface drawn between theplates being Q, the quantity which passesacross the wire; so that if S be a closedsurface which surrounds the plate B, aquantity Q passes out of S along the wireand simultaneously a quantity Q entersS by the displacement which takes placeat all points of the dielectric. In thisrespect electric displacement resembles the displacement ofan incompressible fluid, in that the amount of electricitywithin a closed surface remains constant; if a charge Q isbrought into the region bounded by a surface S, there issimultaneously a displacement of electricity of amount Q out-wards across 8 in the dielectric medium. The direction ofdisplacement is from a positive charge into the dielectricmedium.

In view of what was stated in 2*11 about the nature ofelectricity, it may be as well to remark that the above com-parison of electricity with an incompressible fluid is a highlyartificial one and the resemblance is only to be regarded asholding good in this one particular of 'displacement'.

In Maxwell's theory, now commonly spoken of as the'Classical Theory' of the electromagnetic field, the vectorcalled the electric displacement plays an essential part. It wasdefined by Maxwell as the quantity of electricity displacedacross a unit of area placed perpendicularly to the direction

7-2

100 GENERALIZED FORM OF GAUSS'S THEOREM [5-2-

in which electricity is displaced. I t is thus a measurable directedquantity. It is denoted by the symbol D, but in accordancewith modern usage we shall define D as 4w times the quantityof electricity displaced as above. The reason for so doing isthat, unless we define and adopt a system of 'rational units',we cannot exclude the factor 4TT from many of our formulae,and we get more uniformity in the formulae and a closeranalogy between the formulae of magnetism and electrostaticsif we introduce the factor 4w at this stage.

5*21. Generalized form of Gauss's Theorem. In accord-ance with the definition of D and the explanation given in 5*2,if S be any closed surface in an electric field

)nd8 = 4:7T (total charge within the surface) ...(1),

where the integral on the left represents the total outwardflux of the displacement vector across the surface.

We may now, as in 2-33, apply this theorem to a smallelement of volume 8v enclosing a point P, and shew that, if Dis the displacement and p the density of electricity at P, then

div D = 4^rp (2),

with the consequence that at all points in the field at whichthere is no volume density of electricity

divD = 0 (3).

Vectors whose divergences vanish through a region are calledsolenoidal or stream/ing vectors, or are said to satisfy the solenoidalcondition. This condition is also satisfied by the velocity of an incom-pressible fluid.

Next, if there is a surface of discontinuity of D, by applyingthe modified form of Gauss's Theorem above to a short cylinderextending across the surface, we can shew, exactly as in 2*34,that if Dx, D2 denote the displacement on opposite sides of thesurface, the normal components measured away from thesurface satisfy the relation

where a is the surface density of the charge on the surface; withthe consequence that across every uncharged interface between

5'23] DISPLACEMENT 101

two dielectric substances the normal component of the dis-placement is continuous, i.e.

{DJ^DX (5)

when both are measured in the same sense.

5*22. Relations between D and E. There are thus twovectors D and E whose principal characteristics are that D isa solenoidal or streaming vector through uncharged space,with a continuous normal component across any unchargedinterface, and that E is the gradient of a potential function.The two vectors are interrelated because D only exists whenE exists; but the relation between them depends on thedielectric medium.

It is seen at once that if we assume that in air (or vacuum)D = E (1),

then the relations 5*21 (1), (2), (3), (4) are merely Gauss'sTheorem in its integral and local forms as set forth in 2-3, 2'33and 2-34.

And if further we assume that in an isotropic* dielectric

D = Z E (2),where K is the dielectric constant, this makes the value of Ein a dielectric \jK of what it would be in air as required in 5-11.

5*23. Recapitulation. In the previous chapters we limitedour considerations to electric fields produced by chargedconductors in air, but in the present chapter we broaden thebasis of our considerations by admitting the possibility of theexistence in the field of masses of dielectric substances. Thishas led to the introduction of a new vector D called the dis-placement; but in a static field displacement does not refer toa continuing process but to something which happened in thecreation of the field. In varying electromagnetic fields, how-ever, 'displacement' undergoes a time-rate of change, and'displacement currents' play an essential part in the theoryof the propagation through space of electrical effects. In

* A substance is isotropic if any spherical portion of the substancepossesses no directional property, in contrast with a crystallinesubstance, for which the relation between D and E would be morecomplicated.

102 MODIFICATION OP COULOMB'S THBOEEM [5»23-

a static field the vector D satisfies certain fundamental rela-tions specified in 5-21, and this theory does not invalidate anyof the previous discussion of fields in air but includes them as aspecial case in which K= 1, or D = E.

5*24. Objections. The discussion of the so-called displace-ment in 5*2 is based on the theory explained in Maxwell'sTreatise on Electricity and Magnetism* but it is open to theobjection that there is no experimental evidence that indielectrics there is an actual displacement of electricity of thekind described. According to modern views it is preferable todefine the vector D solely in terms of the conditions whichit has to satisfy; and to this end we assert that a completetheory of the electrostatic field requires

(i) the existence of a single-valued potential function <j>;and, in addition to the electric intensity E which is thenegative gradient of <f>, a vector D which satisfies

(ii) the generalized form of Gauss's Theorem, and(iii) the relation D = RE;

these conditions being sufficient to determine the field of anygiven charges. There is then no need for any further definitionof D apart from the fact that it satisfies these relations.

5*25. Modification of Coulomb's Theorem. Suppose thata conductor on which there is a charge of surface density a isin contact with a dielectric medium of constant K. Then sincethere is no field inside the conductor, in 5*21 (4) we may put

where {D^)n is the normal component of displacement directedfrom the surface of the conductor into the dielectric. Thensince D = KE, we have

or En = ±no\K (1);shewing again that the electric vector is \jK of what itwould be in air.

* 3rd edition 1892, pp. 65-70.

5-3] REFRACTION OF THE LINES OF FORCE 103

5-26. Refraction of the lines of force. When the lines offorce cross an uncharged interface between two dielectrics, theyundergo a refraction in accordancewith a definite law. I t was proved in2*411 that in crossing the commonsurface of two media the tangentialcomponent of the electric intensity iscontinuous, and the argument of 2-4ilis valid whatever the media. It wasalso proved in 5-21 that at an unchargedinterface between two dielectrics thenormal component of displacement iscontinuous.

Let Klt K% be the dielectric constants and let the electricintensities at points close to the interface and to one anotherin the two media be Ex, E2, making angles d1,62 with the normalto the interface. Then from the continuity of the tangentialcomponent of the intensity, we have

Ex sin d1 = JSssind2;

and from the continuity of the normal displacement we have

Kx Ex cos 0X = K% E2 cos 02;

from which it follows that

and this is the law of refraction of the lines of force. I t appliesof course to the case of lines of force entering or leaving a blockof dielectric substance surrounded by air, in which case oneof the K's is unity;

5*3. The potential. The equations to be satisfied by thepotential in a field which contains dielectric bodies may nowbe deduced from 5-21 and 5*22.

In every electrostatic field we have

E=-grad^ (1)with rectangular components

" v' z~~dx' ~dy' ~3s.(2):

104 EQUATIONS FOR THE POTENTIAL [5*3-

so that, if as in 5*22 we put D = KE, we have

Then substituting these values in 5*21 (2), viz.

div D = 4irp,we get

where K is the dielectric constant and p is the volume densityof electricity at the point (x, y, z); and at points at which thereis no volume density of electricity the equation for <f> is

We observe that equations (4) and (5) include the case inwhich the dielectric constant K varies from point to point ofthe region. Also, that in a homogeneous dielectric since K isconstant it may be removed from (5); so that (5) reduces toV2<£ = 0; hence in a homogeneous dielectric where there areno charges the potential satisfies Laplace's equation, just asin empty space.

Using 5*21 (4), at a surface of discontinuity of the medium,if a is the surface density on the interface and <f>lt <f>2 thepotential functions in the dielectrics whose constants areKx, K2 on opposite sides of the interface, we have

where 9%, dn2 are elements of the normal drawn from theinterface into the medium on each side.

5*4. Comparisons. When fields containing dielectrics are comparedwith fields in air, the following considerations sometimes lead to simpleresults. Suppose that the equipotential surfaces are known for a givendistribution of charges in air. Let the space between any two of thesesurfaces be filled with homogeneous dielectric of constant K. The newfield will retain the same surfaces as its equipotential surfaces, thoughthe values of the potential over the surfaces will not be the same asbefore. The conditions to be satisfied in the new field are (i) that thepotential shall be continuous; and (ii) that the normal displacement

5-42] EXAMPLES 105shall be continuous across the interfaces; i.e. that Dn or KEn shall becontinuous. Both conditions can be satisfied by making the gradientof potential in the air regions in the new field the same as in the original,and the gradient of potential in the dielectric region \jK of what it wasin the original field.

We shall illustrate this method by the following example.

5'41. Example. A conductor with a charge e is surrounded by equi-potential surfaces. If the space between the surfaces of potentials fa, fais filled with homogeneous dielectric of constant K, shew that the energy isreduced by

Let <j> be the potential of the conductor A before the dielectric isintroduced, as in fig. (i), so that the energy is \efa After the introductionof the dielectric the charge of the conductor is unaltered, but its poten-tial becomes <fj', and the potentials of the given equipotentials B, Gbecome fa', fa', as in fig. (ii).

Then using the facts that the potential drop between A and B in airis unaltered; that the drop between B and C in the dielectric is 1/K ofwhat it is in air; and that the drop between O and infinity in air isunaltered, we have ,, , , , ,

9 ~fa =?-<t>i>

and ^ ' - 0 = ^ , - 0 .Therefore by addition

But the energy of the new field is \e$', therefore the loss of energy

5*42. Again, suppose that the solution of a condenser problem isknown when air is the medium between the conducting surfaces, andthat we imagine the whole of a tube of force to be filled with a homo-

106 EXAMPLES [5-42-

geneous dielectric of constant K. The equipotential surfaces •willremain the same for they must still cut the lines of force at right angles,and the conditions of the problem require the potential drop along allthe tubes of force to be the same. Hence, if the potentials of theconductors are kept constant, the charge at the end of the dielectrictube must be K times what it was in air; for otherwise the potentialdrop in this tube would fall to l/K of what it was before.

5*43. Example. The space between two concentric conductingspherical shells of radii a and b, a<b, is divided into two parts by adiametral plane, one part being filled with a dielectric of specific inductivecapacity K1 and the other part with a dielectric of specific inductivecapacity K2. Shew that the equipotentials are spheres concentric with thetwo conducting shells, and find the capacity of the condenser which isformed of the two conducting shells. [M. T. 1934]

Let fa, <f>2 be the potentials of the conductors and o- the surfacedensity on the sphere of radius a when the medium between theconductors is air. Then as in 3*4

•(1),

since 4*raV is the total charge.

When the dielectrics are inserted, the equipotential surfaces betweenthe conductors will still be concentric spheres since the lines of forcewill be radial; and in order to maintain the same potential drop asbefore along all the tubes of force, the surface density over one half ofthe inner sphere will need to be K-^a and over the other half Kta,making the total charge 2ird*cr(K1+Ki), and the capacity being theratio of this to the potential difference <j>\ — <j>i is, from (1),

5*5. Examples, (i) A condenser of capacity one microfarad is to beformed by piling 2ra square sheets of metallic foil each of 10 cm. edge,interleaved with sheets of insulating paper 0-1 mm. thick of dielectricconstant 3, the sheets of foil being attached alternately to the two terminals.Find the necessary value of n, given that one microfarad is9x 10s electro-static units. [M. T. 1932]

5-5] EXAMPLES 107Let <f>lf <j>t be the potentials of the terminals, and let the sheets

numbered 1, 3, 5, ... have positive charges and be connected to theterminal of potential <j>t. Neglecting edge effects, the surface densitieson opposing faces of the metal sheets must be equal and opposite andmay be denoted as in the figure.

Denoting the distance between the sheets by t, and the dielectricconstant by K, and considering the field between the sheets 1 and 2,we have .

<f>i — j>z = t]31 = -go1t,

where Et is the electric intensity.

/l -5Oi

-O\

OI01

— 0~m— 2

f 0~m—i

1

3

4

5

6

in-1.

1n-\

In,

But there is a similar relation for each successive pair of sheets, viz.

l f tE *

Therefore, by addition,

! - 0.) = j£ (orj + (7, + ...

•D . ., •. area of a sheet x Scr (2n — 1) K x areaBut the capaoity= 7 j = * '-r— ,

where t=0-01 cm., K = 3 and the area =100 sq. cm.

Therefore the capacity=(2n— l)300/0-04jr electrostatic units; andfor this to be as much as 9 x 105, we find that n will need to be 189.

108 EXAMPLES [5*5

(ii) Two large plane conducting plates each of area A are placedparallel to each other in air at a distance d apart. One is given a charge Qand the other is connected to earth. If a cylindrical piece of dielectric ofspecific inductive capacity K, thickness d, area of section A' (A'<A), isplaced between them with its generating lines perpendicular to the plates,shew that, neglecting edge effects, the mechanical force acting on each plate

** 2*Q*l{A + (K-l)A'}. [M. T. 1924]

Let <7! be the surface density on the part of the positively chargedplate in air, and CT2 that on the partincontactwith the dielectric. Then j ^7

d

But since the fall in potential I A—A'between the plates is the samepalong the tubes of force in the dielectric as along the tubes in air, wemust have _ c-

andtherefore a1 = Q/{A + A'(K-l)}.

The mechanical force per unit area of the part of the plate in air is2w<71

2 and if we assume that on the part in contact with the dielectricit is 2irat*/K, then the total force is

which, by substituting for o and er,,

(iii) Two concentric thin metal spheres, of radii a, b, have between thema concentric shell of dielectric of radii c,c', where a<c<c'<b. The innersphere is earth connected and the outer is given a charge e. Find the ratioin which this charge is divided between the inner and outer surfaces of thesphere, and shew that the potential at a point in the dielectric at a distancer from the centre is

fl 1 1 / 1 l \ W f l 1 1 / 1 1 \ 1 11\a~c + K\c r)]f\a c+K\c~ c')+c'~bf'

eb

where K is the dielectric constant. [I. 1926]Of the charge e let the portion e' reside on the inner surface of the

sphere of radius b and the portion e — e' on its outer surface. By con-sideration of the tubes of force which start from the charge e', we seethat the charge on the sphere of radius a is — e'.

Then remembering tha t a uniform spherical distribution producesno field inside itself, the field in the space between the spheres takenalong the radius in the sense of r increasing has the following values:

— e' — e' —"'for a<r<c, —$-; forc<r<c', -^r^; and for c'<r<b,

5-5] EXAMPLES 109Let fa be the potential of the outer sphere, then since the inner is

earth connected, we have, for the potential difference,

e-e'

Again, the field for r > b is e/r2, so that we also have

= 6 - 1 1 - 2[a c

Therefore « = 6-11-2+* (i_IUi l

6 [a c K\c o) o b)which determines the ratio e': e—e'.

Again, if 0 denotes the potential at a point in the dielectric at adistance r from the centre, and we take the fall in potential betweenthe sphere of radius a and this point, we have

or, , a l l /i i\)= e ' H-^ )y,(a o K\c rj)

and then by substituting for e' in terms of e we obtain the requiredresult.

110

EXAMPLES

1. Two large parallel plates at distance d are maintained at potentials(/>!, </>2; find the foroe per unit area on either of them, and find the effectof interposing a slab of dielectric of specific inductive capacity K andthickness * between the plates. [I. 1903]

2. The plates of a parallel plate condenser are at distance h apart.Prove that, if a slab of uniform dielectric, of thickness t and having adielectric constant K, is inserted between the plates, the capacity is

increased in the ratio 1: | l —| ( l - j t ) \ • tM- T -

3. If the area of a parallel plate condenser is one square metre, thedistance between the plates 0-01 cm., and the dielectric constantK. — 1, find in ergs the energy stored in the condenser when charged to apotential difference of 300 volts, that is, one electrostatic unit.

[M. T. 1934]

4. Two large plane metal sheets are situated parallel to each otherat distance a apart and are connected to earth. A third sheet is placedmidway between them and raised to potential F. Two slabs of homo-geneous dielectric substances of thicknesses *, t' and dielectric constantsK, K' are placed parallel to the sheets one in each gap. Shew that thecentral sheet is attracted towards the former or latter slab according as

*'• i , iV, K'(K-l)- is less or greater than

Find by how much the central plate must be moved to be in equi-librium. [M. T. 1933]

5. If two parallel conducting plates be separated by layers of twodielectrics of thickness ^ and t2 and specific inductive capacities Kx

and Kt and the plates have potentials ^ and ^a > prove that the force

on the first plate is SK*K** ,Jfo~t?*' .., where S is the area of eitherplate. [I. 1892]

6. Two large parallel conducting plates are maintained at potentialsfa and ^2 and the space between them is filled up by slabs of dielectricwhoses.i.c.'sarei^andifa, whose thicknesses are d1 and dt, and whosecommon face is parallel to the plates. Find the potential at any pointbetween the plates.

Shew that the potential everywhere is the same as if the dieleotricswere replaced by an insulated conducting sheet along their commonface, and the charge on this plate per unit of area were

[M.T.1909]

EXAMPLES 111

7. Find the capacity of the condenser C, D described in the followingproblem, making it clear in what units your answer is expressed. Acondenser consists of two plates A, B at a distance a cm. apart. A secondcondenser consists of two plates O, D at a distance b cm., with a slabof paraffin of thickness c cm. and specific inductive capacity K betweenthem. The slab and plates are all of area S sq. cm. The plates A, C areinsulated and B, D are put to earth. A is first raised to potential V inelectrostatic units, then connected to C by a fine wire and finally theslab is withdrawn.

Compare the attractions between the plates A, Bin the three stagesand prove that the work done against electrical forces in withdrawingthe slab is

[It may be assumed in the above that the lines of force are perpen-dicular to the slab and plates and that the condensers are too far apartto affect one another by induction.] [M. T. 1913]

8. Find the capacity of a spherical conductor of radius a, closelysurrounded by a concentric spherical shell of dielectric whose inductivecapacity is K and outer radius 6. Find the energy when the charge is e.

[St John's Coll. 1905]

9. If a spherical conductor, of radius a, with no other conductor inthe neighbourhood, is coated with a uniform thickness d of shellac ofwhich K is the specific inductive capacity, shew that the capacity ofthe conductor is increased in the ratio K(a+d): Ka + d. [ I . I 902]

10. If the space between concentric conducting spheres of radii aand b (a < b) is filled with two dielectrics of specifio inductive capacitiesK, K.', their common surface being a concentric sphere of radius c,find the capacity of the condenser. [I. 1904]

11. Evaluate the coefficients of capacity and induction for twoconcentric spherical conductors of radii a, b, (i) when the interveningmedium is air, (ii) when the inner conductor is closely surrounded by aconcentric shell of a dielectric whose s.i.o. is K, the outer radius of theshell being b-t. [St John's Coll. 1906]

12. A spherical condenser consists of two concentric conductingspheres of radii a, b (b>a). A spherical shell of dielectric K extendsfrom the inner sphere to a distance c (< 6) from the centre. The innersphere is insulated and receives a charge e and the outer is earthed.Prove that the potential inside the dielectric is

112 EXAMPLES

where r is the distance of a point from the centre. Find also the capacityof the condenser. [M. T. 1935]

13. A spherical condenser of radii a, d is modified by the insertion ofa concentric spherical shell of specific inductive capacity K, radii band c (d > o> b > a). Find its capacity as modified. [M. T. 1928]

14. A metallic sphere is surrounded by a thin concentric conductingshell formed by two hemispheres with their rims in contact, the spacebetween the sphere and shell being filled with a dielectric of specificinductive capacity K. If charges e, e' be given to the shell and sphere,shew that if the halves of the shell remain in contact the charges mustbe of opposite sign and the ratio of their magnitudes must lie betweenthe limits 1 ± \\^/K. [M. T. 1908]

15. If half the space between two concentric conducting spheres befilled with solid dielectric of specific inductive capacity K, the dividingsurface between the solid and the air being a plane through the centreof the spheres, shew that the capacity of the condenser will be thesame as if the whole dielectric were of uniform specific inductivecapacity J ( 1 + J K ) . [M. T. 1894]

16. The space between two concentric conducting spheres is filledon one side of a diametral plane with dielectric of s.i.o. K and on theother side with dielectric of s.i.o. K'. The inner sphere is of radius aand has a charge e; shew that the force on it perpendicular to thisdiametral plane is rr r_.

lJL-JCe*tf' U-

17. The space between two concentric conducting spheres of radiia and A is filled with n concentric layers of dielectric of s.i.c. KltK Kn. The radii from a to A inclusive are a series in harmonicalprogression.

Prove that O, the capacity of the condenser, is given by

18. Three thin conducting sheets are in the form of concentricspheres of radii a + d,a,a — c respectively. The dielectric between theouter and middle sheet is of specific inductive capacity K, and thatbetween the middle and inner sheets is air. At first the outer sheet isuninsulated, the inner sheet is uncharged and insulated, the middlecoating is charged to potential V and insulated. The inner sheet is nowuninsulated without connection with the middle sheet. Prove that thepotential of the middle sheet falls to

(a + d)l{Kc(a+d) + d{a-c)}. [M. T. 1895]

EXAMPLES 113

19. A curve is drawn on the surface of a freely electrified ovalconductor surrounded by air, dividing it into two parts on which thecharges are E and E' respectively. The space, enclosed by the portionof the surface of the conductor on which the charge is E' and by thelines of force drawn from every point of the curve to infinity, is filledwith a dielectric of s.i.c. K. Prove that the potential of the conductoris now diminished in the ratio E + E':E + KE'. [M. T. 1906]

20. A very long condenser is formed of a dielectric of given specificinductive capacity bounded by two coaxial cylinders of given radii:if the two surfaces are coated with tinfoil, the outer being uninsulatedand the other raised to a given potential, determine the energy per unitlength. [M. T. 1903]

ANSWERS

1. (i) {h-tt

3. 2-78 xlO*.

6. <><„<*; , „

d <r<ld A-Ad1<x<di, A-A,

7. (S/47T (b — c + ojK) cm. The attractions are as1: (6 - c + clK)*l(a+b - c + cjKy: &»/(« + b)\

\K\a

10. KK'abcl{K'b(c-a)

11. (i) g u = - q12 = q%% - 6 = ab/(b - a).(ii) 9 l l = - g l 2 = ?22-6=Kab(b-t)/{b (b-t-a) + Kat}.

13. Kabcd/{Kcd (b-a) + Kab (d-c)+ad(c-b)}.

20. K<(>2/{41og (b/a)}, where <j> is the given potential and a, b the innerand outer radii.

Chapter VI

ELECTRICAL IMAGES

6*1. Theorem of the equivalent layer. Let She & closedequipotential surface surrounding certain charges £e m afield produced by the charges £e and by certain other charges£e' external to 8. The lines of force cut S at right angleseverywhere, and this would still be the case if we supposeda thin conducting sheet to occupy the position denned by S.But, from considerations of tubes of force, this conductorwould now have a charge — 2e on its inner surface; and,since we suppose its total charge to be zero, the charge onits outer surface would be £e. In the substance of the con-ducting sheet the lines of force cease to exist but otherwisethe original field is unaltered. Also, as explained in 2-31 (iii),the fields inside and outside 8 are now independent of oneanother, and the field inside 8 may be removed while thefield outside 8 remains unaltered. But the field outside 8 isnow due to a charge £e spread over the surface togetherwith the same external charges 2e' as before; and, since 8is an equipotential surface, the charge 2e is in equilibriumon the surface.

The surface density of the charge 2^ is given by Coulomb'sLaw, viz. . E , rsi ii

4TTCT = En = — dcp/dn,where <f> is the potential due to £e and 5X jointly.

Hence if 8 be an equipotential surface in a field due to givencharges, the charges inside 8 can be distributed over S (regardedas a conductor) in an equilibrium layer without affecting theexternal field.

Again, as regards the field inside 8, in fig. (i) it is a variablefield; but in fig. (ii) there is no charge inside 8 and 8 is an equi-potential surface, therefore the potential is constant inside 8and the field is zero. But in fig. (ii) the field inside 8 is due to thejoint effects of the surface layer £e and the external charges

6-2] THEOREM OF THE EQUIVALENT LAYER 115

2X, and since these together produce a zero field, therefore thepart of the field inside 8 in fig. (i) due to the external charges£e ' could equally well be produced by a charge — £e spreadover the surface.

Hence if 8 be an equipotential surface in afield due to charges2e inside 8 and 2e ' outside 8, the field inside 8 will be unalteredif the external charges £e ' are abolished and a charge — ]>> isdistributed over 8 (regarded as a conductor) in an equilibriumlayer.

At first glance it may look as though this result is indepen-dent of what the charges JJe' may be, but this is not so, because8 must be an equipotential in the field due to all the charges,

and the surface density of the layer 4 r r^ depends on the07b

potential due to all the charges. In fact through the remainderof this chapter we shall be concerned mainly with varietiesof the problem—given 8 and 2>', to determine £e.

The theorems of this article are known as Green's Theoremof the Equivalent Layer. They can also be proved analytically.

6-2. Images.* Definition. An electrical image is a pointcharge or set of point charges on one side of a conductingsurface which would produce on the other side of the surface

* The theory of electrical images was the discovery of WilliamThomson, afterwards Lord Kelvin, who communicated his first ideason the subject by letter to M. Liouville in 1845, the year in which hegraduated as Second Wrangler.

8-2

116 PLANE AND A POINT CHARGE [6-2-

thesame electricfield as isproduced by theactual electrificationof the surface.

6*21. Field produced by a point charge in the presenceof a conduotor at zero potential with an infinite plane face.Let there be a point charge e at a point A at a distance a froman infinite conducting plane at zero potential which cuts theplane of the paper at right angles in the line XX'. The tubesof force which start from e fall on the plane and it is requiredto find the surface density of the charge induced at any pointof the plane, and the electric field on the same side of the planeas the point charge e.

Imagine a point charge — e placed at a point A' so situatedthat the plane bisects AA' at right angles. Then the potentialat any point P on the plane, due to e at A and — e at A', is

-pn — -TTD — 0; SO that the plane is an equipotential surfaceAir JL Jrfor the charges e at A and - e at A'. We may regard the planeas part of an infinite sphere 8 whosecentre lies on the left of the figure,thus reproducing the conditions of6*1, viz. a closed equipotential surface8 with a charge e outside it and acharge — e inside it; and by thetheorem of the equivalent layer theinternal charge — e may be spreadover 8 in a state of equilibrium with-out affecting the external field. More-over, by the uniqueness theorem4*21 (i), there is only one way inwhich a charge — e can be distributedover the plane in equilibrium in the presence of the externalpoint charge e. Hence so far as the field in air is concerned,i.e. in the region on the right of XX' in the figure, theelectricity induced on the plane produces exactly the sameeffect as would a point charge — e at A'.

The point charge — e at A' is commonly described as theimage, in the plane, of e at A, by analogy from optical images,

X

6-3] SPHERE AND POINT CHARGE 117

but it is well to bear in mind that the charge — e at A' producesthe same effect at a point on the right of the plane as thewhole surface charge induced on the plane.

If Q is any point in air at distances r, r' from A, A', thepotential of the field at Q is

whereof the first term is directly due to e at A, and the secondterm gives the effect of the induced charge.

To find the surface density induced at a point P on theplane, we note that the electric vector at P is the resultant ofa repulsion ejAP2 along AP and an attraction ejPA'z alongPA'. So that if En denotes the normal component, by Cou-lomb's Theorem

aPA'2'PA'

= -2ea/APs,so that a= -eaj'2mAPz,or the surface density is inversely proportional to the cube ofthe distance from A.

6*3. Sphere and point charge. To find the electric fieldoutside a conducting sphere, due to an external point charge andthe charge which it induces on the sphere; and to determine thesurface density of the induced charge.

(i) When the sphere is at zero potential. Let 0 be the centreof the sphere and a its radius: and let there be an externalpoint charge e at 4 , whereOA-f.

To solve the problem wehave to determine the magni-tude and position of a pointcharge e' inside the sphere, sothat it will be a surface of zeropotentialforthechargeseande'.

Place a charge e' at A' the inverse of the point A, i.e. the

118 SPHERE AND POINT CHARGE [6-3

point on OA such that OA'. OA = a2. Then we can determinee' so that the potential is zero at every point of the sphere.This will be so, if for any point P on the sphere

A'P^AP' "'A'P OP

by similar triangles, since OA. OA' = OP2. Therefore

e'=-ea/f.

Hence the sphere is a surface of zero potential for charges eat A and —ea/f at A', and therefore, by the theorem of theequivalent layer, for a charge e at A and a charge — ea/f dis-tributed in equilibrium over the surface; and since all conditionsare satisfied this must be the solution of the problem proposed(4-21(1)).

The total induced charge is therefore — eajf and this in-dicates the proportion of the tubes of force which start from cand fall on the sphere.

The potential at any external point Q is given by

j e «* mf.A'Q (1)<

For the surface density at any point P on the sphere we have

4™ = K (2),P P/J

where En is the resultant of -j-p-2 along AP and . p .,„ alongPA'.

We know that at the surface of the conductor the resultantintensity is along the normal OP, so that if we resolve each ofthe foregoing components of En in the directions AO and OPthe components in direction AO will cancel and the requiredresult will be the sum of the components along OP. Thus

a n d PA'

6-3] SPHERE AND POINT CHARGE 119

Therefore E =0P {— — \n \AP3 f.PA'3)

ea f alAP\

But AP:A'P=f:a,

therefore a=-~= —-^-—-r^J. (3).4TT 4na.AP3

(ii) When the sphere is insulated and without charge. Weremark that by the uniqueness theorem 4*21 (i) we have onlyto find a solution which satisfies the necessary conditions andthen we know that it is the solution.

Since the sphere is now to be insulated and without charge,the total induced charge is zero, the effect of induction beingto separate equal quantities ofpositive and negative electricity.But the sphere though no longerat zero potential must have aconstant potential. So all theconditions of the problem will besatisfied if we superpose on thefield of case (i) the field due to a charge +ea/f uniformlydistributed over the sphere, for this will make the totalcharge zero and leave the sphere an equipotential surface.

Since the external field due to a uniform spherical charge isthe same as if the charge were collected at the centre of thesphere, we may now say that the external field is due to e at A,— ea/f at A' and ea/f at 0.

To find the surface density we have only to add to that givenin (3) that due to the uniformly distributed charge ea/f, sothat in this case

" A—n A 133 ' A—rtt ' '"

It is clear that at points such that AP3=f(f2 — a2} there isno electrification. This defines a circle on the sphere separatingpositive from negative electrification.

120 SPHEBB AND POINT CHARGE [6*3-

(iii) When the sphere is insulated and has a total charge Q.For the solution in this case it is only necessary to superposeon the field (ii) that due to a /...\charge Q uniformly distributed ^ ''over the sphere. The externalfield may now be represented asdue to e at A, —ea/f at A' andQ + ea/f at 0; and the surfacedensity at any point P is

ff=_^s+ JL .(5).

(iv) When the sphere is maintained at a given potential <f>.Since a uniformly distributed charge fat on the surface of thesphere would give it a potential <j>, it is clearly only necessaryto superpose the field due tothis distribution on the fieldof case (i). The external fieldmay now be represented as dueto e at A, —ea/f at A' and <f>aat 0; and the surface densityat any point P is

a= — .(6).

6*31. The force on the point charge e can be found in anyof the foregoing cases by using the law of inverse squares. Thusin 6*3 (iii), the point charge e is repelled from a sphere havinga total charge Q with a force

e WH—r

where A'A =f—a2jf. Therefore the force

_e© e2a ezaf(/2-a2)2"

Whether this is a repulsion or attraction depends on therelative values. It is clearly a repulsion if Q/e is large enough.

P u t / = a + «, and let the distance x of e from the surface of

6-33] POINT CHABGE INSIDE A SPHERE 121

the sphere be small. Then, if we expand the expression for theforce on e, the principal terms are

eQ e*a?

so that the force is an attraction unless Q > ea*/4x2.

6'32. I t is to be observed that electrical images are all'virtual' or 'imaginary1. We cannot have a field on one sideof a surface due to an image on the same side of the surface.In the cases considered so far the point charge e is a realcharge and the other point charges are all imaginary chargesput in to aid the solution of the problem in question.

6-33. Point charge e inside a sphere at zero potential. Theexternal field hi this case is zero since there is no charge outsidethe surface of zero potential. Let the conducting sphere be ofinternal radius a with a point charge e at a point A at a distance/ (< a) from the centre 0.

As hi 6*3 (i), we can shew that a charge c' at the inversepoint A' together with eat A will make the sphere a surface ofzero potential if c' = — eajf.

Then, as hi the second part of the theorem of the equivalentlayer (6*1), considering the field inside the sphere due to e atA and — eajf at A', the contribution due to the external charge— eajf can equally well be produced by a distribution — e hiequilibrium over the surface of the sphere. But the field whichwe are investigating is the field inside the sphere due to e at Aand the charge — e which it induces on the inner surface of the

122 EXAMPLES [6-33-

conduoting sphere; so that, conversely, this field may be repre-sented as due to e at A and an image — ea/f at A'.

By an argument which is, step by step, the same as that of6-3 (i) we can shew that the surface density at a point P isgiven by

< 7 = — • (1).

The force on the point charge e is an attraction along A A'

equal to -ee'/AA'2 or e2a//(»2 ~/ 2 ) 2 -

6*34. Examples, (i) A thin plane conducting lamina of any shapeand size is under the influence of a fixed electrical distribution on one sideof it. If crj be the density of the induced charge at a point P on the side ofthe lamina facing the fixed distribution, and o that at the correspondingpoint on the other side, prove that al — al = aa, where a0 is the density atP of the distribution induced on an infinite plane conductor at zeropotential coinciding with the lamina.

Let L denote the lamina and R the remainder of the infinite plane ofwhich L is a part. Denote the given fixed distribution by Se. Con-sider an electric field which contains the lamina L insulated and un-charged, but under the influence of ahypothetical rigidly distributed chargeon R, in which the density at everypoint of JR is equal and opposite to thedensity which would be induced at thatpoint of R by the charges Se if the planewere a complete infinite plane at zeropotential; so that in the notation of thequestion — a0 denotes the density at anypoint of R. Then since there is no otherelectricity in the field but this charge onJR and the charge which it induces onthe lamina, the field is symmetricalabout the plane of the lamina, and tubesof force proceeding from the charge onR fall on both sides of the lamina inexactly the same way, so that atcorresponding points on opposite sides the surface density has thesame value, say <r8. This field is represented in section in fig. (i).

Consider a second field fig. (ii) in which the given fixed charges Seinduce a charge of density denoted by a0 on an infinite plane con-ductor, at zero potential, at the same distance from Se as is the lamina.

Now superpose the fields by moving fig. (ii) to the left until the planecoincides with the plane in fig. (i). The charges on R annihilate one

Ri

6'4] EXAMPLES 123

another, and there remain the charges Se and the lamina L with acharge of density at on the side remote from Se and a charge al = ai + <x0

at the corresponding point on its near side; so that we have provedthat CTJ — a2 = <r0 as required.

(ii) A straight wire of length 21 is charged with electricity of amount aper unit length. It is placed in the presence of a conducting sphere ofradius a and the sphere is earthed. The perpendicular distance from thecentre of the sphere to the wire is c and the ends of the wire are equidistantfrom the centre of the sphere. Shew that the sphere receives a charge ofamount o__^_v.-i^C- [M. T. 1934]

It is assumed that the charge on the wire is 'rigid', i.e. that itsdistribution is unaffected by induction.

If a dy be the charge of an element dy of the wire at P, at a distance yfrom its middle point, there is an image at the inverse point P' with ac h a r S e . a _ • gady

~aay~VW¥Hence the total induced charge, being the sum of all such image

charges, „ d

= — I ,, - 2 " a sinh"1 lie.

6-4. Geometrical method of evaluating an induced charge./ / , in the solution of a problem, an electrostatic field in air isexpressed as a field due to a number of point charges, then thecharge induced on any portion of a conducting surface can beexpressed in terms of the solid angles which this portion subtendsat the point charges.

Let dS be an element of area of the conducting surface at Psubtending a small solid angle dco at a point A, at which there

124 THE INDUCED CHARGE [6*4-

is a point charge e; and suppose that it is required to find the con-tribution which this point charge makes to the induced chargeon a finite portion of the surface, on the side remote from A.

If a denotes the surface density,then the chargeon dS is adS, where4ma=En, the normal intensity.

But the point charge e con-tributes to the intensity a com-ponent e/r2 along AP, whereAP = r. And if AP makes an angle6 with the normal, the normalcomponent due to e is ecos0/r2,and its contribution to adS is

ecosO.dS , , cos0. dS; but r——dco,

r2

therefore this contribution is edcaj4=7T.If dS subtends solid angles dcu', dut", ... at the other point

charges e', e", ..., they will make like contributions, so that

a dS = ^ (edco + e'da>' + e"da>" + . . . ) ;

and by integrating we get, as the charge induced on anyfinite area S,

^(ew + e V + e ' V + ...) (1),

where co, a>', on", ... are the solid angles which 8 subtends atthe charges.

It must be observed that we assumed that the inducedcharge lies on the side of the surface remote from the pointcharges, but in general some of the charges lie on one side ofthe surface and some on the other, and it is clearly necessary tochange the sign of any term in (1) if the corresponding chargelies on the same side of the surface as the induced charge.

6*41. Examples, (i) A point charge e is placed at distances a, bfrom two infinite conducting planes at zero potential which meet at rightangles. Find the ratio in which the induced charge is divided between theplanes.

6*41] THE INDUCED OHAEGE 125Let the plane of the paper cut the given planes at right angles in the

lines Ox, Oy and contain the point charge e at A. If ABCD is a rect-angle with its sides parallel to Ox, Oy, it is easy to see that the two givenplanes would be at zero potential under the influence of point chargese at A and O and - e a t B and D.

The charge induced on the plane Ox can therefore be expressed interms of the solid angles which it subtends at A, B, C and D. These aremeasured as in 1'33. Thus the plane through Ox perpendicular to theplane of the paper, having a straight boundary through 0 and otherwiseunlimited, subtends at A the solid angle between two planes, both atright angles to the plane of the paper and one passing through AOwhile the other is parallel to Ox; and the solid angle subtended at Abeing twice the angle between these planes is

2(ir-A0x) or 2(w-tan-16/a).

y

The plane clearly subtends an equal solid angle at D; and at B and Oit subtends a solid angle 20BA or 2 tan"1 bja.

Hence taking account of the fact that the charge induced on theplane is on the same side of it as the charges at A and B and on theopposite side to the charges at C and D, we have for the induced charge

-^- .2 ( w - t an- 1 6/a ) + - . 2 tan"1 - +-£ . 2 tan"1-4w v ' ' 4ir a 4w a

- ~ . 2 (TT - t an - 1 b/a),

where we have written down the contributions of the charges atA, B,C,Din turn.

This makes the total charge induced on the plane through Ox equal

to tan - 1 =-; similarly the charge induced on the plane through Oy

2& bis tan-1 - , the two together amounting to — e.

it a

126 PLANE WITH A HEMISPHERICAL BOSS [6*41-

(ii) A point charge e is placed at a distance f from the centre of aninsulated uncharged sphere of radius a. Shew that the total charge on thesmaller part of the sphere cut ojf by the polar plane of the point is

^ T - 1 9 1 9 ]

In this case, as in 6'3 (ii), the point charges are e at A, — ea/f at theinverse point A' which lies on the polar plane of A, and eajf at theoentre O. The solid angles which the spherical cap PBP' in the figuresubtends at A, A' and 0 respectively are

2 T T ( 1 - COS OAP), 2TT and 2n (1 - cos POA)

or

Therefore the charge induced on this portion of the sphere is

In the same way the spherical cap PGP' subtends at A, A' and Osolid angles 2n-(l — cos OAP), 2ir and 2«-(1 + cos POA), and takingaccount of the fact that the induced charge is on the side of this surfaceremote from all the point charges its amount is

as is otherwise obvious since the total charge is zero.

6*42. Point charge and an infinite conducting plane with ahemispherical boss. Let XX' be an infinite conducting plane witha hemispherical boss DOD' of centre O and radius a, and let a charge ebe at a point A on the axis of the boss, where OA =f (> a), the plane andboss being at zero potential. Then if B be the inverse of A in the sphere

6-42] PLANE WITH A HEMISPHERICAL BOSS 127

and A', B' the geometrical images of A, B in the plane, point chargese at A, — e at A', — ea/f at B and ea/f at B' will have the plane andsphere as a surface of zero potential. Therefore the field in air, i.e. onthe right of the conductor, due to the charge e at A and the inducedelectricity will be the same as the field of the four point charges.

From 6'3 the surface density at a point P on the boss is given by

e(J»-al

'~ 4 A P •(1);

and from 6"21 the surface density at a point Q on the plane is given by(ea/f) OBea

X

Applying the method of 6*4 to find the total charge induced on theboss,, the solid angles subtended by the hemisphere DCD' at both Aand A' are 2w(l -cos OAD) orat B the solid angle is

2TT(1 + COSO.BD) or

since OB^a^jf; and at B' the solid angle is2TT(1 -COSOB'D) or 2w{l-a/-v

/(a l l+/2)}.Hence taking account of the signs of the charges and the side of the

surface on which each lies, we get for the induced charge

/ ) *" fi i a ) i "* fi a

1V t* fi / ) *" fi i a ) i "* fi a }2\ V(<**+P)> 2/ |1V(«2+/ i !)l 2/t V(a2+/a)J

2 1 V(«a+/2)/'

128 ORTHOGONAL SPHERES [6-42-

the four terms being the contributions arising from the charges atA, B, B', A' in order, and the result is

Since the total induced charge is — e, that on the plane must be

6*5. Conductor bounded by two orthogonal spheres.The distribution of an electric charge on the surface of anyconductor and the capacity of the conductor can be found, ifwe can determine a set ofpoint charges which willhave the surface of theconductor for an equi-potential surface. This iswell exemplified by the caseof a conductor bounded bytwo spheres which cut oneanother at right angles.

Let A, B be the centres of the spheres PDD', QDD' and a, btheir radii. We shall call them the spheres (.4) and (B). Iftheir plane of intersection cuts AB in C, then ADB is a rightangle, and we have

and

so that C is the inverse of B in (A) and the inverse of Ain (B).

Let a charge a be placed at A, its image in (_B) is a charge— ab/\/(a2 + ft2) at C, and these charges together have (B) asa surface of zero potential. Similarly a charge b at B and thesame image —ab/-\/(a?+b2) at G would have (A) as a surfaceof zero potential. But the charge a alone would raise (A) tounit potential, and 6 alone would raise (5) to unit potential.Hence for a set of three point charges a at A, — ab/-\/(a2 + b2)at C and b at B the two spheres would be at unit potential; andby the theorem of the equivalent layer, if the total charge

were distributed over the surface PDQD', the external field

6-51] SERIES OF IMAGES 129

would be unaltered, and the potential of the surface wouldstill be unity, so that the capacity of the conductor is

Again the surface density at a point P on (A) due to thecharge a at A is a/iira2, and due to b at B and its image at C is

b (AB2 — a2)— r R p 3 (6*3 (i) (3)), so that the whole surface density

1 / b3 \at P is -— 11 — -5-= I, and in like manner the surface density

4TT» \ JDJT*]

(l - ^ ) .at a point Q on (B) is (

The whole charge on the portion DPD' of the surface may be foundby the method of 6*4. Thus the solid angles which it subtends at A, Oand B are respectively

2ir(l+cos CAD), 2-n- and 2w(l -OOBCBD)

a n d

so that the charge on DPD' isabla U , a \ »

or

and similarly for the charge on DQD'.

6-51. Series of images. The fact that C was a commoninverse point in 6*5 enabled us to get a solution with onlythree point charges, and in any case in which the spheres cutat an angle which is a submultiple of n the problem could besolved by a finite number of point charges. But otherwise, asin the case of two non-intersecting spheres (A) and (B) ofradii a and b, of which ( 4) is at unit potential and {B) is atzero potential, if we begin by placing a charge a at the centre of(A) its image must be taken in (B) and then the image in ( 4)of this image must be taken, and so on indefinitely, so thatthere would be an infinite series of images. The sum of thecharges inside (A) would be its coefficient of capacity qn, andthe sum of the charges inside (B) would be the coefficient ofmutual induction q12. In some cases the distance between theconductors is large compared with their radii, so that approxi-

130 SERIES OP IMAGES [6-51-

mate results may be found by taking a few terms of the seriesof images.

6*52. E x a m p l e . The outer of two concentric spherical conductors ofradii a, b (a<b) is earth con-nected and the inner is chargedto potential <j>. A point chargeQ is then placed in the spacebetween the spheres at a distance i i \ Pc from the centre. Prove that the n

charges on the spheres are as

[I. 1896]

Denote the spheres by A andB. The original charge e on A is given by

, o o ab(j><P = — r> or e— ^——Y " b b—a

ea

.(1).

Let 0 be the centre and P the position of the charge Q; OP = c.Let Px be the inverse of P in A,

Qi be the inverse of Pt in B,P 2 be the inverse of Qx in A,Q2 be the inverse of P2 in B and so on;

and denote image charges by square brackets. Then we have

blc

,=£T=&I, [PJ=-WJ OQt

; and so on.~~^Wc

Similarly, let Qt' be the inverse of P in B,Pi be the inverse of Qx' in A,Qz be the inverse of Px' in B and so on;

then in the same way we may shew that

[V8']= _

J ac b^

} -f-; and so on.

6'6] ELECTRIC DOUBLET 131

Then the total charge on the sphere A is

ab (Q Q

And since there is no external field the total charge on B is minus thesum of the charges it contains, i.e.

ab IQ Q 1 ab (Q Q

and hence the result.

6-6. Field due to an electric doublet. An arrangement oftwo equal and opposite point charges — e and e at an infini-tesimal distance 8s apart constitutes an electric doublet, orbipole, the magnitude of the charges being such that theproduct eSs has a finite limit as e increases and 8s decreases.

2 M cosO

This limit is called the moment of the doublet and denotedby M, and the line joining the charges in the sense from thenegative to the positive charge is called the axis of thedoublet.

In the chapter on magnetism we shall prove that M may beused as a vector to denote both the magnitude of the momentand the direction of the axis of the doublet and that doubletscan be compounded and resolved by applying the vector lawof addition to their moments.

Let a doublet consist of charges — eeXB and e at A, whereAB = 8s, and the limit of e8s as e increases and 8s decreases isfinite and = M.

9-2

132 ELECTRIC DOUBLET [6'6-

Let 0 be the middle point of BA, and let P be a point (r, 0)referred to 0 as origin and the axis BA as axis of co-ordinates.The field is clearly symmetrical about the axis of the doublet,and the potential at P is given by

, e e

e er —J8scos0 r + Sscosd

e / 8s cos 0 \ el 8s cos 0i1 + \ el

"7 r\e8scos0, , „ , .

5— to the first order m

Mcosd

Since is a function of two variables r, 0, therefore the electricintensity at P has components in the two directions hi which

-r and 6 increase; and denoting them by Er and Eg, we have

_ d<j>_2Mcosd)

ld<f> MmnOl * ''

and ^ - _ _ ^ - _ _ jand the electric vector E is the resultant of these two com-ponents.

6-61. Insulated conducting sphere in a uniform field offorce. If hi 6-3 (ii) we imagine the point charge e outside theinsulated sphere to move away to an infinite distance and atthe same time to increase in such a way that e//2 remainsfinite and equal to F, the lines of force hi the finite part of thefield will be lines radiating from a point at infinity, i.e. parallellines and the undisturbed field will be of uniform intensity F.At the same time, as the point A in the figure of 6*3 (ii) movesto infinity the point A' moves up to 0, and the charges — eajfat A' and ea/f at 0, being at a distance a2// apart, form adoublet of moment ea3//2, or Fa3, whose axis is in the directionof the undisturbed field F.

6-61] SPHERE IN A UNIFORM FIELD 133

It follows that when an insulated sphere of radius a isplaced in a uniform field of intensity F the induced chargeproduces the same external field as would a doublet of momentFa? placed at the centre of the sphere with its axis in thedirection of the field.

Take the undisturbed field F from left to right as indicatedin the figure; then the total field at a point (r, 6) outside thesphere is made up of three components, viz. the given field Fand the two components of intensity due to the doublet Fas,. 2Fa3cosd ,. .. . Fassia6 , , . 1V

i.e. j radially and ^— transversely m the sensein which 6 increases. It is easy to see that at a point on the

surface of the sphere, these three are components of a singleresultant 3.Fcos0 along the normal. Therefore the surfacedensity induced on the sphere is given by

It is positive over one half of the sphere and negative overthe other half, the circle 6 = ^TT being a line of no electrification.

The total positive charge on the hemisphere can be expressedas an integral, viz.

fJo

(2),

and there is an equal and opposite negative charge.It should be noted that if we take an axis of a; in the direction

of the uniform field F, the potential of the field must be — Fx

134 DIELECTRIC PROBLEMS [6-61-(in order that the negative gradient may be F), and thepotential of the doublet is by 6*6 Fasooa8jr2, so that thewhole potential of the external field is

.Fa3 cos 0.(3).

6*7. Dielectric problems. A point charge in front of ablock of uniform dielectric with an infinite plane face.Let e be the charge at A, XX' the plane face of the block ofdielectric and A' the point in the dielectric such that XX'bisects A A' at right angles. Theplane XX' divides space into tworegions, air space numbered 1 anddielectric numbered 2. There is anelectric field in both regions, butwith different potential functions faand (/>2. Though there is actuallya small constant difference ofpotential (contact difference) be-tween the surface of the dielectricand the air in contact with it, it issmall enough to be negligible, so

yyy/ / /

iA,' 1

X

V

0 A

1

X'

we assume that on the plane XX' we have <j>1 = 4'2other condition to be satisfied is the continuity of normaldisplacement (5*21 (5)), since there is no charge on theinterface.

We will shew that we can satisfy these conditions by meansof potentials, in air ,

9i=z+Zi (!)>

where r, r' denote distances from A, A'; and in dielectric

. e"(2).

We remark that we cannot have a term of the form e'"jr'in <j>2, because it would make the potential infinite at A', andthere is no singularity in the dielectric which could accountfor such an infinity in (f>2.

6*71] DIELECTRIC PROBLEMS 135

Then since <f>i = <f>2 on XX' (where r = r'), we must havee + e' = e" (3).

And since the normal displacement is got by multiplyingthe normal intensity by the dielectric constant, the conditionof continuity of normal displacement requires that

e-e' = Ke" (4),where K is the specific inductive capacity of the dielectric,the minus sign arising from the fact that e' is on the oppositeside of the interface to e and e".

Thus all conditions are satisfied by. l-K , .. 2e

e = : e and e" = -—-^ .(5).

That is to say, the field in air is identical with a field createdby the given charge e at A together

I jr

with a charge -—= eat A'; and the1 + A

field in the dielectric is such as wouldbe produced by a single charge

2eat A. I t follows that the lines

of force in the dielectric are straightlines radiating from A as shewn inthe figure.

The force on the charge e isee'/AA'2 or ee'/4a2, if a is the distanceof the charge from the block. This means that the chargeis attracted to the dielectric block, or vice versa by a forceK— 1 e2

r=—-—, representing a tension in the lines of force..ft. -p JL 4fl-

6*71. Dielectric sphere in a uniform field of force. Leta sphere of radius a of homogeneous dielectric of constant K beplaced in a uniform field of electric intensity F. To find howthe field is modified by the presence of the sphere. The surfaceof the sphere divides space into two regions—air, numbered 1and dielectric numbered 2.

Take an axis Ox through the centre 0 of the sphere in the

136 DIELECTRIC SPHERE [6-71-

direction of the undisturbed field F. Since the negativegradient of the potential is F, the potential of the undisturbedfield must be given by

<f>=-Fx=-Frcoa8 (1).

To solve the problem we have to find potential functions, forregions 1 and 2, i.e. solutions of Laplace's equation, con-tinuous across the surfacer = a, making the normal dis-placement continuous at r = a,and tending to the form (1) \ ° J x

at a great distance from thesphere.

Further we notice that in 6*7 the field in the dielectric is adefinite multiple of what the field would be if the dielectricwere not present; and the field in air due to the presence ofthe dielectric is a definite multiple of what the field would beif the dielectric were a conductor. So we endeavour to obtaina solution of this problem in analogous fashion, rememberingthat if the sphere were a conductor it would produce in region 1the same effect as a doublet at its centre.

We assume therefore that the potentials in regions 1 and 2are given by A a

^ (2)

and </>2 — Brcosd (3),

where A, B are constants to be determined.It can easily be verified that these are solutions of Laplace's

equationin polar co-ordinates, butitisnot necessary to performthe verification because the separate terms are functions whichhave previously occurred in our work as potential functions.

As r increases, <£x tends to the form — Fr cos 8, which is oneof the requisite conditions to be satisfied. Also, when r = a, wehave <f>x = <j>2, so tha,t .

2 = Ba (4);

and ^ Z ^ , so thatdr dr

F (5).a3 v '

6-72] DIELECTRIC SPHERE 137

Therefore all the conditions are satisfied by (2) and (3)provided that R_1 Q F

Fa3 and B = —K + 2 •(6);

i.e. the field in the dielectric is uniform and 3/(1? + 2) of whatit would be if the dielectric were not present. But we mustremember that it is Dn or KEn which is continuous, so that the

number of unit tubes of force which cross a plane area perpen-dicular to the field in the dielectric is BKj(K + 2) of what itwould be if the dielectric were not present; this means that, forlarge values of K, there are roughly three times as many tubesof force in the dielectric as there would be in air.

Also, in air the dielectric sphere produces the same effectjr i

as would a doublet of moment -^—- Fa3 placed at its centre.

6'72. Converse problems. We may use the case just consideredto illustrate a converse problem: What distribution of electricity in airwill produce a given field of potential?

The solution is found from the fact that at every point of space atwhich there is a volume density p, it is determined by Poisson'sequation 2-41 (3)

and on every surface on which there is a surface density a by 2'411 itis given by a, R,

8^^

where 8nx, dn% are directed away from the surface.Now apply this process to the field given by

(3),

It is easy to verify that V2 x = 0 andvolume density of electricity anywhere.

2 = 0, so that there is no

138 TWO-DIMENSIOXAL IMAGES [6-72-

Also 8nx = 8r and fina = —8r, so that by differentiating and putting= a in (2),

or

K + 23 K-\

K + 2

FcoaB (5).

Hence a surface distribution of density a, given by (5), on a sphere ofradius a in air, would produce the same field inside and outside asexists in the case of the dielectric sphere in the uniform field. This andcorresponding surface distributions in other problems are sometimescalled 'fictive layers'.

6*8. Two-dimensional images. In 2-53 and in 3*43 weconsidered some two-dimensional fields, and in 2-532 we sawthat two equal and opposite parallel line charges of amount± e per unit length produce a field of potential

<£ = C l-2elogr + 2elogr' (1),

where r, r' denote distances from the lines.It follows that the equipotential curves in the two-dimen-

sional field are the family of coaxial circles with limiting pointsA, A' at the charges.

The plane XX' bisecting A A' at right angles is also anequipotential surface, and will be the surface <f> = 0 if we take

6-81] TWO-DIMENSIONAL IMAGES 139

Hence the image of a line charge through A with regard toa parallel conducting plane XX' at zero potential is an equaland opposite line charge through A'.

The surface density induced at any point P on the plane is

or o-=-ea/77-.4P2 (2),where a is the distance of A from XX'.

6*81. Image of a uniform line charge in a parallel circularcylinder. Continuing the same argument, if 0 be the centreof one of the coaxial circles which are equipotential curvessurrounding the point A', the limiting points A, A' are inversepoints with regard to this circle. Let a be the radius of thecircle and let OA =/.

The potential due to line charges + e through A and A' is<£ = C - 2e log r + 2e log r\

and at a point P on the circle this takes the form

so that the cylinder of which the circle is a cross-section willbe at zero potential if we take Cf = 2elog(//a). We can thenshew as in 6-3 (i) that the surface density at P is given by

To deduce the case of the cylinder insulated and uncharged

140 CAPACITY OF A TELEGRAPH WIRE [6-81

since the charge per unit length when the potential is zero isthe image charge — e, we must superpose the field due to acharge e uniformly distributed, and that is equivalent to aline charge e through 0.

6*82. Capacity of a telegraph wire. Reverting to 6*8,let XX' be the surface of the earth andconsider the field of horizontal linecharges e, — e through A, A', whereA A' is bisected at right angles by XX'.

The potential

X'will be such that XX' is a surface of —zero potential if (7 = 0. Take a smallcircle of centre 0 and radius a and let itbe one of the coaxial family of whichA, A' are limiting points. The potentialat a point P on the circle is

, o . A'P „ _ OA'4> = 2 e log i p = 2e log ——.

tv

A'

Now let the circle be regarded as the cross-section of atelegraph wire, then its capacity in the presence of the earth

OA'is the value of c when <$> is unity, i.e. 1/2 log . But if A is

the height of 0, OA + OA' = 2h and OA.OA'=a2, so thatOA' = h+ \ /(^2 —a2); and the capacity

Since a is small compared to h, the value is approximately

141

EXAMPLES1. A positive point charge at A is fixed in front of a large conducting

plane which is earthed. Shew that the tubes of force, which start onthe point charge and make an angle <j> or less with the normal from Ato the plane (not with the normal produced), end on a circular regionof the plane whose radius subtends an angle 0 at A, where

sinJ^=V2sin £0. [M. T. 1930]

2. Two equal charges are condensed at points distant a and b froman infinite conducting plate which is to earth: the charges are on thesame side and on the same line perpendicular to the plate. Shew thatthe lines of force from the nearer charge fall on a circular area of theplate of radius r, where

ri) = l. [I. 1907]

3. Two equal point charges of opposite signs are placed at pointsA, B whose distance apart is 2a, in front of an infinite plane conductor,the distance of each from the plane conductor being a. Shew that theforce required to maintain either charge in its position is inclined tothe plane at an angle of 45°, and is less than it would be if the plane wereabsent in the ratio (2^2 - 1): 2. [St John's Coll. 1906]

4. Two equal charges are placed at a distance a from an infiniteconducting plane at zero potential, and at a distance 3a/2 from eachother. Shew that, if these charges have the same sign, the resultantmechanical force experienced by either charge makes an angle tan-1 § f §with the normal to the plane. [M. T. 1915]

5. A positive charge e is placed at distances a and b from two semi-infinite planes at zero potential intersecting at right angles. Find thesurface densities of electrification on the nearest point of each plane.

[M. T. 1927]

6. Two point charges A, B, each of amount Q, are placed at equaldistances from each other and from an infinite conducting plane sheetat potential zero. Find the charges induced on the portions of thesheet cut off by the plane through A perpendicular to AB. Sketch thelines of force in the plane through AB perpendicular to the sheet.

[M. T. 1924]

7. A conductor is formed of two planes inclined to one another at anangle Jn-; the planes are bounded by their line of intersection and areof very great area; the conductor is maintained at zero potential andis influenced by a charge e at a point P in the space between the twoplanes, such that the perpendicular from P to the common edge makes

angles a, ^—a with the planes. Prove that the charges induced on the

two planes are e ( — 1H ), — e. — respectively. [I. 1902]\ it J IT

142 EXAMPLES

8. Shew that , if the inducing charge is a t a distance from the centreof a sphere at zero potential equal to double the radius, the surfacedensities at the nearest and the most remote points of the sphere arein the ratio 2 7 : 1 . [St John's Coll. 1907]

9. A point charge e is held at a distance / from the centre of aninsulated spherical conductor of radius a, which carries a charge Q.Prove that the surface density at the point of the sphere most remotefrom the charge e will be zero if

+o)/{/( /+o)«}.[St John's Coll. 1906]

10. An insulated conducting sphere of radius a is under the influenceof a point charge e at a distance / ( > a) from its centre. What is theleast positive charge that must be given to the sphere in order thatthe surface density may be everywhere positive. [M. T. 1893]

11. A charge e is placed at a distance c from the centre of an insulateduncharged sphere of radius a, c being greater than a. Shew that thepart of the sphere which is positively charged is separated from thatwhich is negatively charged by a circle of points a t distance from e

S1™11 fcy r» = c (c - a) (c + a). [M. T. 1933]

12. A point charge e is placed at a point A a t a dis tance/ from thecentre of an insulated conducting sphere of radius a ( < / ) . Determinethe charge that must be placed on the sphere in order that the line ofno electrification may be the intersection of the sphere by the polarplane of A; and shew that the numerical ratio of the greatest negativeand positive surface densities is in this case {(/+ a)j(f— a)}*. [1.1932]

13. A sphere is insulated and carries a charge equal to that placedat an external point; prove that if the ratio of the distance of the pointfrom the centre to the radius is greater than 2-62:1, the surfacedensity on the sphere will be nowhere negative. [St John's Coll, 1909]

14. The centre of an insulated uncharged conducting sphere of radiusa is at the middle point of the straight line joining two equal pointcharges of electricity, which are a distance 2c apart. If ale is small,shew that the force on either charge would be increased in the approxi-mate ratio

if the sphere were removed. [M. T. 1931]

15. An insulated uncharged conducting sphere is placed centrallybetween two charges of equal magnitude. Shew that if they are of likesigns the repulsion between them is diminished; but if of unlike signsthe attraction between them is increased. [M. T. 1921]

EXAMPLES 14316. If a sphere of radius a is earthed and positive charges e, e' are

placed on opposite sides of the sphere, at distances 2a, 4<x respectivelyfrom the centre and in a straight line with it, shew that the charge e' isrepelled from the sphere if

e'<25e/144. [M. T. 1935]

17. A field is produced by a point charge e in the presence of anuninsulated spherical conductor of radius a whose centre is at a distancec from the charge. Find the force with which the charge is attractedto the sphere; and shew that the force is changed to a repulsion byinsulating the sphere and connecting it to a large distant conductor atpotential <f> provided that

<f> > jp^pjk • P t John's Coll. 1915]

18. A conducting sphere of radius a at zero potential is under theinfluence of a charge at an external point 0. Shew that the fractionof the induced charge which lies within a right circular cone withvertex 0 and axis through the centre of the sphere is (a — c)/a, where2c is the intercept made by the sphere on any generator of the cone,and both intercepts made by the cone on the sphere are included.

[I. 1911]

19. A point charge e is placed at a distance 6a from the centre of aninsulated sphere of radius a, which has a charge Q. Shew that if thereis a point of equilibrium in the field, whose distance from the centre is4a taken towards e, then Q = 4-gfip r e.

Give a rough drawing to indicate the nature of the equipotentialsurfaces. [M. T. 1915]

20. Prove that, if a small charged conductor be at a distance r fromthe centre in a space bounded by a spherical conducting surface ofradius a at zero potential, the force repelling it from the centre isare2/(a2 - r a ) 2 , where e is the charge. [M. T. 1923]

21. Within a spherical hollow, in a conductor connected to earth,equal point charges e are placed at equal distances / from the centre,on the same diameter. Shew that each is acted on by a force equal to

ea [*, t j C \ a + 47>1 • tS t John's Coll. 1905]

22. A point charge e is held at a distance / from the centre of aninsulated uncharged spherical conductor of radius a; prove that theforce required to hold the point charge in position is e2a//(/2 — a2)2 ore2a8 (a2 — 2/2)//3 (/2 — a2)2, according a s / is less or greater than a.

[St John's Coll. 1907]

23. In a spherical cavity of radius a within a conductor a charge eis placed at a point P distant/ from the centre G.

Shew that a line of force from e making initially an angle a with PC

144 EXAMPLES

will meet the surface of the cavity at a distance (a2 — / 2 ) / (a — / c o s a )from P . [I. 1906]

24. A charge e of electricity is situated a t a distance c from thecentre of a conducting sphere of radius a. If the sphere is insulated anduncharged, find the rise in its potential due to the presence of thecharge, and shew tha t there is an attractive force between the sphereand the charge equal to

c8 (c2 — a2)2 L J

25. A point charge e is placed inside a hemispherical hollow ofradius a in an uninsulated conductor. The point charge is on the axisof symmetry a t a d is tance / from the plane boundary. Shew tha t it isin equilibrium if / /a is equal to a root of the equation

x* - 8a;7 - 2xl - 8a:3 + 1 = 0.

Shew also tha t this equation has one and only one root between 0and 1. [M. T. 1932]

26. An infinite plane conductor a t zero potential with a hemi-spherical boss of radius a is under the influence of a point charge + eat a point on the axis of symmetry distant 2a from the plane. Find thepotential a t any point and sketch the lines of force. Determine theslope at the point charge of the critical line of force separating lines offorce going to the boss and to the plane. [M. T. 1928]

27. An infinite plane conducting sheet has upon it a hemisphericalconducting boss of radius a and the whole is a t potential zero. A chargee is placed at a distance c from the centre on the radius of the spherewhich is normal to the plane. Shew tha t the density a t any point ofthe plane is e

where r, r' are the distances of the point from the charge e and its imagein the sphere.

If a\c is small, shew tha t a is greatest at a distance ai c$ approximatelyfrom the centre. [M. T. 1934]

28. An infinite plane has a hemispherical boss, and a unit pointcharge is placed in front of the boss on the common normal to the planeand the hemisphere. Shew tha t the charge induced on the par t of theplane external to the boss is — cos2 6 sec 8, where 8 is the angle subtendedat the charge by any radius of the base of the hemisphere.

[M. T. 1916]

29. A dome of conducting material is built in the form of a hemi-sphere on the ground; a small charged conductor is situated midwaybetween the dome and the ground, on the vertical through the centreof the dome. Obtain the system of images, and prove that the mechan-

EXAMPLES 145

ical force on the small conductor is upwards, and equal to 2%e*/aa>where e is the charge on the conductor and a is the radius of the dome.

[M. T. 1914]

30. A point charge e at P is at a distance PO = c from an infiniteplane conductor, which is uninsulated. Shew that, when a conductinghemisphere of radius a (< c) is placed on the plane with its centre at 0and its vertex towards P, the number of unit tubes of force which fallon the plane outside a circle of centre O and radius r (> a) is decreasedby ea*/c (a4 + cV) i [M. T. 1909]

31. A hollow conductor has the form of a quarter of a sphere boundedby two perpendicular diametral planes. Find the image of a chargeplaced at any point inside it. [M. T. 1897]

32. A conducting surface consists of two infinite planes which meetat right angles and a quarter of a sphere of radius a fitted into the rightangle. If the conductor is at zero potential and a point charge e issymmetrically placed, with regard to the planes and the sphericalsurface, at a great distance / from the centre, shew that the chargeinduced on the spherical portion is approximately — SeaF/trf*.

[M. T. 1903]

33. A thundercloud, which may be regarded as an electric doubletwith its axis vertical, is moving uniformly along a horizontal straightline and directly approaching an observer on the ground who is record-ing the electric intensity close to the surface. Shew that the rate ofchange of the electric intensity vanishes when the elevation of thecloud is tan-1 J. [M. T. 1923]

34. A sphere of radius a has its centre at a distance c from an in-finite plane conductor at zero potential. Shew that its capacity is,approximately, o + aa/2c, where ajc is regarded as small. [M. T. 1911]

35. A condenser is formed of two concentric conducting spheres ofradii o and b (o<6). If a point charge e is placed between the spheres,at a distance c from the centre, and the spheres are connected by afine wire, the charge on the inner sphere after connection will be

-ea(b-c)/c(b-a). [I. 1907]

36. A charge e is placed midway between two equal spherical con-ductors which are kept at zero potential. Shew that the charge inducedon each is

neglecting higher powers of m, which is the ratio of the radius of aconductor to half the distance between their centres. [I. 1908]

37. If a particle charged with a quantity e of electricity be placedat the middle point of the line joining the centres of two equal spherical

146 EXAMPLES

conductors kept at zero potential, shew that the charge induced oneach sphere is _ 2e,n (1 - m + m2 - 3m3 + 4m*)neglecting higher powers of m, which is the ratio of the radius to thedistance between the centres of the spheres. [I. 1893]

38. A spherical conductor of diameter a is kept at zero potential inthe presence of a fine uniform wire, in the form of a circle of radius c ina tangent plane to the sphere with its centre at the point of contact,which has a charge e of electricity; prove that the electrical densityinduced on the sphere at a point whose direction from the centre of thering makes an angle iji with the normal to the plane is

(a2 + c2 sec2 i/r - 2ac tan </r cos[I. 1892]

39. A conductor is formed by the larger segments of two spherescutting at right angles; A, B are the centres of the two spheres, O anyexternal point; D, E are the images of O in the spheres with centres atB, A respectively and AD, BE meet in F. If the conductor be unin-sulated and any charge be placed at G, shew that the charge inducedon the sphere bears to the inducing charge the ratio

-^{AABD) -^(AABE) +^(AABF) -.^/(AABO),where AABC represents the area of the triangle ABO. [I. 1893]

40. A point charge is placed in front of the plane face of an un-limited mass of dielectric; find the potential function, and sketch thelines of force. Shew that the number of unit tubes of induction whichcross the face is K/(K + 1) of those emitted by the charge. [I. 1910]

41. Shew that the capacity of a spherical conductor of radius a is

increased in the ratio 1:1+ ^—= 7i=: by the presence of a large mass ofJ\. + 1 Zo

dielectric with a plane face, at a distance b from the centre of thesphere, if a/b is so small that its square may be neglected.

[St John's Coll. 1907]42. A long fine straight wire carrying a charge e per unit length is

parallel to and at a distance / (> a) from the axis of an insulateduncharged long conducting cylinder whose cross-section is a circle ofradius a. Shew that the attraction per unit length between the cylinderand the wire is 2e2a2//(/2 — a2). Also shew that the surface density onthe cylinder vanishes at the points of contact of tangents drawn frompoints on the wire. [I. 1928]

43. A charged sphere of radius a, with its centre at height h^ abovethe ground, large compared with a, is at potential Fx: determineapproximately the potential to which it will attain, when it is raisedto a height h3. [M. T. 1917]

EXAMPLES 147

44. The polar equation of a surface of revolution is(r2 + c2 - 2cr cos 6) (r - 6)2 = o2r2.

If a conductor be made of this form, insulated and electrified, find thelaw of distribution of electricity on the conductor and the potentialdue to it at any point of external space. [I. 1897]

ANSWERS

5 « fi °3 )' 2*\ tJ ' \ (

6. - iO , -fQ. 10. ea2(3/-a)//(/-a)2.

12. eaj^^^-jj. 17. e*ca/(c* - a*)*.

24. e/c. 26. cos-H^

44. Let 4 be the point (c, 0) and r' denote distance from A. When thepotential of the conductor is unity, the potential at any point ofexternal space is ajr' + bjr; and the electrification at a point Pon the conductor is given by 4T7O-=resultant of a/r'3 along APand 6/r2 along OP.

Chapter VII

ELECTRIC CURRENTS

7*1. Current strength. When two conductors at differentpotentials are connected by a wire there is a gradient ofpotential along the wire and equilibrium no longer exists, butthere is a transfer of electricity called a current in the wirefrom one conductor to the other. The strength of thecurrent in the wire may be measured by the rate of increaseof the positive charge on one of the conductors, say G — dQ\it,where C is the current and Q the charge.

Such a current would be transient unless the conductorsare connected with the terminals of a battery or 'cell' bywhich they are maintained at different potentials.

The presence of the current is manifested by heating of thewire and the creation of a magnetic field in its neighbourhood.

The field of a steady current is said to be a stationary field,because, although there is a flow of electricity, the conditionsremain the same at all points as time progresses. We mustpoint out however that directly the two conductors aboveare connected the field ceases to be electrostatic; we havetherefore no longer any ground for asserting the existence ofa single-valued potential function, and the fact that the fieldof a steady current is derivable from a potential function(not single-valued) is a new empirical assumption.

In the electron theory metals contain a number of electronsmoving freely between the molecules. If there be no externalelectric field, the velocities of the free electrons are in randomdirections and there is no tendency for a collective motion inone direction rather than another. But an electric field directsthe flow and establishes a drift of electrons in the direction ofthe field, though held in restraint by collision of electrons withone another and with metal molecules.

For our purposes it is sufficient to measure the current in theway indicated above; or, by considering a cross-section of the

7-12] THE VOLTAIC CELL 149

wire and stating that if, in a short time 8t, n units of positive elec-tricity cross the section in one direction and n' units of negativeelectricity cross it in the opposite direction, then the strengthof the current is (n+n')/8t, so that it is immaterial whetherwe regard the current as a flow of positive electricity in onedirection or a flow of negative electricity in the opposite direc-tion, or partly one and partly the other.

7*11. The voltaic cell. The simplest form of cell consistsof a plate of zinc and a plate of copper immersed in a vessel ofdilute sulphuric acid, with copper wires attached to the plates.When the wires are joined the 'circuit is completed' and acurrent flows from the zinc to the copper in the liquid and fromthe copper to the zinc through the wire. Achemical action takes place in the cell; the acidattacks the zinc plate, zinc sulphate beingformed and hydrogen liberated from the acid.But the tiny bubbles of hydrogen are carriedacross by the current and adhere to the copperplate where with their positive charges placedopposite to negative charges on the zincplate they act like a parallel plate condenser and producea field opposed to the current in the liquid and consequentlyreduce the effectiveness of the cell by increasing what is knownas its resistance. This process is called the polarization of thecell; its effect is to reduce the potential difference of the ter-minals A, B, so that such a cell is of no use when a permanentpotential difference is required, and more elaborate forms ofcell have been devised to get rid of polarization.

7'12. Electromotive force. We made reference in 2*411to the possibility of a difference of potentialbetween two substancesin contact. Considera closed ring—fig. (i)—formed of threemetals A, B, C joined in sections. In equi-librium the potential is constant throughouteach metal, but there may be differencesat the contacts which may be denoted by

B

Zn

H2SO4

A

Cu

150 ELECTROMOTIVE FORCE [7-12-Such a discontinuity in the potential is associated in a

static field with the existence of a 'double sheet' of electricityover the surfaces in contact, i.e. surface charges of opposite

The contact differences will be such that in equilibrium

4>A = <f>C ~ ^ A

But if the circuit is formed of two metals A, B, and an acid Cas in a voltaic cell, there is not equilibriumbut a flow of electricity. Consider this arrange-ment when A and B are separated by an airgap, and suppose that a small piece of metalA' of the same kind as A has been soldered onto B so that the gap is an air gap between twopieces of the same metal A, A' (fig. (ii)). Thenwe have

A',

B

C

A

00

This potential difference <f>A. — <f>A can be tested by an electro-meter. We define the electromotive force of the cell as thispotential difference when the cell is ' open' and the terminalsare of the same substance; and we see that it is the sum of thecontact differences of potential of the elements which composethe cell.

7'121. It should be noticed that contact difference of potentialdepends upon temperature. Thus if we havea ring formed of two metals A and B andthe junctions are kept at different tempera-tures t, t', there will be a current round thering since [A<f>B]t*[-B<t>A]f

Also the passage of a current through ajunction is always accompanied by theproduction or absorption of heat, in such away that if heat is produced when thecurrent passes in one way it will be absorbedwhen the current is reversed.

7-2. Electrolysis. In addition to the production of heatand of a magnetic field, the passage of a current through

7-2] ELECTROLYSIS 151

various substances in solution frequently results in a chemicaldecomposition and this process is called electrolysis.

When a current passes through a liquid from one metalplate to another, the plates are called electrodes, that ofhigher potential is called the anode, and that of lower potentialthe cathode. The substance which is decomposed is called anelectrolyte and the constituents into which it is decomposedare called ions; that which is deposited at the anode is theanion, and that which is deposited at the cathode is thecation.

When a current passes between platinum electrodes throughacidulated water, the water is decomposed into oxygen andhydrogen. In the process molecules of water in contact withthe anode are decomposed, the oxygen atoms form into mole-cules which collect on the anode and the hydrogen atoms attackneighbouring molecules of water seizing and combining withtheir oxygen atoms and liberating hydrogen atoms, which inturn attack their neighbours. This process takes place acrossthe whole space between the electrodes and in the end liberatedhydrogen atoms assemble on the cathode.

The Laws of Electrolysis are due to Faraday who alsooriginated the terms we have just defined. Faraday foundthat there is always a constant ratio between the amount ofan ion deposited and the quantity of electricity which passes.

To every substance a number is attached called its electro-chemical equivalent. I t is the number of grams of the substancedeposited during the passage of a unit of electricity. The Laws ofElectrolysis may be summarized in the statement that thenumber of grams of a substance deposited at an electrode isequal to the number of units of electricity which pass through theelectrolyte multiplied by the electrochemical equivalent of thesubstance. In the case of elementary ions the electrochemicalequivalent is the atomic weight divided by the valency, and,in the case of compound ions, it is the molecular weightdivided by the valency.

An apparatus in which the quantity of a cation depositedcan be measured—e.g. the weight of silver deposited from asolution of silver nitrate—may be used to measure the quan-

152 OHM'S LAW [7-2-

tity of electricity which passes in a given time. Such anapparatus is called a voltameter.

7*3. Ohm's Law. The current between two points of a circuitvaries directly as (he electromotive force and inversely as theresistance.

We have already defined the electromotive force of a cell asthe difference of potential of its terminals on open circuit, andwe may in the same way speak of the electromotive forcebetween two points as the difference of potential betweenthem. I t also represents the work that would be done, orthe energy used up, as a unit of electricity passes from onepoint to the other. Ohm's Law* is really a definition of theresistance of a conductor between two points; for it makes theresistance the ratio of the electromotive force to the currentwhich it produces.

It must be observed that electromotive force is not a goodname for what it is intended to connote, because its physicaldimensions are not those of force but of potential; at the sametime it is the physical entity which results in the motion ofelectricity.

If 0 is the current from P to Q, R the resistance and <f>P, <f>Q

denote the potentials of P and Q, Ohm's Law may be expressedby the equation j . j .

(1),

assuming there to be no source of energy in the conductorbetween P and Q.

We observe that if an electric field has a single-valuedpotential function j>, then in any closed circuit for which Qcoincides with P we have <f>Q = <j>p, so that from (1) there canbe no current in the circuit. Hence the existence of a currentimplies a source of energy such as a cell or battery.

Let A, B be the terminals of a cell [fig. 7*11], R the resistanceof the wire joining A, B, r the resistance of the liquid in thecell and O the current. Then if E is the electromotive force ofthe cell, since it drives the current C through a total resistance

ave E = CR + Cr (2).

* G. S. Ohm (1787-1864), German physicist.

7-31] OHM'S LAW 153

But from (1), when the circuit is closed, we have(3);

and from 7*12 E is the value of </>A — <f>B when the circuit isopen. It follows from (2) that the potential difference of theterminals is less when the circuit p

is closed than when it is open bythe amount Cr, needed to drivethe current through the liquidin the cell.

Now let P, Q denote any twopoints on the closed circuit andconsider the flow from P to Q in the sense PBAQ. If rlt r2

denote the resistances of PB, AQ, we have

and cf,A-cf>Q = Cr2;so that <f>P-cl>Q + E=C(r1+r + r2) (4).

Consequently we may express Ohm's Law in this rather moregeneral form: / / P, Q are any two points connected by a wire;<f>P, <f>Q the potentials at P, Q; EPQ the internal electromotive forcein PQ, i.e. the electromotive force of any battery located in PQand driving current in the sense from P to Q; GPQ the currentfrom PtoQ, and RPQ the total resistance of the wire including thatof the battery, then

==GPQRPQ (5).

7-31. Resistance of a set of conductors, (i) When arrangedin series.

Let AB, BC, CD,... LM be the conductors, rx, r2,... rn theirresistances and C the current flowing from A to M when the

B M

potentials of the ends of the conductors are </>A, $B,... <j>M; andlet R be the whole resistance between A and M. Then byOhm's Law

154 SETS OF CONDUCTORS [7-31-

so that by addition

B u t j>A-<}>M ,

therefore B=r1 + r2 + ...+rn (1).

(ii) When arranged in parallel.Let ALB, AMB, ANB, ... be the conductors, r1; r2, ... rn

their resistances and Cx, C2,... Gn the currents passing throughthem when </>A, <f>B are the potentials at A, B. Then if C is thetotal current from A to B and R the total resistance,

But C=O1 + Ca + ... + On,

therefore -= = — H (-... + — (2).K r± r2 rn

It follows that if n. conductors of the same resistance r arejoined in series the resistance of the compound conductor is nr,but if they are joined in parallel the resistance of the com-pound conductor is rjn.

7-32. Specific resistance and conductivity. The specificresistance of a substance is the resistance of a unit cube of thesubstance to a current passing parallel to one of its edges.

If s denotes the specific resistance of a substance, the resist-ance of a wire of length I whose cross-section is the unit of areais Is, by 7*31 (i). And if the cross-section be a units of area, thewire may be regarded as made up of a wires of unit cross-section arranged in parallel, so that, by 7*31 (ii), the resistanceis ts/oc.

Conductivity. The reciprocal of the specific resistance of asubstance is called its conductivity.

7'34] UNITS 155

7-33. Units. In the form in which we have stated it, Ohm'sLaw requires that a unit of electromotive force should producea unit current through a unit resistance. The names of thepractical units are the volt, the ampere* and the ohm. Thusan electromotive force of one volt produces a current of oneampere through a resistance of one ohm.

In terms of the absolute electrostatic units defined in 3-9:

1 volt = J Q- absolute electrostatic c.G.s. unit of E.M.F.,

1 ampere = 3xlO9 absolute electrostatic c.G.s. units ofcurrent,

and 1 ohm = 3~2 x 10~u absolute electrostatic c.G.s. unit ofresistance.

But the reader must note that these physical quantitiesoccur most often in connection with electromagnetic fields sothat it is important to know their values in terms of absoluteelectromagnetic units, the basis of which we have not yetexplained. The values are as follows:

1 volt = 108 absolute electromagnetic c.G.s. units of E.M.F.,

1 ampere = 10"1 absolute electromagnetic c.G.s. unit ofcurrent,

and 1 ohm = 109 absolute electromagnetic c.G.s. units ofresistance.

7*34. Arrangement of cells. Cells are said to be ammgred in seneawhen the zinc of the first is joined to the copper of the second, the zincof the second to the copper of the third and so on. The electromotiveforce of such an arrangement being the sum of all the contact differ-ences of potential is clearly the sum of the electromotive forces of theseparate cells; and as in 7*31 the resistance is the sum of the separateresistances.

If on the other hand the cells are joined in parallel, i.e. if all the zincplates of a number of like cells are joined and also all the copper plates,this is merely equivalent to increasing the size of the plates withoutaltering the nature of the cell and thus will not affect its electromotiveforce; but by 7*31 if there are n cells their resistance in parallel will be

reduced to - th of that of a single cell.nLet mn equal cells be arranged in m parallel sets each containing n

cells in series. Let r be the resistance of a cell, E its electromotive forceand R the external resistance which completes the circuit.

* After A. M. Ampere (1775-1836), French physicist.

156 JOULE'S LAW [7-34-The E.M.F. of each series of n cells is nE and its resistance nr, and

when m such series are joined in parallel the E.M.J1. remains nE and theresistance becomes nr/m. Hence if C be the total current, we have

nE

This may be written

C=

VV m~V ~n) + \J \mn

Hence the current will be a maximum when — = — , or when tir/m = B,m n

i.e. when the internal resistance is equal to the external resistance,provided this arrangement is possible; and the maximum value of thecurrent is seen to be

Cmax ==

7-4. Joule's Law.* Let R be the resistance of a lengthAB of wire, E the electromotive force from A to B and G thecurrent. Then in time ht an amount CSt of electricity is drivenfrom A to B by the potential difference E, so that an amountof work EGSt is done, or BC28t units of work. This work istransformed into heat and is made manifest in the heating ofthe wire. The rate of heat production is therefore, in mechanicalunits, RC2 per unit time; and this is true for any portion of thecircuit provided that R denotes its resistance and C the current.

7*41. The heating effect, to which reference was made in7-121, when a current passes through the junction of twometals is known as the Peltier effect.^ In this case the quantityof heat emitted or absorbed is directly proportional to thecurrent, and not to the square of the current as in the Jouleeffect.

7-5. Kirchhoffs Laws % for steady currents in a networkof wires.

1. In a state of steady flow the algebraical sum of the currentswhich meet at a junction of wires is zero.

* J. P. Joule (1818-1889), English physicist,t J. C. A. Peltier (1785-1846), French physicist.t G. R. Kirchhoff (1824r-1887), German physicist.

7-51] KIRCHHOFF'S LAWS 157

This law means that the term 'steady flow' implies thatelectricity does not accumulate at any point of the network.

2. In any closed circuit in a network the algebraical sum of theproducts of the resistance and current in each conductor is equalto the algebraical sum of the electro-motive forces of the batteries in thecircuit.

Let L, M, N,... K be the junc-tions in the circuit. Adopting thenotation of 7-3, viz. RPQ denotesthe resistance of the conductorPQ including that of any batteryit may contain, CPQ is the currentfrom P to Q, EPQ the electromotive force of the batterydriving current from P to Q, and <f>P, (f>Q are the potentialsat P and Q, we have, as in 7*3 (5)

so that by addition -vGB^ Y.E (1)

It follows that, in any closed circuit in the network in whichthere is no battery, we must have

0 (2).

7*51. E x a m p l e s , (i) A system, of conducting wires, all of equalresistance, form the edges of a cube. Find the currents carried by the wiresif a battery is connected without extra resistance to two adjacent verticesof the cube. [M. T. 1930]

Let E be the electromotive force and B the resistance of the battery,r the resistance of each edge of the cube. We have to find the currentsin twelve wires, but the solution of the problem will become tediousif we use more unknown quantities than is necessary; so we makewhat use we can of Kirchhoff's first law from the outset and also makeuse of symmetry.

Referring to the diagram let the current enter the cube at A and letx denote the current in AF. By symmetry the currents in AB, ADmust be equal, so we denote them by y. We also observe that if the

158 EXAMPLES ON K I R O H H O P F ' S LAWS [7-51

battery were reversed, the currents would also be reversed, so that thecurrents in OF and KF will also be y. At B the current in AB divides,let 2 denote the current in BO, then y — z must be the current in BG.In like manner the currents in DK, DG are z and y — z. At C thecurrents in BG, DG unite and give a current 2 {y — z) in CH, whichdivides at H into y — z in both HO and HK. Then the currents z in BOandy—zin.HOunite to give y in GJP1,and similarly y ia.KF as previouslystated.

Also the total current through the battery is the current whichdivides at A, i.e. x + 2y. Therefore, from the circuit formed of thebattery and AF, j? ^_,_ a,,\ a.™ = rc (1).

2{y-z)

y K

From the square A B OF, the resistances being equal and there beingno battery in the circuit,

x — y—z—y = 0 or x — 2y—2 = 0 (2).Similarly, from the square BOHO

2-4(2/-z) = 0 or 4y = 6z (3).

Hencey_z _ x _ E5~4~"l4~24fi+14r .(4).

And thence the currents in each of the twelve wires can be writtendown.

(ii) Two long straight parallel wires are joined by cross wires of thesame material at equal distances, forming an infinite ladder of equalsquares, the resistance of a side of a square being r. A current enters andleaves the network at the ends A, B of one of the cross wires. If the currentsin successive segments of one of the long wires, measured from A, areut, Ma, us shew that

Shew also that un = u1(2— -v/3)"-1, and that the equivalent resistanceof the network is r/^/3. [M. T. 1932]

Let O be the current which enters at A and leaves at B, x the currentin AB, vltvt,vt, ... the currents in successive cross wires. We notice

7-51] EXAMPLES ON KIECHHOFF'S LAWS 159

that there is symmetry about the line AB, and also a symmetry madeevident by making the current enter at B and leave at A.

From Kirchhoff's first law

«n = V l + »«. a n d «n+l = ««+« + ««+liand from the second law

«»+l + ««+i + «n+l - «» = 0.Therefore, by eliminating vn and un + 1 , we get

.(1).

To solve this linear difference equation, we note that the roots of thequadratic z2 — 4z + 1 = 0 are 2 + y/3 and find that (1) has a solution ofthe form un-A (2+V3) n + -B(2—v/3)" (2),

where A and B are constants to be determined.

But as n -> oo it is evident that un -> 0, so that A must be zero, and

Therefore u1

and Mn

Again, from the central squares, we have0 = x + 2u1

(3).

(4),

and

By eliminating vt we get

x = 3MX — «2 >

and from (3) this gives x = 3% — ut (2 —\/3)

Therefore from (4), by eliminating ut, we get

But if B is the total resistance of the network

so that, since 0=V3x, therefore B = r/Vs.

160 WHEATSTONB S BRIDGE [7-52-7*52. When a network consists of a number of meshes it is often

convenient to use symbols for the ctirrents round the meshes insteadof currents in the separate wires. Then provided the symbols denotecurrents travelling round the meshes in the same sense, the actualcurrent in any wire which is common to two meshes is the differencebetween the currents in these meshes.

Thus let A B O be a triangular mesh, the resistances of whose sidesBO, OA, AB are a, b, c, and let x denote the current in the mesh ABO

and u, v, w those in adjacent meshes as in the figure. The actual cur-rents in BO, OA, AB are then x — u, x — v, x — w and Kirchhoff's secondlaw gives a(x- -v) + c(x-w)=:E,where E is the total electromotive force in the wires of the mesh; andthis may be written

(a + 6 + c) x — cm — bv — cw = E;

i.e. the mesh current is multiplied by the whole resistance and theproducts of adjacent mesh currents and corresponding resistances aresubtracted. The only advantage of this method is that it enables us towrite down equations in a form immediately ready for solution.

We shall make use of the method in the following article.

7#6. Wheatstone's bridge.* This is an apparatus used forcomparing resistances. It consists of four conductors BA, BD,GA, CD connected as in the figure, the points B, C beingconnected to the terminals of a battery and the points A, Dto the terminals of a galvanometer—an instrument whichindicates and measures a current passing through it.

Let the resistances of BC, CA, AB, DA, DB, DC, includingthose of the battery and galvanometer in BC and DA, be

* Sir Charles Wheatstone (1802-1875).

7-6] WHEATSTONE'S BRIDGE 161

a, b, c, a, )8, y. Let x, y, z denote the currents round the meshesDBG, DCA, DAB, and E the electromotive force of thebattery.

Then by applying Kirchhoff's second law to the threemeshes in turn, we have

(a + p + y)x-yy-pz = E,

and

These equations may then be solved to find the currents.The current through the galvanometer is y — z; where

a+p+y —y— y b + y + cf.

-J8 — a

- p— a

C + O. + P

y= P + y-y

-P

E0

0

-P— a

C + OL + P

or = E (<x.B+ By + y<x. + cy) (1),

where A stands for the determinant on the left. Similarly

(2).

Hence y = z, or there will be no current through the galvano-meter if , o ,„.

bp = cy (3).Further, we notice that A is symmetrical in the resistances,

and the expression on the right of (1) for y, the current in OA,does not contain the resistance of BG or CA, so that if thebattery were transferred to CA, this same equation wouldthen determine the current in BC. This is a particular case ofa general theorem which will be proved later.

162 WHEATSTONE'S B B I D G E [7-6-

When a battery in BC produces no current in DA, these arecalled conjugate conductors and the bridge is said to be balanced.

The condition (3) for the balancing of the bridge may beobtained independently as follows:

BIf no current flows along AD, the points A and D must be at

the same potential, so that if u, v denote the currents long BA Gand BDG, we have

where <f>A = <f>D. Hence

cu = flv and bu = yv,

so that 6/J = cy.

7*61. A Wheatstone's bridge may be used for obtaining apotential difference which is a small fraction of that of thebattery. For, reverting to the general case

^ - & > = « ( y - * )= aJfi?(cy-&j3)/A, from 7-6(1), (2),

and by adjusting the resistances of the arms so as to makecy — bfi small, the ratio of<f>A — </>DtoE can be made as small aswe please.

7*62. By using standard coils of known resistance, andadjustable arrangements for varying the resistances of threeof the four conductors BA, BD, CA, CD until a balance isattained, it is clear that the relation 6j8 = cy enables a fourthunknown resistance to be determined.

7-7] CURRENTS IN A NETWORK 163

7-7. Currents in a network. Reciprocal Theorems. LetAx, A2, ... An be the n junctions of a network of wires. Wecan include the possibility of there being two or more wiresdirectly joining Ap, Aq by supposing another and differentA to be inserted between Ap and Aq on each of these wires.

Let Bpq = the resistance of ApAa,Epq = the electromotive force of the battery in the wire

ApAa driving current from Ap to Aq,Cpq = the current along ApAq,<$>p = the potential at Ap,Qp = the current, if any, entering the network at Ap;

and let dashes be used to denote any other set of values of thelast four quantities. The resistances of course are invariable.

Then as in 7-3 (5)

Opq Bpq=Epq + <f>p-<f>q

a n d Ka + i>p'-h'=G'pAa-Multiply these equations together, cancel out the factor

Rpq and add similar equations for all the wires, giving

where 2 means a summation over all different pairs p, q,{p¥=q) such that Ap, Aq are directly joined by a wire; i.e.there is no further junction on the wire between Ap and Ag.And we may suppose the summation extended to all differentpairs p, q, (p¥=q) if we use the conventions Opq = 0, O'pa = 0when Ap, Aq are not joined by a wire.

Now

ButQp=C

Therefore

164 RECIPROCAL THEOREMS [7-7-

Similarly SG^ (*,-*„) = 2 * , «,',and therefore, from (1),

^E'pgCpq + ^p'Qp = %Epq C'm + 2 # , Qp' ...(2).

We now consider two cases:(i) When no currents enter the network at the junctions, the Q's

are all zero and the theorem (2) becomes

TiE'pqCpq = ^EpqG'pq (3).

Now consider two states:(a) when all the E's are zero except Epa which is equal to E;

and (]8) when all the E"& are zero except E'rs which is equalto E; then (3) gives

ECrs = EO'pa or Grs=C'pq,

which means that the current in ArAs due to an electromotiveforce in ApAq is equal to the current in ApAq due to an equalelectromotive force in ArA8. We had a special case of thistheorem in 7%6.

(ii) When no batteries are present, i.e. when all E's are zero,the theorem (2) becomes

X<t>PQP = X<f>PQP (4).Now consider two states:(a) a current Q enters the network at Ap and leaves it at

Aq, i.e. Qp= Q, Qq = — Q and all other Q's are zero; (jS) a cur-rent Q enters the network at Ar and leaves it at As, i.e. Qr' = Q,Qs'= —Q and all other Q"& are zero; then (4) gives

which means that the potential difference between Ap, Aq

when a current enters at Ar and leaves at A3 is equal to thepotential difference between lAr, As when the same currententers at Ap and leaves at Aq.

7-71. Minimum rate of heat production, (i) To shew thatwhen there is a steady flow in a network of wires containing nobatteries, of all possible distributions of current in accordancewith Kirchhojf's first law that one which is also in accordancewith the second law gives the minimum rate of heat production.

7-71] MINIMUM BATE OF HEAT PRODUCTION 165

With the notation of 7*7 let Cpq denote the true current inthe conductor of resistance Rpq connecting the junctionsAp, Aq; then by 7-4 the rate of heat production is

H=YtRmG\a (1),

where the summation may be taken over every pair p, q,(p¥=q), if we make a convention that Cm = 0 when Ap, Aq

are not directly joined by a wire.In any other distribution of current in steady flow let the

current in ApAq be Xpg, and let Xpq = Cpq + epq with a likeconvention. Then the corresponding rate of heat production is

orNow 2

H' = H + 2^R

-fa) + e13(fa-<f>.+ e23 (</>* —<

• o e + y\i

^3) + e24(^2~ 4

? e2

) + . . .+e l n (^.4) + . . . + €2re(

(2).

>n (enl + ««2 + • • • + 6«, n-l)= 0,

from Kirchhoff's first law, since the total additional flow awayfrom each junction must vanish. Therefore (2) reduces to

and as the terms in the expression ^Spqe2

pq are positive,therefore

(ii) To shew that when there are batteries in the wires and, the,circumstances are in other respects the same as above, then theactual distribution of currents is such as to make H — 2^Epq Cpq

a minimum, where H is the rate of heat production.

Let F = H

or F = XRpqC\q-2XEpqCpq (3).Then if Cpa + em denote the current in ApAq in any other

state of continuous flow, the corresponding function is

F' = 2£«r (Om + eM)2 - 2^Epq (Gpq + epq)

epq . . . (4) .

166 TELEGRAPH WIRE WITH A ' FAULT' [7-71-

But S ( 5 M C « , - » M ) « K = S W P - ^ ) « M

= 0, as above.

Therefore F' = F + ^Epae\g,and F < F'.

7*72. Example. We will apply the minimum method of 7*71 to7'51 (i). The distribution of currents depicted in the figure is inaccordance "with Kirchhoff's first law, so the function to be made aminimum is

We have therefore

and

~ = 2R (x + 2y) + 2rx - 2E=0,

8F8y =

8F_~Bz:=

These equations are equivalent to 7*51 (i) (1), (2) and (3) and there-fore lead to the same solution.

7*8. Telegraph wire with a'fault'. A current is sent alonga telegraph wire by a battery, one terminal of which is con-nected to the earth; and in the same way one terminal of thegalvanometer at the receiving end is also put to earth, so thatthe earth forms the return half of the circuit and its resistanceis negligible. A leakage from the wire is called a fault. Examplesof wires with specified faults are solved by simple applicationof Kirchhoff's laws, others may be solved by expressing thecurrents as linear functions of the potentials.

(i) A fault of given resistance develops in a telegraph wire. Shew that,with a given battery transmitting signals, the current received is least whenthe fault is at the middle of the wire.

M

r

C-C

U

R-a:

W

Let B be the resistance of the wire LN between the battery and thegalvanometer, r the resistance of the fault at M, x that of the length

7-8] TELEGRAPH WIRE WITH A 'FAULT' 167

LM of the wire, so that R — x is the resistance of the remainder MN.Let E be the electromotive force of the battery, 0 the current trans-mitted and C" that received, a current 0—0' leaking through the fault.

From the circuit LMUV in the figure, we have

Gx + (O-O')r=E;

and from the circuit MNWU

G'(R-x)-(C-C')r = 0.

By eliminating O we get

{R(r+x)-xi}C'-Er,

so that C is least when Rr + Rx — x* is greatest, i.e. when

Rr+iR*-(iR-x)* is greatest,

and that is when x=$R, or when the fault is in the middle of the wire.

(ii) A, B are the ends of a telegraph wire with a number of faults. R and8 are the resistances to a current from A when Bis to earth and insulatedrespectively ; and R', 8' are the corresponding resistances to a current sentfrom B. Prove that RS'-R'S. [I. 1894]

Let C be the current emitted from A and C the current receivedat B; and let <f>A, <f>B denote the potentials of A and B. Then the currentsmust be linear functions of the potentials, say

C=k<f,A + l<t,B (1)and O' = m^>A+ntj>B (2),

where k, I, m, n are constants which depend only on resistances, sothat the same relations hold whichever way the current passes.

Then we have the following cases:

(a) B to earth; resistance is R; so that <f>B = 0,<f>A = RO and (1) givesO = h RG, so that h = 1/R. •

(b) B insulated, so that 0" = 0; and resistance is S, so that $A = SO,the whole current going to earth through the faults.

Hence (1) gives 0= g G+l<f>B

and (2) gives Q = mSO+n<j>B;

therefore, by eliminating the ratio O:<f>B, we get

(S-R)n = lmRS (3).

Next let O' denote the current emitted from B and C the currentreceived at A; then

(c) A to earth; resistance is R'; so that <f>A = 0,<f>B = R'C and (2) givesG'=nR'O', so that n=l/R'.

(d) A insulated, so that (7=0; and resistance is 8', so that <£B = S'C.

168 EXAMPLES [7-8-

Hence (1) gives 0 = ^ + IS'C

and (2) gives C = m$A + ;

therefore by eliminating the ratio 0":<j>j,, we get

(4).

Then from (3) and (4), and putting n = •=-,, we get

p = -§, or B8'=R'8.

7*9. Examples, (i) A current is led to and from an electric lamp ata distance of 40 feet by a pair of wires, each of resistance 13 ohms per1000 yards. If the difference of potential across the lamp is 25 volts andthe current supplies energy to the lamp at the rate of 30 watts, at what ratein watts is energy lost in the leads ? {A watt is the rate at which work isdone by a current of 1 ampi/re working through 1 volt.) [M. T. 1909]

The potential difference across the lamp is 25 volts, and energy isbeing supplied to the lamp at the rate of 30 watts, therefore thecurrent is f% or J ampdres; and this is the current in the leads.

But the leads consist of 80 feet of wire of resistance 13 ohms per1000 yards, i.e. of resistance ff ohm.

The rate of heat production RC* in the leads is therefore f f x § $ = 0-5watt, nearly.

(ii) A galvanometer has a resistance of 1 ohm and each graduationrepresents 1 milliampe're. Find the resistance with which it must beshunted to make it read amp&res on the same scale. It is required to use thesame instrument as a voltmeter. Find the least resistance which must beconnected in series with it, to make its readings accurate within 1 per cent,if the resistance of the batteries whose voltage it is required to measurenever exceeds 2 ohms but is otherwise unknown. On what scale will it thenread volts? [M. T. 1921]

The theory of galvanometers will be considered in a later chapter.For present purposes it is sufficient to know that a galvanometer is an in-strument which measures the strengthof a current passing through it: it isoften called an ammeter in the sense thatit measures the number of amperes ofcurrent; but if the resistance betweenthe terminals is known, then the in-strument can clearly be used to measurethe difference of potential of the terminals and in this sense it con-stitutes a voltmeter. In order to be able to use the same instrument formeasuring wide ranges of current or of electromotive force otherresistances are coupled with it, in the former case in parallel and in the

7-9] EXAMPLES 169

latter case in series. A resistance fixed across the terminals in parallelwith the instrument is called a shunt. This example illustrates theuse of a galvanometer both as an ammeter and as a voltmeter.

Let A, B be the terminals of the galvanometer and r its resistance.Let A, B be joined by a ' shunt' of resistance R. Let G be the currentto be measured and suppose that a current x passes through thegalvanometer and G — x through the shunt. Then, by Kirchhoff'ssecond law, R(G-x) = rx (1).

In the example above r = 1 ohm, and the passage of a milliamperethrough the instrument is to measure the passage of an ampere in thecircuit as a whole, so that if G = 1 then x = y^u- By substituting thesedata in (1), we find that R = v fo ohm, a shunt of small resistance takingthe greater part of the current.

When used as a voltmeter the insertion of a resistance in series withthe galvanometer serves to reduce the potential difference of theterminals, so that by the substitution of different resistances a widerrange of voltages can be measured by the same instrument.

Let E be the electromotive force of a battery in volts, r' its resistancein ohms, and S ohms a resistance joined in series with the battery andgalvanometer. Then

E=C(r + r' + S) (2),

where O is the current through the whole circuit.But r = l and 0<r '<2 , so to reduce the possible variation in

r' 1r + r' + S to 1 per cent, we must have .——r—H<TTTS> o r <S>99r' — 1,

1 +r + o 100where r' may be as great as 2, so that the least value of S is 197.

Then from (2) E = 200(7 approximately, and one graduation meansthat 0 = 1 0 ^ amp. and therefore E = J volt.

(iii) Current is supplied by a motor working at rate W to two circuitsinparallel, one of which contains accumulators of voltage Eo and resistancer on charge, and the other lamps of total resistance R. Prove that the currentthrough the lamps is

If r/R is very small explain why disconnecting the accumulators mayresult in damage to the lamp circuit. [M. T. 1929]

The work done by the motor is partly converted into heat in thecircuits and the lamps, and partly used in driving current through theaccumulators which are absorbing energy at the rate Eo for every unitof current which passes.

If A, B are the common terminals of the circuits, <j>^, <f>B theirpotentials and x, y the currents through the lamps and the accumula-tors, we have , I !!„

170 EXAMPLES [7-9

for the voltage of the accumulators must be regarded as opposed tothe passage of the current y; and

(2).

By substituting for y in terms of a; from (1) in (2), we get a quadratict0TX

giving 2(R+r)and since x is positive the positive sign must be taken.

If r/R is very small, it is evident from (1) that the lamps only receivea small portion of the current from the motor; and if the accumulatorsare disconnected, the whole current will pass through the lamps andthey may break as the result.

(iv) Two condensers having capacities Gt and G3 are arranged asshewn, and O is a galvanometer. The resistances R3 and Rt are adjustedso that on closing the battery circuit not even a transient current flowsthrough O. Treating the transient charging currents as steady, shew that

RzC^RtCt. [M. T. 1928]

Let us consider the effect of inserting a condenser of capacity C in awire LN in which a current x is flowing from L to N. If the wire meets

7-9] EXAMPLES 171

the condenser plates at X and T and B is its total resistance, by Ohm's

fa — fa +4>Y— <t>N— Rx-But if Q is the charge on the positive plate of the condenser,

so that Rx = <f,L-<l>N- Q/C.

Hence the presence of the condenser diminishes the electromotiveforce in the wire by QjG. And the current x is the rate at which Q isincreasing, or x = Q.

In the particular problem there is no current in MN so that <fof = i£y.Let x, y denote the currents along LNK and LMK, and let Bx, Bt bethe resistances of LN and LM.

Then, as shewn above,

B1x = <j>L-^,N-QilG1 and B2y = </>I/-<l>M-QiICi,where Qt, Qa are the positive charges on the condenser plates. Also

Hence, since <j>M = <j>N, we have< <j^ a n d

But in this problem the currents are regarded as steady so that thecharges which accumulate are proportional to the currents, and wemay write Q1 = ta. Q% = ty.

Then by eliminating x and y, we get

and this relation must hold throughout the whole process of chargingand is therefore true for all values of t. Consequently we must have

EXAMPLES

1. If the resistance of a copper wire 1 metre long and weighing1 gram be 0-15 ohm, find the length of a wire of the same materialwhich weighs a million grams and whose resistance is 6000 ohms.

[I. 1891]

2. The resistance of 1 mile of copper wire whose diameter was0-065 inch was found to be 15-6 ohms. The resistance of a wire of purecopper 1 foot long and 0-001 inch in diameter is 9-94 ohms. Comparethe specific resistance of the copper used in the first wire with that ofpure copper. [I. 1905]

3. Two cells whose E.M.F.'S are 2 and 3 volts and internal resistances2 ohms are connected in parallel with a resistance of 1 ohm. Find thecurrents through the external resistance and each cell. [M. T. 1908]

172 EXAMPLES

4. Three cells, each of E.M.F. 1-8 volts and internal resistance 1-1ohms, are used to send a current through a resistance of 1 ohm. Findwhether the current is greater when the cells are joined in series or inparallel. [M. T. 1915]

6. Find the least number of Grove's cells, each having an E.M.F. of1-87 volts and a resistance of 0-17 ohm, which will send a current of16 amperes through a resistance of 1-5 ohms, and shew how the cellsshould be arranged. [M. T. 1919]

6. Shew that the least number of Leclanch6 cells (E.M.F. 1-55 volts,internal resistance 0-7 ohm) by which an incandescent lamp requiringa current of 2 amperes and a potential difference of 10 volts can beworked is 24. [M. T. 1893]

7. A battery of mn equal cells is such that when it is arranged in mparallel sets of n cells in series the maximum current G is produced fora given external circuit. Shew that when the cells are arranged in nparallel sets of m cells the current is 2mnC/(m2 + n2). [I. 1902]

8. A current of not less than 1 ampere has to be sent through a wirewhose resistance is 100 ohms by a number of cells (B.M.F. 1-5 volts,internal resistance 4 ohms). Shew that it can be done by using 715 cells,and find the current which passes. [I. 1894]

9. A certain kind of cell has a resistance of 10 ohms, and an electro-motive force of 0-85 of a volt. Shew that the greatest current whichcan be produced in a wire whose resistance is 22-5 ohms, by a batteryof five such cells, arranged in a single series of which any element iseither one cell or a set of cells in parallel, is exactly 0-06 of an ampere.

[M. T. 1901]

10. A certain kind of cell has a resistance of 8 ohms and an B.M.F.of 1-2 volts. Twelve cells are arranged in three elements, each elementconsisting of four cells in parallel, and the elements being arranged inseries. What current will such a battery produce in a wire whoseresistance is 20 ohms? [St John's Coll. 1906]

11. A battery whose internal resistance is 5 ohms is connected bya wire of resistance 20 ohms with a galvanometer whose resistance is80 ohms; the galvanometer is shunted with 20 ohms and the currentthrough it is 0-1 ampere. What is the B.M.F. of the battery? [I. 1910]

12. A battery, whose B.M.F. is 3 volts and internal resistance 8 ohms,is connected through a resistance of 92 ohms with a galvanometer ofresistance 80 ohms. What is the current through the galvanometer,and what would it be if the galvanometer were shunted with 20 ohms ?

[St John's Coll. 1907]

13. In an overhead-wire tramway the resistance of the wire is 0-3of an ohm per mile. A battery at one end gives 500 volts and the

EXAMPLES 173

current passes along the wire and through three tramcars, returningto the battery along a rail of negligible resistance. The cars are atdistances 1, 2, and 3 miles from the battery and each take a current of60 amperes from the circuit. Find the difference of potential betweenthe wire and the rail at the farthest car. [I. 1907]

14. ABC is the trolley wire of an electric tramway 2000 yards long.B is its middle point and O is the position of the generating station.Current is supplied along three feeders, OA, OB, OC, the resistancesof these being 0-35, 0-25, 0-10 ohm respectively. The resistance of thetrolley wire is 0-80 ohm. A car situated at a point 500 yards distantfrom A is taking 100 amperes from the trolley wire. Find the current ineach of the three feeders.

If the resistance of the return circuit is negligible, and the potentialdifference at the generating station is 500 volts, determine the potentialdifference at the car. [M. T. 1912]

15. It is required to light 40 electric lamps arranged in parallel.Each lamp needs a potential difference of 100 volts between its ter-minals and uses 0-16 ampere. If the resistance of the leads to thedynamo is 2 ohms, calculate the E.M.I1. required for the dynamo, theenergy lost per second in the leads, and that used in the lamps.

[M. T. 1922]

16. The points A, B, C, D are joined by five wires AB, BC, CD, DAand BD. The resistances in these wires are respectively 5, 5, 5, 3 and8 ohms. Find the equivalent resistance of the network for a currententering at A and leaving at C. [M. T. 1915]

17. The arms AB, AD, AC of a Wheatstone's bridge are formed ofwires of resistance 1 ohm, and BC, BD by wires of resistances 10 ohmsand 9 ohms respectively. The points A, D are also connected by a wireincluding a galvanometer of total resistance 9 ohms. If CD containsa battery of B.M.F. 1 volt and has a total resistance of 3 ohms, shew thatno current will pass along AB, and determine that through thegalvanometer. [Trinity Coll. 1898]

18. A cable AB, 50 miles in length, is known to have one fault andit is necessary to localize it. If the end A is attached to a battery andhas its potential maintained at 200 volts, while the other end B isinsulated, it is found that the potential of B when steady is 40 volts.Similarly when A is insulated the potential to which B must be raisedto give A a steady potential of 40 volts is 300 volts. Shew that thedistance of the fault from A is 19-05 miles. [Trinity Coll. 1896]

19. An electric current passes along a uniform cable from A to B.The resistance from A to B is B, the potential at A is V, and the cableis earthed at B. If, owing to a leak at some intermediate point X,

174 EXAMPLES

current of amount C reaches B when current of amount O is sent outfrom A, find the ratio of the length AX to the length AB.

Shew that the resistance of the leak is

(O_O')» (CfJ*~F)- P*- T - 1 9 3 1 ]

20. A cable AB of length I of uniform wire develops a leak at acertain point. To locate the fault two observations are made. Theresistance between A and the earth through the cable when B isearthed is found to be that of a length a of the wire, and that betweenB and the earth when A is earthed that of a length 6. Shew that thepoint atwhichthe fault exists divides AB in the ratio {a (I — b)/b (I — o)}i,and find the resistance of the leak. [M. T. 1934]

21. A telegraph wire joining two places A, B drops from one of itssupports at a place G and rests on another wire which is earthed at bothends. If A is the ratio of the current strength at A to that at B when thecurrent in AB is sent from A, and ft is the ratio when the current issent from B, shew that C divides AB in the ratio (/JT1— 1): A— 1.

[M. T. 1908]

22. AB is a telegraph wire. Faults whose resistances measured inlengths of wire are r and s develop at distances y + z and z from theendJB. Currents are sent from A. Shew that the difference of the resist-ances to the current from A according as £ is insulated and put to

earth is r^l(y+r+s)(y8 + yz + rz + sz + rs). [1.1922]

23. A, B, G are three stations on the same telegraph wire. Anoperator at A knows that there is a fault between A and B, and observesthat the current at A when he uses a given battery is i, i' or *" accordingas £ is insulated and G to earth, B to earth, or B and G both insulated.Shew that the distance of the fault from A may be obtained in the form

{/ca - K'6 + (6 - a)*(*a - K'6)4}/(/c - K')>

where AB = a, BG - b - a, K = J^r, and K' = .,^7,. [M. T. 1904]

24. A 'fault' occurs in a telephone wire, AB, due to accidentalconnection to earth at an unknown point C, by a conductor of appre-ciable, but unknown, resistance r. The wire AB is uniform and has aknown resistance R. Measurements of the resistance between the endA and earth are made by a Wheatstone bridge, (1) when the end B isinsulated, (2) when the end B is earthed, the results being Bt and B2

respectively. Shew that, if the resistance of the earth is negligible,

AC_Bt-V(B1-Bt)(B-Bt)AB~ B

and find the value of r. [M. T. 1925]

EXAMPLES 175

25. The 'universal' shunt galvanometer is arranged as follows:between two points A and B there are two wires, one of which containsthe coil of the galvanometer, and the other admits of a connectionbeing made at a variable point G. If a current enters at A and leavesat G, shew that the fraction of this current measured by the galvano-meter is proportional to the resistance of AC. [M. T. 1909]

26. A cell of electromotive force E and internal resistance r is con-nected by wires of resistance rx, r t to the terminals of a galvanometer ofresistance O, there being a shunt of resistance R across the terminals.Find the current through the galvanometer. [M. T. 1933]

27. Two arms AB, AG of a Wheatstone's bridge have resistanceslOohmseach, BD is a standard coil of 5 ohms, and OD a coil of approxi-mately 5 ohms, whose exact resistance is required. The galvanometer(in BO) has a resistance of 50 ohms and can detect lO"6 ampere. Ifthe maximum current that may be passed through the resistance BDis ^ ampere, find to two significant figures the smallest differencebetween the unknown coil and the standard which can be detected.

[M. T. 1922]

28. Two conducting circuits OPQ, O'P'Q' are connected from P toP' and Q to Q' by wires of resistances r and r' respectively. A currententers the circuit at O and leaves at O'. Shew that, if the resistances ofthe lengths OP, PQ, QO are A, B, G and of the lengths O'P', P'Q',Q'O' are a, b, c respectively, the currents inPP ' and QQ' are in the ratio

B 0 • b0 4-r' A B • "ft • -- + r -A + B+C+a+b + c+r-[M. T. 1916]

29. A network is made up of the sides of a regular tetrahedron inwhich the mid-points A and B of a pair of opposite sides are connectedby a wire whose resistance is the same as that of any one of the sides ofthe tetrahedron. A current is led in at A and out at B. Shew that theequivalent resistance of the network between A and B is f r, where ris the resistance of one of the sides of the tetrahedron, [M. T. 1934]

30. A BCD is a uniform circular wire of resistance 4 ohms, and A OG,BOD are two wires forming diameters at right angles, each of resistance2 ohms; prove that if a battery be placed in AD, the resistance of thenetwork is y- ohms, and if in AO, ^- ohms. [I. 1909]

31. A,B,O,D are four points in succession at equal distances alonga wire; and A, C, and B, D are also joined by two other wires of thesame length as the distances between those pairs of points measuredalong the original wire. A current enters the network thus formed at Aand leaves at D; shew that \ of it passes along BG. [M. T. 1909]

32. A square ABCD is formed of a uniform piece of wire, and thecentre is joined to the middle points of the sides by straight wires ofthe same material and cross-section. A current is taken in at A and s

176 EXAMPLES

drawn off at the middle point of BC. Prove that the equivalent resist-ance of the network between these points is f § of that of a side of thesquare. [M. T. 1906]

33. The sides of a hexagon ABCDEF are formed of wires each ofresistance R, and wires each of the same resistance JR join BF and BE;shew that the equivalent resistance of the conductors between A andD is |gi?. [I. 1901]

34. A network is made up of 13 feet of uniform wire, placed so as toform four equal squares side by side. A unit current enters at one extremecorner and leaves by the diagonally opposite corner. Shew that thetotal resistance is equal to that of 2/ff feet of the wire, and find thecurrent in each of the cross pieces. [I. 1902]

36. A network is formed of uniform wire in the shape of a rectangleof sides 2a, 3a with parallel wires arranged so as to divide the internalspace into six squares of side a, the contact at points of intersectionbeing perfect. Shew that if a current enter the framework by onecorner and leave it by the opposite, the resistance is equivalent tothat of length ^ a of the wire. [Trinity Coll. 1895]

36. An octahedron is formed of twelve bars of equal length andthickness and of the same material; a current enters the system at oneend of a bar and leaves at the other end of the same bar; shew that theresistance of the octahedron is i% of that of a single bar. [I. 1902]

37. Five points are connected by ten wires, each pair being joinedby a wire of the same resistance B. Shew that the resistance to a currententering at one of the points and leaving at any other point is fJB.

[I. 1926]38. Six equal and uniform wires are connected so as to form a

regular tetrahedron ABCD. A current enters at the middle point ofAB, and leaves from the middle point of CD. Shew that the resistanceis Jr, where r is the resistance of one of the wires. [I. 1907]

39. A cubical framework consists of 12 equal wires, each of resist-ance JR. Shew that, if a current enters at any corner of the cube, andleaves at the opposite corner, the equivalent resistance of the cube is{i?. [I. 1908]

40. A cube is formed of twelve uniform wires of the same resistancer, the opposite corners are connected by wires of resistance r' whichare otherwise insulated. Prove that the resistance to a current whichenters at one corner of the cube and leaves at the opposite corner is

41. A tetrahedron frame ABCD is formed by six wires, the resist-ances of opposite edges being equal. Prove that the whole resistanceof the frame for a current entering at A and leaving at D is

EXAMPLES 177

where rt is the resistance of A B or OD, rt that of A 0 or BD and ra thatof AD or BO. [1.1902]

42. If every pair of n electrodes are connected by a conductor ofresistance R, shew that the equivalent resistance of the networkbetween any pair of electrodes is 2R/n. [M. T. 1894]

43. In Wheatstone's bridge the resistances of the conductors AB,BO, AD, DO are Rx, JR2, i?s, RA respectively. There is a galvanometerin the conductor BD and a battery in AC. The resistances Rt and JR4being unknown, and the bridge being not quite in adjustment, it isfound that it can be brought into adjustment by shunting R± with alarge resistance X, or by shunting Rt with a large resistance Y. Shewthat approximately

where it is assumed that X and Y are large compared with Rlt RitR3,Rt. [M. T. 1924]

44. A cell of resistance r is connected to the ends of a wire AB. Thecell is then replaced by two different cells, of resistances R, R', arrangedin parallel, producing the same current in AB and having the com-bined resistance r when in parallel. Shew that the total heat productionis greater in the second case than in the first, by the amount whichwould be produced in the circuit of the two cells if the wire AB werebroken. [M. T. 1914]

45. A network consists of a rectangle ABOD of wire, in whichAB = CD = 2a, BO = DA = a, and a wire 1 7 of the same material oflength a, containing a battery of B.M.F. E and negligible resistance;X can slide along AB and Y along DO so that X F remains parallel toBO. Shew that the heat evolved is 6aE*/r{15a? — 4£a}, where r is theresistance of unit length of wire and f is the distance of X from themid-point of AB. [M. T. 1935]

46. Three wires APB, AQB, ARB are arranged in parallel betweenthe points A, B; their resistances are p, q, r ohms respectively. Bat-teries of negligible resistances and of electromotive forces E, F voltsare now inserted in the branches APB, AQB; the negative pole ofeach battery is connected to A. If the rate at which electrical energy isexpended by the two batteries together is W watts, shew that

pq + rp + rq

47. Coils of resistances i?x and i?a are connected, in parallel, to abattery of internal resistance r. Shew that the rate of expenditure of

178 EXAMPLES

energy in the two coils will exceed that in either of the coils if con-nected separately to the battery, provided

where R is the smaller of Rx and i?2. [M. T. 1923]

48. A Wheatstone's bridge is represented diagrammatically by aquadrilateral ABCD, in which A, C are the battery and B, D thegalvanometer terminals, and the resistances of AB, BC, CD, DA areP, Q, R, S. A balance is obtained by putting resistances a, /? in parallelwith R, S. The latter pair are then interchanged and a new balanceobtained by putting resistances a' in parallel with R and j8' in parallelwith S. Prove that

( i ) ( J 4 ) ( H ) [L 1914]

49. A quadrilateral ABCD is formed of four wires and the diagonalAC is formed of another wire. In the wire AB is a battery of electro-motive force E in the direction BA, in AD another of E.M.F. E' indirection AD and in AC a galvanometer; if the total resistances in A B,AD, BC, CD are c, 6, /8, y and 6jS = cy, prove that there is no currentthrough the galvanometer provided Eb = E'c. [I. 1912]

50. An electric circuit contains a galvanometer and a batteryof constant electromotive force V. The resistance of the galvanometeris Q and that of the rest of the circuit including the battery R. Shewthat on shunting the galvanometer with resistance S the currentthrough the galvanometer is decreased by

VRG(R+G)(RS + RG+GS)'

A circuit contains two lamps, each of resistance R, in parallel onleads each of resistance S. The resistance of the rest of the circuit,including the battery of constant voltage V, is r. Shew that if onelamp is broken, the heat emitted in unit time by the other is increased

61. The circuit of an electric battery is completed through agalvanometer of high resistance G, and its observed deflection is D.The poles of the battery are then connected by a shunt of resistance 8,and the deflection falls to D'. Prove that the resistance of the battery is

V CM. T. 1911]

62. A battery is sending current through an external resistance R.The terminals of the battery are connected to a condenser and thecharge communicated is found to be Qt. On repeating the experiment

EXAMPLES 179

with an infinite external resistance, the charge given to the condenseris Q2. Prove that the internal battery resistance is

B(Q2-Qi)IQi- [M. T. 1912]

53. A tramcar takes its current from a trolley wire of total resistancer whose terminals are both kept at potential V. If the power taken bythe car is assumed constant and equal to H, shew that the minimumpotential difference between the trolley and the earth is

and that this occurs when the car is midway between the terminals.Shew that the waste of power in the wire is then a maximum, and isequal to H (V/V- 1). [St John's Coll. 1915]

54. A, B are the ends of a long telegraph wire with a number offaults and C is an intermediate point on the wire; the resistance to acurrent sent from A is R when C is earth connected, but if G is notearth connected the resistance is S or T according as the end B is toearth or insulated. If R', S', T' denote the resistances under similarcircumstances when a current is sent from B towards A, shew that

T'(R-S) = R'(R-T). [M.T. 1903]

ANSWERS

1. 200 kilom. 2. As5:4approx.

3. 0-375, 0-875, 1-25 amps. 4. Less in series in the ratio 41:43.

5. 75 arranged in three parallel sets each containing 25 cells in series.

8. 1-00046 amps. 10. 0138 amp.

11. 20-5 volts. 12. ff\j amp., jf^ amp.

13. 392 volts. 14. 40, 40, 20 amps.; 478 volts.

15. 112-8 volts; the leads consume at the rate of 81-92 watts and thelamps at 640 watts.

16. 4j*fj ohms. 17. -fa amp.

20. That of a length ^ M ' ^ i ' V ^ l ^ r 1 - of wire.6 y/{b(l-a)}+\/{a(l — b)} b-a

26. RE/{RG + {R+G)(r+r1+ri)}. 27. 0-0085 ohm.

Chapter VIII

MAGNETISM

8*1. The peculiar properties of the oxide of iron Fe3O4 wereknown to the ancient world. It was called lodestone or leadingstone because of its power of attracting iron, and its ancientname XiOos Mayvfjns was attributed by Lucretius to Magnesiain Thessaly, where the ore was found in large quantities. Themodern science of Magnetism may be said to date from WilliamGilbert of Colchester who published his great work De Magnetein 1600.

If a body such as a steel needle has been rubbed with a lode-stone, and so acquired magnetic properties, and is then sus-pended so that it can turn freely about its centre of gravity, itsets itself in equilibrium so that a definite line in the body—called the axis of the magnet—is parallel to a definite directionin space—called the direction of the earth's magnetic field,roughly north and south. If the ends of the axis are marked onthe magnet, it is found that, when the magnet is free to move,one end always points in a northerly direction and the othersoutherly. The azimuth, or direction of the magnetic field westof north, is called the variation or the magnetic declination, andthe angle between the direction of the earth's magnetic fieldand the horizontal plane is called the magnetic dip. The earth'smagnetic field is completely specified by its intensity, declina-tion and dip.

8'ii. Magnetic poles. I t is found that if the axes of severalmagnets are marked and the magnets are freely suspended,ends which both point to the north or both point to the southrepel one another and an end which points to the north attractsan end which points to the south.

The ends of a long thin magnet are called its poles. Likepoles repel and unlike attract. The two kinds of poles may becalled positive and negative. They act as centres of force. Ina long thin bar magnet the rest of the bar appears to be almostdevoid of magnetic properties; but, if the bar be broken in two,

8-12] THE FUNDAMENTAL VECTORS 181

each portion appears to have poles similar to those of theoriginal magnet. This and repeated experiments of the samekind lead to the conclusion that a uniformly magnetized bodyis composed of a great number of small magnets with their axesin the same direction and like poles all pointing the same way.

It is impossible however to isolate a magnetic pole or obtaina body in which the total sum of magnetism is not zero. Thisfact may be verified by a simple experiment, viz. that if amagnet be placed floating on water supported on a cork,though the earth's magnetic field exerts on the magnet acouple which causes it to rotate until it settles in the plane ofthe magnetic meridian, yet the magnet then remains at rest.Whereas if it contained an excess of either positive or negativepoles, the earth's field would exert on the magnet a resultantforce which would give it an acceleration.

8"12. The fundamental vectors. Magnetic fields can beexplored by small magnets such as compasses, and the couplesexerted on the same magnet in different positions in the field,or on different magnets in the same position, can be comparedby observing their periods of oscillation. Such experimentslead to the conclusion that for a given small magnet at a fixedpoint in a given field there are two vectors M and H, the formerin the direction of the axis of the magnet and the latter in afixed direction, such that the couple on the magnet is the vectorproduct* [MH], i.e. MHsiad, where M, H denote the magni-tudes of the vectors M, H, and 9 is the angle between them,and the vector M is invariable for the same magnet. M iscalled the moment of the magnet and H the intensity of themagnetic field or the magnetic force.

Hence a small magnet of moment If in a field of intensity His acted upon by a couple of moment MHsia 9, where 9 is theangle between the positive directions of the axis of the magnet

* The vector product of two vectors A, B inclined at an angle 8 is avector G at right angles to the plane of A and B, of magnitude AB sin 6and in such sense that G is along the axis of a right-handed screwwhich would turn from A to B through an angle 0 less than two rightangles. The reader will remember that a couple can be represented bya vector perpendicular to its plane. The vector product of A and Bis variously denoted by [AB], A A B and A x B.

182 FIELD OF A MAGNETIC BIPOLE [8-12-

and the line of force, and the couple tends to reduce the angle0, i.e. to set the axis of the magnet along the line of force.

If we regard a thin magnet +m m H

as composed of poles ofstrength + m at a distance Iapart and set its axis at aninclination 0 to the direction ^of the field H, we see that mH ~m

the hypothesis that a field of intensity H exerts a force mHon a pole of strength m implies that the field exerts on thesmall magnet a couple mHlainO, or [mlH], so that M mustbe identical with ml.

8«2. Field of a magnetic bipole. Experiment confirms thefact that the forces exerted in a magnetic field are consistentwith a law of force between magnetic poles of the same form asthat between electric charges, i.e. that like poles of strengthsm, m' at a distance r apart repel oneanother with a force mm'/r2. We there-fore assume that each pole which goesto compose a magnet produces a fieldof potential like that of a point chargeof electricity.

To find the potential of the field pro-duced by a small magnet or magneticbipole. Regard the magnet as composedof two poles of strengths — m at B and mat A, where BA = 8s,and m8s = M, the moment of the magnet.

Take an origin 0 at the middle point of BA, as in 6*6, andlet P be the point (r, 0) referred to OA as axis. Then thepotential at P is given by

, _ m m _ m ^ m^ = = ~ 5 P + Z P = = ~ r + |Sscos

8s cos 0 \ ml, 8s cos 0-"HmSscos0 , . . „

= j — *° * n e ft™* power of os,, i f cos 0

or <j> -a— (1).

8-22] FIELD OF A MAGNETIC BIPOLE

This result may also be written. (Mr)

183

•(2),

where (Mr) or Mr denotes the scalar product* of the vectorM and the position vector r of the point P relative to thecentre of the magnet.

Again, since in the figure cos 8 = — dr/ds, another form forthe same result is a / i \

*-<® <3>-8-21. We may now establish the vectorial property of the

moment of a magnetic bipole thus:

Since

therefore the potential of a magnet of moment Mx + M2 isthe sum of the potential due to Mx and M2 separately. Hencethe ' moment of a magnet' obeys the vector law of compositionand resolution.

8*22. Components of magnetic force due to a magneticbipole. The magnetic force H is the negative gradient of thepotential so that its components alongand perpendicular to the radius vector are

2Jfcos0

H H,

and

a'~ dr

__ltye~ rdd~~

r3

Memdr3

(1).

The resultant H of these two componentsis a tangent to the line of force at P. Letit make an angle ifi with the radius vector.Then we have, along the line of forcethrough P,

so that

r j - =dr

dr

H

Hr

2cos0— = . x d d ,r sin 6

or r = C s i n 2 0 (2),

* The scalar product of two vectors A, B is defined to be A B cos 9,where 0 is the angle between their positive directions.

184 FIELD OF A MAGNETIC BIPOLE [8-22-

where C is an arbitrary constant, is the equation of the linesof force.

The lines of force in any section of a magnetic field may beexhibited by the help of iron filings. Thus in any such case asthat just considered, if iron filings are scattered on a piece ofcardboard laid on the magnet, and the cardboard is gentlytapped, the filings become magnetized by induction and setthemselves along the lines of force.

8*23. Other expressions for the field due to a bipole.Taking the expressions Hr and He of8*22 for the components of H, weobserve that Hg can be resolved

, , . , . . „ J „ Mcos9 .obliquely into Hgootv or ^—• in

H,

the direction of r, and — HgcoaeodM

or —-g in the direction of M; and

„ , . 2Jfcos0. x, ,. x. „Hr being — m the direction of

r, it follows that the field H is theresultant of two oblique components, viz.

3ifcos0.

and

3— m the direction of r

M f—5- in the direction of M

r3 I

.(1).

This result may be written in vector form thus:

M , 3(Mr)r

r3 .(2).

Alternatively, we may take a small arbitrary displacementSr and equate the work done by H to the loss of potential.Thus

H3r = - 8 = - S {i=p J from 8-2 (2)

MSr 3 (Mr) §rr3 + r* *

8-25] OSCILLATIONS OB1 A SMALL MAGNET 185

But, if a is the angle between r and 8r, Sr — | Sr | cos a = r8r/r,

8 O t h a t

which gives the result (2).

8'24. Oscillations of a small magnet. Let a small magnetof moment M be free to turn about its centre in a magneticfield whose intensity is H at the centre of the magnet. Inequilibrium the axis of the magnet coincides in direction withH. Let the magnet be turned through an angle 6 in any planethrough the direction of H. The couple tending to restore themagnet to its equilibrium position is MHaia.9 (8*12), thereforeif K is its moment of inertia about its centre, its equation ofmotion is „« , , „ . Q

K6= —MHaw.8,

and for small values of 9 this represents harmonic oscillations

of period 2nV(K/MH).We note that though the magnet has two degrees of freedom

and a body with two degrees of freedom has in general twoindependent principal oscillations, yet in this case its period ofoscillation in every plane through the direction of the resultantforce at its centre is the same, so that oscillation in any suchplane may be regarded as a principal oscillation, and the twoperiods of principal oscillations are the same.

8*25. Examples, (i) Two smallmagnets of moments m, m' are fixedat two corners of an equilateral trianglewith their axes bisecting the angles. Athird small magnet is free to turn aboutthe other angular point. Shew that itsaxis makes with the bisector of thethird angle an angle

, A/3 m-m'\tan"1 2= . .\ 7 m+m'JThe magnets m, mf at B, O produce

at A fields with components — "^B„ _2mcos30°_m\/3 „ _msin30°_jm

/_2m/cos30°_my3 m' sin 30° m'

186 POTENTIAL ENERGY OF A BIPOLE [8-25-

in the directions indicated in the figure, where r is the length of a sideof the triangle.

Resolving along and perpendicular to the bisector of the angleBAG, the components are

or

and (ff1-ff1 ')sin30o + (ff2 '-Ha)sin60°, or (Hence the line of force at A, which is the direction that the thirdmagnet takes up in equilibrium, makes with the bisector of the angle

BAD an angle tan6 -1

VV 7 m + m'J

(ii) A small magnet is suspended by a torsionless fibre so that it canoscillate about the magnetic meridian in a horizontal plane, when it isobserved to execute N vibrations per minute. A bar magnet is placed withits axis horizontal and parallel to the meridian and passing through thecentre of the small magnet, at a distance dt measured from its mid-pointand large compared to its own length. The small magnet now executes ntoscillations per minute, and when dx is changed to d%, n± becomes n2.Shew that nt*-N*_d1*

n1t-Nt~dJ''

The oscillations may be supposed of small amplitude. [M. T. 1926]

As in 8*24, if K is the moment of inertia of the small magnet, andM its moment, its period of oscillation in a horizontal plane in theearth's field is 27T^/{KjMH), where H is the horizontal component of the

earth's magnetic force; so that JV = — ^ / ( - = - ] .ir \ \ K J

The effect of a bar magnet of moment M' with its axis along themagnetic meridian is to produce a force 2M'/d1

s along the meridian atdistance dx; so that H is increased by ZM'jdj3, and therefore

30 /(Mf 2Mr

Therefore n > l"S! = 7i-

8*3. Potential energy of a magnetic bipole in a given field.Let the small magnet be represented by a pole — m at B and apole m at A, where BA = hs is drawn in a definite direction.

8-31] MUTUAL POTENTIAL ENERGY 187

Let <£ be the potential of the given field at B, then <f> + - 8s iscs

the potential at A, and the potential energy of the two polesor the work which an external agent would have to do in orderto place them in their assigned positions in the given field isgiven by - ,

^= -M.H, (1),

where Hs is the component of the given field in the directionof the axis of the magnet; or

W= -MHcoaB (2),

where 6 is the angle between the positive directions of M andH ; ° r W= - (MH) (3),

where the formula on the right denotes minus the 'scalarproduct' of the vectors M and H.

8-31. Mutual potential energy of two small magnets;i.e. the potential energy of one in the field of the other. Themutual relations of two small magnets lendthemselves to very simple treatment byvectorial methods. But we shall first set outthe arguments in such a way as to be in-telh'gible without vector algebra.

Let M, M' be the moments of the magnets,r the distance between their centres, 0, 8' ,--.the angles which the line joining the centres /M +makes with the positive directions of the axes 'and e the angle between the axes (the magnets are not ingeneral coplanar).

Then, from 8-3, the potential energy of M' in the field ofilf is given by W=.M'HM,,

where HM, denotes the component along M' of the field dueM

to M. But by 8'23 the field H due to M has components g

188 TWO SMALL MAGNETS [8-31-

in the direction of M and ;— in the direction of r. So byr8

resolving these in the direction of M', we get

(1).

Using vectors, we write, from 8*3,

TF=-(M'H),where from 8-23 (2)

M 3(Mr)r

so that ^ (MMQ_3(Mr)(M'r)IT r°

or if we put r = rrx, so that rx is a unit vector in the directionof r, the result may be written

Pf = {(MM')-3(Mr1)(M'r1)}/r3 (3),

and this is clearly equivalent to (1).

8-32. Couple exerted by one small magnet on another.The couple exerted on the magnet M' by the magnet M maybe represented, as in 8*12, by the vector product [M'H], whereH is the field due to M; i.e. the couple is M'H sin ip,i£ipia theangle between M' and H. But by 8*23 H is the resultant of

. 3Jf cos0 . i t . ,. X. , , M . ,,components -—=— in the direction oi r and —=• in the

direction of M; therefore the couple exerted by M on M' is3M'3I

compounded of a couple — 3 — cos 6 sin 6' in the plane of

M'MM' and r, and a couple 3— sin e in a plane parallel to M'

and M. Or vectorially, the couple required is [M'H], whereby 8-23 (2) M

so that the couple is

[M'M] 3[M'r](Mr)^3—+ — 7 5 '

8-33] TWO SMALL MAGNETS 189

or, if mx, m / , rx denote unit vectors in the directions of M, M'and r, the result may be written

/rJ^r,)} (1).

8-321. We observe that the components of the couple in8*32 can also be deduced from the expression for the potentialenergy of M' in the field of M, viz.

MM'W = =-=- (cos e - 3 cos d cos 0').

We suppose that M is fixed in position and that M' is freeto turn about its centre. Then the direction of If' is determinedby the angles e and 6', for 6 is a fixed angle.

If for example we draw radii OP, OP', OR of a unit sphereparallel to M, M' and r, then P and R are fixed points and thearcs PP', RP' equal respectively to e, 6', determine P'.

Hence M' is acted upon by a couple tending to increase e of., , dW MM'saxe , ,, . . , . .

magnitude — -=— or § , and this must act m a planeparallel to M and M'; and a couple tending to increase 6' of

., , dW 3MM' a . ,, . ,, . _ ,magmtude — -^^ or s— cos 0 sin 0 in the plane ot r and

M'. As before, these two couples can be written as in 8*32 (1),paying due regard to signs and remembering that

8-33. Force exerted by one small magnet on another.The force exerted by M on M' and tending to move the latterbodily without rotation must be deduced from the potentialenergy. This may be written

^ MM'

where p, p' denote the projections of r on the axes of the mag-nets. In a general displacement in which M remains fixed andM' alters its position but not its direction, e remains constantand r, p, p' may be regarded as independent variables on whichthe position of the centre of M' depends.

Thus if through the centre 0 of M we draw lines OP, OP'

190 TWO SMALL MAGNETS [8*33-

of lengths p, p' in the directions of the magnets, and throughP, P' take planes at right angles to OP, OP', these planesintersect in a line, and the centre 0' of M' is a point on thisline at a distance r from 0.

The force tending to move M' parallel to itself therefore hascomponents 7iW *MW , 5w\

- — j — I COS € 5- Ip \ r5 /dr ~

ZMM'

in the direction of r;dW_ ZMM'p' BMM' cos 6>'

in the direction of p or M; anddW ZMM'p ZMM'cosO

(cos e — 5 cos 6 cos 6') (2)

(3)

(4)

in the direction of p' or M'.It is easy to see that the force may be written as a vector

sum in the form (

where p = 3 {(MM') - 5 (Mrx) (M'r^},

/t=3(M'p1)and /*' = 3(Mr1).

Alternatively, as in 8*23, if F denotes the force exerted byM on M', we have

{ ^ 3 ( M y M ' r > } from 8-31(2)_ 3 (MM') hr 3 {(M 8r) (M'r) + (Mr) (M'Sr)}- ^i + 5

_ 15 (Mr) (M'r) 8r

(3(MM')r 15 (Mr) (M'r) r- b T \ ft r7

3M (M'r) 3M' (Mr))+ ^ + rs j ;

so that F = (prx+JJM + fi'M'where p, p, p! have the meanings above.

8-34] COPLANAR MAGNETS 191

8*34. Goplanar magnets. The formulae for coplanarmagnets can be obtained independently. Thus, as in 8*3, thepotential energy of a small magnet M' in the field of a magnetMia W=-M'HM,,where HM. is the component in the direction of M' of the fieldH due to M. But, as in 8-22, H has components 2M cos 0/r3

along r and M sin0/r3 perpendicular to r, and M' makes anangle 0' with r (see the figure of 8*31), therefore

MMW= _ ^ L (2 cos 0 cos 0 ' - sin0 sin 0') (1).

This result follows of course directly from 8*31 (1) by puttinge = 0-0'.

Let us now find the force and couple exerted by M on M'.The couple can be written downat once as [M'H], but we shallalso find it by a general methodwhich will also serve to deter-mine the force. Let R, T denotethe forces tending to move themagnet M' without rotationalong and perpendicular to r,and let O denote the couple y^-6'\ \d\tending to turn M' about its — M fcentre. Keeping M fixed, let M' undergo a small displacementof a general type which alters r, 6 and 6'. The work done in thissmall displacement is equal to the loss of potential energy so that

Rdr + Trdd + Gd (6 - 6') = - dW

Since the displacements are arbitrary, we must have

or

^ ^ | ^ sin0') ...(3)

and # = | ^ = ^-(2cos0sin0' + sin0cos0') (4).do r

192 OOPLANAR MAGNETS

From (3) and (4) we get

[8-34-

(5).

The forces R, T and the couple O can be compounded intoa single force.

In like manner the magnet M is subject to forces exerted byM' equal and opposite in direction to R and T, and to a couple

...(6).cv r

The couple 0 and 0' are not equal and opposite, but nocontradiction is involved, because the transverse forces T arenot in the same straight line, but T and 0 acting on M'compound into a single force which is equal and opposite to— T and 0' acting upon M.

8*35. E x a m p l e s , (i) A magnetic parUcle'of moment /x lies in amagnetic field at the point (rx, 8t) ina plane. The axis of the particle liesin the plane and makes an angle <f> with the direction given by 0=0 . Thepotential of the magnetic field at the

point (r, 8) in the plane is . Find

the potential energy of the particle anddeduce that, if its centre is fixed, it canrest in equilibrium, if <l> = 281 or if<£ = 77 + 20!. [M. T. 1934]

Since the potential of the field is

, therefore at a point (r, 8) the

magnetic force has components

This makes the force at the centre of the magnet /j. in direction of its axiscos 0X 008(^—0!) 008(^ —

Hence the potential energy of the magnet in the given field isW= - H.eoa(<l,-281)lr1*.

The couple O tending to turn the magnet about its centre is given bya_ 8TT_

and since this vanishes if <f> — 20!, or if $ = IT + 20!, these give possiblepositions of equilibrium.

8-4] GAUSS'S VERIFICATION 193

(ii) Three small magnets of moments m1, ma, mt are placed alongrectangular axes with their centres at equal distances a from the origin.Shew that if a small magnet be free to turnabout its centre at the point (2a, 2a, 2a) it willtake up a position of equilibrium in which thedirection cosines of its axis are proportional to

and m1 + mi-ma. [I. 1927]Let A, B, G, P be the positions of the four

magnets and A, (i, v the direction cosines ofthe fourth.

The direction cosines of AP, BP, GP arei . f i f; f.i.f; hhi respectively. Thereforeif M be the moment of the fourth magnet itspotential energy is

For a position of equilibrium W must be stationary in value for allsmall changes in A, p, v subject to A2 + /xa + v8 = 1.

Therefore SW=0,or2M

- m, - m8) 8A + ( - - ms) 8/x + (-»% - ma+m3) Si-} = 0,

for all small values of 8A, 8/t, Sv subject to

and this requires that

8'4. Gauss's verification of the law of inverse squares.Comparison is made of the couples exerted by a small magnet M on asmall magnet M' in two particular positions.

We note that in the special cases (i) when M is 'end on' to M', sothat in 8'34 9=0 and B' = fa, the couple exerted on'ilf' by M is 2MM'lr*;and (ii) when M is broadside to M', so that 6 — fa and 6' = 0, the coupleis MM'jr3; i.e. the couple in fig. (i) is twice the couple in fig. (ii) formagnets at the same distance apart. We shall now shew that if the lawof force were a force oc 1/r" at distance r, then the couple in fig. (i)would be p times the couple in fig. (ii).

Let us regard the magnet M as composed of two poles — m at B andRBM 13

194 GAtTSS S VERIFICATION [8-4-m at A, where BO = OA = a, and let 00', the distance between thecentres of the magnets, be r.

Then in fig. (i) the force at O' due to M=-.

to the first power of a

= Mp/rp+1, since M = 2ma along 00'.Also in fig. (ii) the resultant force at O' is clearly parallel to AB and

mama, ., „ , -= — - to the first power of a

r* r r'r

so that the couple in fig. (i) is p times the couple in fig. (ii).

M'

B O A 0 '- M

M'

0) 00

o'

Now let M' be suspended so that it can turn freely about a verticalaxis; and let H be the horizontal component of the earth's magneticfield.

Firstly, let M be placed pointing east and west with its centre duewest of 0' at a distance r; M' will in its equilibrium position make anangle 8 with the magnetic meridian, so that the couples produced bythe two fields H and Mp/rp+1 balance one another; i.e. so that

M'H sin 6 = cose • ( I ) -

Secondly, let M be placed with its centre due south of 0' still pointingeast and west and so that 00'= r as in fig. (ii). The deflection of M'will now be 6', where

M'H sin 6'=M' cos 6' •(2).

8'5] MAGNETIC MOMENT 195

Therefore tan 8 =p tan 8'; but 8 and 8' can be measured and thus itcan be confirmed that p = 2.

8'41. Experimental determination of the magnetic momentof a small magnet. We saw in 8*24 that the period of oscillation of asmall magnet free to turn about its centre in a given magnetic field is2nV(KIHM), where K is its moment of inertia, M its magneticmoment and H the intensity of the given field.

If therefore the magnet is free to turn about a vertical axis, aboutwhich its moment of inertia is known, by observing the time ofoscillation we can determine the product HM, where H is now thehorizontal component of the earth's magnetic force.

Further, if we place the given magnet pointing east due west ofanother small magnet M' as in 8-4, where M' is free to rotate about thevertical, the deflection 8 of M' will be given, as in 8*4 (1), by

M'H sin 8 = ^ ^ cos0

so that

Hence by measuring r and 8 the ratio M/H can be determined, andwe have seen above how to determine the product MH; so that bythis process both M and H can be found.

8*5. To determine the possible positions of relative equili-brium of two small magnets free to turn about their centresin the same plane. Referring to the figure of 8*31, the mutualpotential energy is given by

so that

and

™ = ^ ^

In equilibrium W must be stationary so that the first derivativesmust vanish. There are four solutions whichgive different relative positions

(i) 0 = 0, 0' = O;(ii) 0=0, 0' = TT;

(iii) 8= in, 8'= in;(iv) 0= in, 8' = In.

The conditions for a true maximum orminimum of W are that the discriminantW9eWpe—We6? must be positive and WM

negative for a maximum and positive fora minimum, where the suffixes denotepartial derivatives.

13-2

196 EXAMPLE [8-5-

But WM = Wg.0. = —jg- (2 cos 0 cos 8' - sin 8 sin 6')

MM'and Wee.= -—j—(-2sin0sin0' + cos0cos0').

Hence the values in the four cases are

(i) (ii) (iii)TJ7 „ 2MM' 2MM' MM'

, __. MM' MM'and ww=—^-, jjp-

and the signs of the discriminant in the four cases are(1) +, (ii) +, (iii) - , (iv) - .

Also in (i) Wge is +, so that W is a minimum and the equilibrium isstable, but in (ii) W$g is —, so that W is a maximum and the equilibriumis unstable. But in positions (iii) and (iv) W is neither a true maximumnor a true minimum, so the equilibrium must be regarded as unstableas it is certainly not stable for all small displacements.

8*51. E x a m p l e . Two small magnetic needles of equal moment Mare pivoted on vertical axes which are fixed at a distance c apart, the lineof centres being perpendicular to the magnetic meridian. Obtain anexpression for the potential energy of the needles in terms of the anglesbetween the Une of centres and the axes of the needles, allowing for theearth's horizontal field H.

Shew that if H< SM/c3, there is a stable position of equilibrium inwhich the needles are parallel and make an angle 6 with the line of centres,given by

3M sin 0=He3 . JH

Prove also that, if H>3M/ct,the position with the axes of theneedles parallel to H is stable.

[I. 1915]The potential energy is ob-

tained by adding to the mutualpotential energy of the magnetsthat of each magnet in the earth's field, so that, by 8'3 (2) and 8*34 (1),we have

MlW=--^(2ooa8ooa8'-ain8sia8')-MHain8-MHam8' ...(1).c

Therefore, using suffixes to denote partial differentiation,

.„ M*

-r

M%

We. = —3- (2 cos 0 sin 8' + sin 6 cos 8') - MH cos 8'.

8-6] FINITE MAGNETS 197

Hence Wg, Wg- both vanish for a position in which 8' = 8 and

3Msin8=:Hc3 (2),

giving a real angle 8itH< SM/c3.Mi

Further Wee = —^ (2 cos 8 cos 8' - sin B sin &') + MH sin 8,

Wg-g. - —j- (2 cos 8 cos 8' - sin 8 sin 0') + J f H sin 0',

Af2

Wfl(), = r (2 sin 8sin 8' — cos 5cos 0').

Hence, in the position given by (2),

W<w Wg-g. - TFV = - r [{(2 cos2 6 - sin2 8) + 3 sin2 0}2 - (2 sin2 0 - cos2 0)2]

==-

Since this expression is positive and since Wee is also seen to bepositive, therefore 'FT is a minimum and the position is one of stableequilibrium.

If however H>3M/c3, (2) does not give a real value for 0, but W9,Wg. both vanish for 8—8' = \-n, and in this case

and Wgg.=

so that Wgg We.g. - W*gg.=M*{(H- M/c3)2 - AM2/c6}

= M2 {{H - 3M/c») (H + M/c3)},

and this is positive and so is Wgg, so that W is a minimum and theposition in which the axes are parallel to H is stable.

8-6. Finite magnets. Intensity of magnetization. A mag-net of any form consists of an agglomeration of small magnetsor bipoles. A small element of volume contains a number ofthese, and we assume that the positive poles can be regardedas collected at their mean centre and the negative poles attheir mean centre giving a single bipole of a definite moment.Each element of volume of a magnetized body may there-fore be regarded as a small magnet and the ratio of themoment of the small magnet to its volume is called theintensity of magnetization and usually denoted by I. Inmost magnetic bodies the vector I varies from point to point,

198 UNIFORMLY MAGNETIZED SPHERE [8-6-

but a body in which I is constant in magnitude and directionat every point is said to be uniformly magnetized.

8-61. Uniformly magnetized sphere. We may regard abipole as constructed by taking coincident equal positive andnegative poles m and — m and pulling them apart a shortdistance Sx. In the same way if we consider a small elementof volume 8v to contain magnetic matter of positive density pand at the same time that there is coincident with it a volumeSv of magnetic matter of negative density — p and we make arelative infinitesimal displacement 8x of the two elements, weobtain a resultant bipole of moment pSv.Sx, wherein p8x isby 8*6 the intensity of magnetization.

In the case of uniformly magnetized bodies of finite size, wemay suppose their magnetic condition to have been producedin the same way. Consider, for example, a sphere of radiusa uniformly magnetized to intensity I. We regard this asproduced by taking a sphere of radius a of magnetic matterof uniform density p with a like sphere of density — p, the twocoinciding in position, and then giving their centres 0, 0'a small relative displacement Sx.

It was shewn in 3*2 that the potential at an external pointdue to a uniform spherical charge is the same as if the chargewere collected at the centre of the sphere, and if we regard auniform solid sphere as stratified in concentric shells it isevident that the same result holds good for a uniform solidspherical distribution.

8-62] THE EARTH 199

Hence the potential at an external point P, whose co-ordinates are r, 0, with 0 as origin and O'O as axis, regardedas due to the positive and negative spheres is

f •nd?p %vUzp _ ira3p Sx cos 6r r + 8xcoa8 r2 '

to the first power of 8x.Now putting p8x equal to a finite number / , no matter how

small 8a; may be, the potential becomes

9 = (1).

Hence the external potential due to a uniformly magnetizedsphere of radius a, magnetized to intensity I, is the same as ifa small magnet or bipole of moment |i7a3l were placed at thecentre of the sphere.

8*62. The earth regarded as a uniformly magnetizedsphere. Since the positive pole of a compass points to themagnetic north, the axis of the bipole equivalent to the earth'smagnetism must point south. If a be the earth's radius and/ its intensity of magnetization regarded as uniform, theequivalent bipole is of moment M=^nazl pointing to thesouth pole. Therefore in a place in the northern hemisphere ofmagnetic latitude I, the components of magnetic force are2M M— j - cos ( TT +1) radially and -»- sin (|TT +1) tangentially, i.e. a& (Jb

force frr/sin^ vertically downwards, and a force frr/cosZhorizontally towards the north. Amagnet free to turn about its centreof gravity will therefore dip throughan angle tan"1 (2 tan V) below thehorizontal

It must be remarked that the earthis very far from being a uniformlymagnetized sphere, the earth's mag-netic poles are not at opposite endsof a diameter and their positions areundergoing continual change, and

N

200 UNIFORMLY MAGNETIZED SOLID [8-62-

there are disturbances of the earth's magnetism associatedwith the appearance of sun spots.

8*63. Uniformly magnetized solid. The following methodmay be adopted to find the potential at an external point dueto a uniformly magnetized solid of any shape.

We may regard the magnet as composed of a large numberof elementary magnets of moment Idv, where I is the intensityof magnetization and dv an elementof volume. Let x, y, z be the co-ordi-nates of a point Q of such an element,let P be an external point (£, -q, £), andlet QP—r make an angle 8 with thedirection of I. Take the axis of a; in thedirection of I. Then the potential atP due to the elementary magnet is

ri / I \

Idvcoad/r2 or /dv^- l - l . Hence the potential at P of the

whole solid is given byH"7*;®* w>integrated through the volume.

But r2=(£ — x)2 + (r) — y)2 + (£ — z)2,

«. * a i a iso that 5--=~5?--

ox r of rTherefore <£ = — / ^ . I - 1 dv

But represents the potential at P due to a uniform dis-tribution of matter of density / , so that the potential of themagnet can be found by applying the operator — djdg to thepotential of this uniform distribution of matter.

8*7. Magnetic shells. An iron sheet of any form, plane orcurved, magnetized normally at every point is called a mag-netic shell. The strength of the shell at any point is denned

8-71] MAGNETIC SHELLS 201

to be the intensity of magnetization multiplied by the thick-ness of the shell at the point.

If the strength is constant at all points, the shell is called auniform or simple shell.

Such a shell acts like two parallel layers of positive andnegative magnetism at a short distance apart, for it can beconceived to be an agglomeration of small magnets all directedalong normals to the sheet, with their positive poles on onesurface and their negative poles on the other.

8*71. Potential due to a magnetic shell of uniformstrength T. Let I be the intensityof magnetization at a point Q onthe positive face of the shell, t thethickness and dS a small areasurrounding Q. The moment ofthe small magnet of which d8is the cross-section is, by 8'6, Itd8 = rdS, and its axis isnormal to the surface.

Let P be a point at a distance r from Q, and d the anglewhich QP makes with the normal at Q drawn from the negativeto the positive side of the surface. Then the potential at P due

to this small magnet is ^ or rdw, where da> is the solid

angle which dS subtends at P. Therefore the potential at Pdue to the whole shell is TW, where to is the solid angle which theboundary of the shell subtends at P; and this formula is adaptedto the case in which linesdrawn from P to the shellmeet it first on the positiveside, and the sign must bechanged if such lines meetthe shell first on the negativeside.

We notice that the ex- "*"pression for the potential of the shell does not depend on itsform provided that its boundary is fixed. It is therefore thesame for all shells having the same boundary.

202 MAGNETIC SHELLS [8'71-

If the shell is closed, e.g. a spherical shell, the potential atall external points is zero, and at all internal points it is — 4TTT.

Let P be a point close to the surface on the positive side ofthe shell, and a> the solid angle which the boundary subtendsat P, so that rco is the potential at P. Then at a point P' on thenegative side and as near as possible to P the solid angle sub-tended by the boundary differs by little from 4,v — to, and inaccordance with the rule of signs established above the poten-tial at P' is , . ,

— T (4TT — to) = rto — 4TTT.

Therefore in passing from the positive to the negative side ofthe shell along a curve which goes round the edge the potentialdecreases by 4TTT, and conversely if we pass through the shellfrom the negative to the positive side the potential increasesb y 4TTT.

The theory of the magnetic shell is of great importancebecause, as will be seen in the next chapter, there is a closeanalogy between the field of a magnetic shell and the fieldproduced by an electric circuit.

8*72. E x a m p l e . A uniform magnetic shell of strength T is bounded bya circle of radius a. Find the magnetic force at a point on the axis of theshell at a distance zfrom the centre of the circle, and also find the mechanicalforce exerted on a small magnet placed along the axis.

8-72] EXAMPLES

The potential of the shell is given by

203

where <o is the solid angle subtended by the shell.At a point P on the axis at a distance z from the centre O of the

boundary, on the positive side of the shell,this becomes

where OA is any radius of the circle;

or

Hence the magnetic force at P is

directed along the axis.Now let a small magnet of moment M be

placed along the axis at P and suppose it to consist of a pole — m atz and a pole m at z + Sz. Then the forces on the two poles are —mH

H+-g— 8z J; so that the resultant force on the small magnet is

the negative sign implying an attraction.

EXAMPLES

1. Two magnetic molecules lie in one plane, their axes ma,Wingangles 6 and 8' with the line joining the molecules. Shew that thecouple on the second molecule due to the action of the first will vanishiftan0+2tan0' = O. [I. 1908]

2. The centres of two small horizontal magnets M, M' are at adistance r apart and their axes make angles 6, 0' with r. Reducing theforces which the magnet M exerts on M' to a force through the centreof M' and a couple, find the magnitude of the couple.

If the two magnets are fastened to a board which floats on water,explain why, although the couples exerted on each other are unequal,the board is not set in rotation. [I. 1907]

3. A small compass needle can swing about a vertical axis, and abar magnet of length 2 and moment M is placed horizontally east andwest with its centre vertically below the needle at a distance c. Shewthat if Ijo is small, the deviation of the needle from the magneticmeridian is given approximately by

M /. 3 22\

where H is the horizontal component of the earth's magnetic field.[M. T. 1915]

204 EXAMPLES

4. Two small magnets of moments Mx, M% are placed with theircentres at the corners B, (7 of an equilateral triangle ABG and theiraxes fixed along BA, CA. A third small magnet is free to turn aboutits centre which is at A. Determine the angle 0 which the magnet atA makes with the bisector of the angle BAG in equilibrium.

Which of the two positions is stable ? [I. 1933]

6. Magnets of moments /*, ju,', p' are pivoted at the vertices A, B,Grespectively of a triangle in which the angles B and G are equal. Theyare in equilibrium when they lie along the internal angle bisectors withtheir positive poles inside the triangle. Shew that

// fi=8coseciB.cos3B(2-3cosJB). [M. T. 1930]

6. Two small magnets of moment m are placed at the corners A, Bof a square A BOD with their axes in the direction AB and BA respec-tively. If two other small magnets of moment m' free to turn abouttheir centres at C, D take up positions of equilibrium with their axesalong A C and BD, find the ratio of m to m'. [I. 1928]

7. A magnetic needle is mounted so as to turn freely about thecentre of a graduated circle, which can be turned into any verticalplane. In two perpendicular positions of the plane, the needle isobserved to make angles Sj., Ss with the horizontal. Shew that theinclination of the earth's field to the horizontal is 8, where

cot2 S = cot2 St + cot2 8a,and obtain a formula giving the position of the magnetic meridianwith respect to the two positions of the circle. [M. T. 1916]

8. Three small magnets of equal moment are placed at equal dis-tances from the origin along rectangular axes pointing away from theorigin. Two are fixed and one is free to turn about its centre. Shew thatin equilibrium the inclination of the latter to the line joining it to theorigin is tan-1 (A/2/6). [I. 1922]

9. Two doublets, each of moment M, are mounted with theircentres at fixed points at a distance r apart. They can rotate in thesame plane and are constrained by a frictionless mechanism to bealways parallel and in the same sense. Find the positions of equilibrium,discuss their stability, and shew that the work which must be done toturn the system from a position of stable to one of unstable equilibriumis 3M*/r3. [M. T. 1932]

10. The axis of a small magnet makes an angle j> with the normalto a plane. Prove that the line from the magnet to the point in the planewhere the number of lines of force crossmg it per unit area is a maximummakes an angle 0 with the axis of the magnet, where

2 tan 0= 3 tan 2 (<f>-6). [I. 1905]

EXAMPLES 205

11. Prove that if there are two magnetic molecules of momentsM and M' with their centres fixed at A and B, where AB=r, and oneof the molecules swings freely, while the other is acted on by a givencouple, so that when the system is in equilibrium this molecule makesan angle 8 with AB, then the moment of the couple is

%MM'ain28/ra(3ooaid+l)i. [M. T. 1901]

12. A small magnet of moment M is suspended with its centre at apoint equidistant from the poles of a fixed bar magnet of length 2aand pole-strength m. The distance of the small magnet from thecentre of the bar is b. If the small magnet be turned through two rightangles from its equilibrium position, shew that the work performed is

4Mma/(a3+ &*)*. [M. T. 1923]

13. Prove that the potential energy of a small magnet of momentp pivoted at a distance r from another small fixed magnet of moment/i' and in the direction of the axis of ft' is

when the axis of /x makes an angle 8 with the axis of p.A uniform field H now acts in addition from the magnet /x towards

the fixed magnet. Prove that the previously stable position of equi-librium becomes unstable when

H>^4-. [M. T. 1929]

14. M, M' are the moments of two small magnets, e the angle betweentheir axes and 8, 6' the angles their axes make with the line r joiningtheir centres. Prove that, if 8= \TT, there is a couple on the magnet M'whose moment is MM' sin e/r3 and whose axis is parallel to the line ofintersection of the planes perpendicular to the axes of the magnets.

[I. 1913]

15. Two small magnets of moment m are fixed with their axes inthe same straight line pointing in opposite directions. A small magnetof moment m' has its centre fixed at a point in the plane bisecting atright angles the join of the other two magnets. Prove that the periodof the small oscillations of the magnet m' in this plane is

2w (Ir3/3mm' sin a)*,where I is its moment of inertia, r its distance from either of the othertwo magnets and a is the angle subtended at m' by the line joining theother two magnets. [I. 1915]

16. The centres of two equal small magnets are fixed at a distanced apart. One magnet is fixed at an angle tan-12 y/Z to the line of centresand the other is free to move in the plane through its own centre andthe first magnet. Find the positions of equilibrium of the second magnetand distinguish between the stable and unstable ones. [M. T. 1935]

206 EXAMPLES

17. A small magnet of moment p is held fixed at the origin of co-ordinates, with its axis in the direction (I, m, n); another small magnetof moment p' has its centre fixed at the point (x, 0,z), and is free toturn so that its axis moves in a plane parallel to the plane 2 = 0. Findthe position of stable equilibrium of p', and shew that the period of itsfree oscillations about this position is

2w I* fTV~* («a + z2)i [{I (a;2 + 22) - 3a; (Ix + nz)}* + m* {x2 + za)ar*»

where I is the moment of inertia of // . [M. T. 1913]

18. Two magnetic needles of equal moment M are pivoted at A, Band can turn in a horizontal plane, AB being horizontal. There is auniform field H parallel to AB. Shew that if in equilibrium the needlesmake angles 8, $ with AB, where 8, <f> are not zero, then either

6 = <l>, 3 c o s 0 = — K, or 8= —(/>, cos6= —K,

where K = and a=AB. [I. 1924]

19. Two small magnets of moments M and 6M are so mounted thatthey can turn about their centres, in the same horizontal plane, theline of centres being perpendicular to the earth's horizontal field H.Prove that the position of equilibrium in which their axes are in thedirection of H is stable if 9M < c3H, where c is the distance between thecentres. [I. 1925]

20. If two magnetic doublets are of equal strength m and are freeto turn about their centres, which are fixed, and if they are in a field ofuniform force H parallel to their line of centres, prove that the positionin which the direction of each doublet is opposite to the direction of His a position of stable equilibrium if H < m/r3, where r is the distancebetween the centres. [I. 1914]

21. A compass-needle is in equilibrium in the magnetic meridian.Find the deflection (8) produced by bringing up a small steel magnet toa known distance from the needle, in the direction of magnetic eastand west.

In the above circumstances, shew that the period of small oscillationsof the needle is diminished in the ratio

1. [I. 1908]

22. Obtain the equation T~2Ttiy{IjmH) for the time of smalloscillation of a magnet of moment m swung about its centre of inertiain a horizontal plane in a uniform field of horizontal intensity H, themoment of inertia of the magnet being I.

Assuming the earth to be a sphere uniformly magnetized parallel tothe axis of rotation with intensity M, shew that the time of smallhorizontal oscillation in latitude <j> is -\/(3IirlmM cos <f>). [M. T. 1914]

EXAMPLES 207

23. A compass needle, which may be treated as a magnetic particleof moment p, is free to turn about a vertical axis; and a long bar magnetis placed so that one end, which may be treated as a north-seeking poleof strength m, is in the horizontal plane passing through the needle, ata distance a from the centre of the needle, in a direction between northand west making an angle a with the magnetic meridian. Prove that,if the effect of the other pole of the bar magnet may be disregarded,the needle will rest in a direction between north and east making anangle /? with the magnetic meridian so that

H sin j8 = ma-2 sin (a + /3),where H denotes the horizontal component of the earth's magneticforce.

Prove also that, if the needle is slightly disturbed from this position,it will oscillate in the period

1rt*J{I sin (a + P)/fiH sin a},where I denotes the moment of inertia of the needle about the verticalaxis. [M. T. 1912]

24. Prove that, if the earth be regarded as a sphere of radius auniformly and rigidly magnetized to intensity / , the dip of the magneticneedle in latitude A is tan"1 (2 tan A).

Find the intensity of the uniform field parallel to the earth's axisthat would destroy the dip, and shew that when this field is appliedthe period of oscillation of a small magnet in any plane through the lineof force is \/{irmk2jIM cos A}, where M is the moment of the magnetand mk2 its moment of inertia about its centre. [I. 1932]

25. The portion of a sphere of radius o included between circles oflatitude a, f3 forms a magnetic shell of uniform strength #c. Shew thatthe magnetic force at the centre of the sphere is

— (cos 2a - cos 2/3). [Trinity Coll. 1898]

2 6. The periphery of a segment of a uniformly magnetized sphericalshell of strength T is a circle of radius a and centre A. 0 is the centre ofthe sphere and B is any point on A O or A 0 produced. A small magnet.free to swing in a horizontal plane, is placed at B. If AB is perpen-dicular to the magnetic meridian and 6 is the angle between AB andthe axis of the magnet when in equilibrium, prove that

2wa2T = (a2 + bi)$H cot 8,where 6 is the distance of B from A and H is the horizontal componentof the earth's field. [M. T. 1933]

27. If the earth were a sphere and its magnetism due to two smallstraight bar magnets of the same strength, situated at the poles, withtheir axes in the same direction along the earth's axis, prove that thedip 8 in latitude A would be given by

8 cot (8 + £A) = cot iX - 6 tan £A - 3 tan3 £A. [M. T. 1903]

208 EXAMPLES

28. Treating the earth as uniformly magnetized, and supposing thehorizontal force to be 0-18 in magnetic latitude 52°, find the intensityof magnetization and the dip in this latitude. [St John's College,1913]

29. A uniform plane magnetic shell is bounded by concentric circlesof radii a, b. Shew that it exerts no force at a point distant o*6*/\/ (a* + 6*)from its centre along its axis, and that the total work done in bringinga small magnet to this point from infinity is also zero. [I. 1929]

ANSWERS2. See 8-34.

4. tan-1 {{Mt ~ Ma)/Vs (Mx + M3)}. The stable position is the one inwhich the axis of the magnet is in the direction of the resultantfield of the two small magnets.

6. 2 : 2 V 2 - 1 .7. The meridian plane makes an angle tan"1 (tan 8X cot 8a) with the

first position of the circle.

9. Stable when in the line of centres, unstable when at right anglesto it.

16. Stable when inclined at 60° to the line of centres on the oppositeside to the first; unstable when reversed in direction.

17. The axis of p makes an angletan-1 m (a:2 + za)/{Z (xi + z*)-3x(lx+nz)}

with the axis of x.

21. tan-1-\/{2-M'/r8J5r}, where r is the distance of the magnet M fromthe compass needle and H is the earth's horizontal field.

28. 0-0698; 68° 39'.

Chapter IX

ELECTROMAGNETISM

9*1. It is found that an electric current produces a magneticfield in which the lines of force are closed curves surroundingthe current. If the current is steady, so is the magnetic field.

9*11. Magnetic field due to a current in a long straightwire. If a cardboard disc is suspended with its plane hori-zontal, so that it is free to turn about a vertical wire as anaxis, and a current passes along the wire, it is observed thatwhen a small magnet is placed on the disc, with a suitablecounterpoise to keep the disc from tilting, no motion ensues.

It follows that the forces exerted on the poles of the magnetby the magnetic field due to the current have equal andopposite moments about the axis.

It can further be shewn by sprinkling iron filings on thedisc that the lines of force are circles round the current, for thefilings become magnetized by induction (see 10-3) and setthemselves along the lines of force.

Let the poles of the magnet be of strength — m and m atdistances r', r from the wire, and let H', H denote the magneticforce at the points where the poles are situated. We assume

REM I 4

210 MAGNETIC FIELD DTJE TO A CURRENT [9*11-

that these forces are at right angles to the current and to theradii r', r, regarding this as a fact demonstrated by experi-ments with small magnets and with iron filings. Then, bytaking moments about the axis, we have

so that Hr = HY;or the magnetic force varies inversely as the distance from thewire. The value of the constant Hr must depend on thestrength of the current. This result provides another basis forthe measurement of current, viz. that the current is propor-tional to Hr, where the symbols have the meanings assignedto them above. This introduces a new system of units, whereincurrents are directly related to the magnetic fields theyproduce, and therefore called electromagnetic units. In thissystem of units the product Hr is equal to twice the current inthe wire, or, denoting the current by,?,

H = 2jlr (1).

9*12. Potential. Let us now examine whether the field dueto a long straight current can bederived from a potential function.

Let the wire be taken as the a"xisof z, positive in the sense in whichthe current is flowing. If we takea right-handed set of axes, it canthen be shewn by experimentingwith a small magnet that thepositive sense of the magnetic forceH is from x towards y, or the senseof the field bears to the direction ofthe current a right-handed screw relation.

Then the components of H parallel to the axes areHx=-2jylr\ Hv = 2jx\r\ Hz=0,

so that

i.e. a perfect differential.

9-121] POTBKTIAL DXTB TO A CURRENT 211

Putting 2jta,n-1y/x= —<f>,

we have

so that

and therefore

where

Hx,

tdx + Hyd

H H -Izy -"a

H =

ly + Hzdz =

d<f> d<j>~dx' ~dy

-g rad^ ,

6 = — 2j tan~x yjx +

—d<p,

' dz''

c (1),

and G is a constant; and we note that the inverse tangent isa many-valued function.

9#121. It will be remembered that, when the potentialfunction was first discussed (2#4), we postulated the existencein an electrostatic field of a single-valued potential function,and we have seen no reason to question the existence of sucha function in a magnetostatic field, i.e. a field due to per-manent magnets, but at the outset here of our investigationof the magnetic field produced by an electric current we arefaced with the fact that the field under consideration has nosingle-valued potential function.

Further, an electrostatic field appears to be conservative,in that no work is done by the field on a unit charge whichdescribes a closed path, but that is not so in the case we arenow considering. Thus, let a unit magnetic pole travel rounda circle of radius r with its centre on the wire and in a planeperpendicular to it. The magnetic force is constant and equalto 2j/r and tangential to the circle at every point of the path,

so that the work done = — x 2TTT = 4mj. We shall see that if wer

attempt to measure potential in terms of work we cannotassign to it a single value at a particular point, because thework done in bringing a unit pole up to that point in the fieldwill depend on the path by which it travels, 4TT? being addedto the amount every time the path makes a circuit round thecurrent.

Putting 8 for tan"1?//*, the formula 9*12 (1) can clearly bewritten . o.fl , . ,,,

£ 2jd + 4mj (1).14-2

212 AMPERE'S CIRCUITAL R E L A T I O N [9-121-

Again, in any magnetic field the work done on a unitmagnetic pole moving along any path is given by

J<and, for the field due to a long straight current, this is equal to

^ or 2j(dd.

For a closed path, which embraces the current once,9 increases by 2n and therefore the work done on a unitmagnetic pole is 4nj; but when a pole travels round a closedpath which does not embrace the current the total" increasein 6 is zero, and therefore no work is done.

The theorem here established for a current in a straight wireis a special case of one of the fundamental theorems of electro-magnetism ; viz. In a magnetic field due to steady currents, thework done on a unit pole as it travels round a closed path is 4TTtimes the sum of the currents embraced measured in the right-handed screw sense. This theorem is often called Ampere'sCircuital Relation.

9*13. Let a current j flow in a straight wire whose cross-section is a circle of radius a. If we assume Ampere's circuitalrelation we can find the magnetic field inside or outside thewire. In a plane perpendicular to the wire draw a circle ofradius r with its centre on the axis of the wire. Then, if r < a,this circle embraces current to the amount jV2/a2,* so that, ifH denotes the force tangential to the circle,

2TTrH = 4jTJrilai or H = 2jr/a* (1);

but, if r > a, then

2TTTH = 4:7TJ or H = 2j/r, as before (2).

We notice that the force due to a straight current cannotbecome infinite, because to carry a finite current a wire mustbe of finite cross-section and the formula for H changes from(2) to (1) as we pass from the outside to the inside of the wire.

* We assume that the current is uniformly distributed over thecross-section of the wire. It can be proved that for a steady currentthis is necessarily the case.

9-2] COMPARISON WITH MAGNETIC SHELL 213

9*2. Comparison with magnetic shell. Consider an infiniteplane magnetic shell of strength,?' with a straight edge along theaxis of z and occupying the half of the plane x = 0 for whichy is negative. Let the direction of magnetization of the shellbe the positive direction of the axis Ox.

The potential at a point P on the positive side of the shell isj times the solid angle subtended by the shell at P, and this isthe solid angle between two planes through P, one of whichpasses through Oz while the other is parallel to the plane of the

shell (in order to reach its boundary at infinity). If $ is theangle xOP, the solid angle is 2 (Jrr—6), so that the potential ofthe shell is given by

= -2j0 + const (1).

By comparison with 9*12 we see that the current produces thesame magnetic field as would be produced by a magnetic shellof the same strength having the current for boundary.

When we consider the field of a current in a long straightwire we must bear in mind that all currents are closed circuitsso that we assume that the rest of the current (apart from thestraight part under consideration) is at a great distance, and

214 CURRENT AND EQUIVALENT SHELL [9-2-

this corresponds to the part of the plane shell at a greatdistance.

9*21. It can be shewn experimentally that the magneticfield of a small plane circuit at distances which are largecompared with the dimensions of the circuit is the same as thatof a small magnet whose axis is normal to the plane of thecircuit, and whose moment is equal to the area of the circuitmultiplied by the strength of the current. For if dS be the areaof the circuit, j the current strength, dw the solid angle whichdS subtends at a point (r, 6) referred to the centre and axisof the magnet, the potential of a magnet of moment jdS isjdS cos 8/r2 OTJdu). Small circuits can be produced by makinga small loop on an insulated wire and then doubling the wireback upon itself, so that the flow and return in the wire destroyone another's effects save round the loop.

Then for a circuit of finite size, by drawing lines across it,it can be divided up into a large number of small meshes, and,if a current j is supposed to flowround each mesh, there will be equaland opposite currents in every linecommon to two meshes so that thetotal effect is the same as if a singlecurrent j flows round the boundarycircuit. But, as above, the currentin each small mesh produces the J

same field as a small magnet, and the aggregate of these smallmagnets is a shell of strength j whose boundary coincides withthe given circuit.

We have now arrived at the point at which we can enunciate,as the experimental basis of our theory of the magnetic fieldsproduced by electric currents, the correspondence which wehave established in 9*2 in a particular case; viz. that a currentflowing in a closed circuit produces the same magnetic field as auniform magnetic shell having the circuit for boundary and ofstrength equal to that of the current, and the direction of the currentbears to the direction of magnetization of the shell a right-handedscrew relation.

y

9-3] PARALLEL STRAIGHT CURRENTS 215

The reader will notice that we have left for later considera-tion the question of the field inside a magnet, and the equi-valence of the fields due to a current and a shell must for thepresent be taken to refer to space outside the shell, but in-asmuch as the field due to a uniform magnetic shell only de-pends upon the position of its boundary—in this case thecurrent circuit—we are not imposing a serious limitation.

9*3. Magnetic field due to parallel straight currents.The field due to two or more parallel currents is found bysuperposition.

(i) Flow and return current in parallel wires. Let twostraight parallel wires carrying currents,?' and — j cut the planeof the paper at right angles atA and B. Then if radii fromA, B to a point P make angles6lt 62 with the line BA, thepotential is given by

<l>= - 2jdx + 2j9z + const. (1).

Therefore the equipotentialcurves are

61 — 92 = const.,

or APB = const.;

i.e. the family of coaxial circles passing through A and B. Andsince the lines of force cut the equipotential curves at rightangles, they must be the orthogonal family of coaxial circleswhich have A, B as limiting points.

Hence if C is the centre of the circle APB, PC is a tangentto the line of force at P; and the actual force at P is

H=^-sinAPG- ^'sin BPGri r2

= — COS0O-— COS0,

_2j.AB

where rlt r2 denote the distances AP, BP.

216 PARALLEL STRAIGHT CURRENTS [9*3-

(ii) Equal and parallel currents in the same sense. Withthe same notation, the potential is given by

<f>= -2j(61 + 02) + const (2),

so that the equipotential curves are 61 + 62 = const.; and asin 2-532 it can be shewn that these are rectangular hyperbolaspassing through A, B, with Cassini's ovals r1ri = const, forlines of force.

9'31. Examples, (i) The significant part of an electric circuit consistsof a current C flowing in the sense Oz in a wire along the line x = 0,y = a,and in the sense zO in a wire along the line x=0,y= —a, the co-ordinateaxes being right-handed. Shew that the magnetic potential at a point(x, y, z) due to the circuit is

where A is a constant, and that, if there is in addition a uniform field ofstrength H, parallel to Oz, a magnetic needle at (k, 0,0) will rest in equi-librium making an angle 6 with Ox, where

tan0= 4aC[M. T. 1930]

Let P be the point (x, y, z). Let a plane through P at right angles tothe wires cut them in M and N and let PL be perpendicular to NM.

Then by superposing the fields due to the two currents and using9-12, we may write down the potential as

<f,=A — 2C(MPL) + 2C(NPL), where A is a constant;

or <j>=A -

= A + 2(7 tan-1 %ax•(I)-

Alternatively, we may regard the wires continued to infinity asforming a closed circuit. The mag-netization of the equivalent shellwill be in the positive direction ofOx, in accordance with the right-handed screw rule, so that P is onthe positive side of the shell. Thensince the solid angle which the shellor the circuit subtends at P is twicethe angle NPM, the potential is

N

giving the same result as before.

9'4] CIRCULAR CURRENT 217

At the point (k, 0,0), the field due to the currents is found, from (1),to be flji

and on this is superposed a uniform field of strength H parallel to Oz,so that the resultant at this point makes with Ox an angle

tan-1 H/Hx = tan"1 {H (a2 + Jc*)/4aC}and this is the direction in which a magnetic needle at the point willrest in equilibrium.

(ii) A current j flows in a long straight wire. Find the force and coupleexerted on a small magnet and its period of oscillation about its equilibriumposition.

Let C be the centre of the mag-net, M its moment and K itsmoment of inertia about G.

Take the axis Oz along the wire,the axis Oy through O and letOG = y.

Using right-handed axes, thefield at C is

Hm=-2j/y, ff, = 0, H, = 0.Hence, if the axis of the mag-

net makes an angle 8 with theline of force, its potential energy in the given field is

W=-MHcoaO (8'3)_ 2MjOQB 6" V '

and the force tending to increase y is_8W_ 2Mjcoa6.

~8y~ y1 ''which means that a magnet whose axis makes an acute angle withthe direction of the field at its centre is attracted towards the wire.

Again the couple on the magnet is [MH] by 8*12, or in this case(2Mjsin 6)jy, and if it is free to turn about its centre its equation ofmotion is K8+(2Mjmn6)ly = 0,

and for small oscillations this gives a period

As explained in 8*24 the magnet has two independent principaloscillations but they both have the same period.

9*4. Magnetic field due to a circular current. Let therebe a steady current j in a circular wire of radius a. The potentialat any point P is given by

<[>=jo} + const.,

218 CIRCULAR CURRENT [9-4-

where to is the solid angle which the circuit subtends at P.This takes a simple form when P is on the axis of the circle,but for points off the axismore analysis is required thanwe have at our disposal.

Let Oz be the axis of thecircle and 0 its centre. Thenat a point (0,0, z) on thepositive side of the circuit inrelation to the right-handedscrew rule

<f> = 2TTJ (1 — cos a) + const.,

where a is the angle which aradius of the circle subtends at the point. Therefore

• \+ const (1).

The resultant magnetic force at the point is directed alongOz and is given by

H=—=- = — —a (2),

and this is independent of the sign of z, so that at all points ofthe axis the direction of the force bears to that of the currentthe right-handed screw relation.

At the centre of the circle the force is therefore

2iTrjj(t (3)

acting at right angles to the plane of the circle in the right-handed screw sense in relation to the current. This result isof great importance in connection with galvanometers.

9*41. We may deduce from 9-4 (2) the mechanical force ona small magnet of moment M placed at a point P on the axisof the current with its axis along that of the current.

Suppose that the magnet consists of a pole - m a t (0,0, z)and a pole in at (0,0, z + Sz), so that M = m Sz. The force on the

/ riff \

pole — m is — mH, and the force on the pole m is m I H + y - Sz I.

9-42] CIRCULAR CURRENT 219

Therefore the resultant force on the small magnet is mSz dHjdzor MdHjdz; and taking the value of H given in 9*4(2), thisgives —0>TTJa2zMI{a2 + z2y, the minus sign indicating that asmall magnet, whose axis is directed along the axis in theright-handed screw sense, will be attracted towards the centreof the circle.

The lines of force are closed curves linked with the circuitand lying in planes through the axis, since the field is clearlysymmetrical about the axis. It will be seen in 10*12 that thecross-section multiplied by the force is constant along a tubeof force, and since the tubes of force will be of narrower cross-section in the plane of the circle and inside it than anywhereelse, so the field is strongest inside the circle.

9*42. Example. A pair of circular coils of radii a, b and of m and nturns respectively carry the same current i in the same sense and are placedparallel on a common axis at a distance \ (a + b) apart. Find the magneticforce they produce at a point on the axis distant Jo, %bfrom the two coilsand ishew that on the axis near thispoint the field is so uniform that itsfirst two differential coefficients vanishat the point if mja? = nib2 and its thirdin addition ifa = b,m — n.

[M. T. 1928]

\J

OP

V

Let the point on the axis distant Ja,Jb from the two coils be taken as theorigin O, and let P be a point at adistance x from O on the axis regardedas the axis of x.

The coils produce the same effect as single turns of wire in whichcurrents mi, ni are flowing. Then as in 9*4 the force at P is

H=-

-J.HWIW I i *X JO "X *

= ~V-5a + 5, +

(m nAt O, where a: = 0,

and, at P,TT_^>ri (m / . , 6»

V*5 l« \ 6o 26a3

Hence, by taking the derivatives of H, we find that, when x = 0, if

220 SOLENOIDS [9-42

m/ai=n/b\ both 8H/dx and d^H/dx3 vanish; and further thatvanishes if m/o4 = n/64, which is satisfied when mja2=n/b2 if m=nand a = b.

9*43. Solenoids. A solenoid is a closely wound spiral coilgenerally of flat pitch. Its cross-section may be of any shapeand it may form a cylinder or a ring.

A current,?" in a single plane circuit is equivalent to a planemagnetic shell of strength j whose boundary coincides withthe circuit, and this is equivalent to two sheets of positive andnegative magnetic matter of surface density a and — a, say,at a small distance d apart such that ad =j.

Consider a pile of equal plane circuits in which equal currentsj are flowing in the same sense. The pile of equivalent shellswill be such that the positive and negative sheets neutralizeone another's effects save at the top and bottom where thereare a positive and a negative sheet.

Now consider a closely wound spiral coil of flat pitch andcylindrical form in which a current is flowing. This may beregarded as a set of parallel currents togetherwith a drift along the cylinder, and if the coilis long compared to its breadth this drift couldbe represented by a current along the axis andcould be compensated or balanced by leadingthe current back along the axis. This isAmpere's solenoid as shewn in the figure, con-sisting of a spiral coil of wire connected with awire along its axis, through which the currententers and leaves the coil. The drift of current up the coil iscompensated by the flow down the axis, so that so far asconcerns its external field the solenoid is the equivalent of apile of magnetic shells of the same strength, i.e. of a uniformlymagnetized cylinder.

If the ends are so far distant that their effect is negligible,then the field outside the solenoid is negligible. To find thefield inside the solenoid it is evident that there cannot be acomponent directed from or toward the axis; nor can there bea component at right angles to the radius in a plane perpen-dicular to the axis, for if there were such a component then

9-44] SOLBKOIDS 221

work would be done on a unit pole describing a circle roundthe axis, though such a circle would embrace no current. Hencethe force at every point inside the solenoid must be parallelto the axis. Let A BCD be a rectanglewith short sides AB, DC parallel to theaxis inside the solenoid. Since the fieldhas no component in the directions AD,BC, the only work done on a unit poledescribing the rectangle arises from thesides AB and CD, and if H, H' denotethe strength of the fieldalongu4.Band.DC,since no current is embraced, we have

H.AB-H'.CD = Q, or H = H';

so that the field inside the solenoid is uniform. Next let theside AB of the rectangle be inside and the side CD outside thesolenoid. Then H' is zero, but the rectangle embraces an amountof current AB. nj if n is the number of turns of wire per unitlength, so that H.AB = ±nAB.nj

or H = krrnj

gives the strength of the field inside the solenoid and its direc-tion is parallel to the axis.

9*44. A ring-shaped solenoid. In this case the magneticshells equivalent to the separate turns of wire are boundedby planes passing through the axisof the ring and as the coil returnsinto itself there is no free polarityand no external field. To find theinternal field, let H be its value ata point P at a distance r from theaxis; then considering that an axialplane through P divides the solenoidinto pairs of turns of wire symmetri-cally situated with regard to thisplane and that any such pair would give a resultant force atright angles to the axial plane, it follows that H is at right anglesto the axial plane. Then if a unit pole travels round a circle of

222 POTENTIAL ENERGY [9-44-

radius r with its centre on the axis and its plane perpendicularto the axis, the total current embraced is Nj, where N is thetotal number of turns of wire in the coil and j the current inthe wire. Therefore

2TTTH = ±TTNJ or H = 2jjr.

9*5. The potential energy of a uniform magnetic shellof strength j in a given magnetic field. The potential atany point due to the shell is jw, where co is the solid anglesubtended by the shell at the point, provided that linesdrawn from the point to the shell meet it first on its positiveside. This expression^ is therefore also the potential energyof a unit magnetic pole placed at the assigned point in thefield of the shell, because it is the workwhich an operator would have to performin order to place the unit pole in position.

Now the total flux of force which pro-ceeds from a unit pole is 4TT and the amountof it which falls within a cone of solidangle w, with its vertex at the pole, is w;therefore the potential energy jca of theunit pole is j times the flux of force fromthe unit pole which passes through theshell from its positive to its negative sideor —j times the flux of force from the unit pole which passesthrough the shell from its negative to its positive side.

Conversely this is also the potential energy of the shell ofstrength j in the magnetic field produced by the unit pole.

Then by superposition the potential energy of the shellin a field due to any given distribution of magnetic poles is

-jN,

where N is the total flux of force of the given field which passesthrough the shell from its negative to its positive side.

From the equivalence of a current circuit and a magnetic shell, weinfer that the same expression would represent the potential energyof a closed circuit carrying a current j in a given magnetic field, Ndenoting the flux of force of the given field which threads the circuitin the right-handed screw sense. But if we use this expression in

9-6] MECHANICAL FORCE ON A CIRCUIT 223

order to deduce the mechanical force exerted by the field on thecircuit, we must be careful to note that the current must be supposedto remain constant during a virtual displacement, for it is only onthis hypothesis that — jN represents the potential energy of theequivalent shell.

9*51. We may also obtain the result of 9*5 by considering theuniform magnetic shell to be a number of small magnets with their axesnormal to the surface. Consider a small element of the shell of area SS.If I is the intensity of magnetization,the element of the shell is equivalent -V___± i i*? *t Ito a small magnet whose poles are of C^r—rr—~—— ' J~^ZJstrength — ISS and ISS at a distancet apart, where t is the thickness of the shell; i.e. a magnet of momentItSS orjSS, since It=j.

By 8-3 the potential energy of this small magnet is —jHnSS, whereHn is the component of the force of the given field in the direction ofthe axis of the magnet, i.e. normal to the shell from its negative to itspositive side. Therefore the potential energy of the whole shell

= -jJHndS=-jN,

where N is the total flux of force of the given field which traverses theshell from its negative to its positive side, or in the right-handed screwsense in relation to the equivalent current.

9'6. To find the mechanical force on a circuit carryinga current in a given magnetic field. Since the potentialenergy of such a circuit is given by

W=-jN (1),therefore a displacement which increases N decreases thepotential energy.

Let the circuit be displacedthrough a small distance Sx in anydirection, an element PP' of lengthSs tracing out a small parallel-ogram PQQ'P'. If *, s' denote thetwo positions of the circuit and8, 8' surfaces of which they are theboundaries, then 8, 8' and theaggregate S" of the small parallelograms such as PQQ'P' bounda closed region of space and, as will be seen in 10*11, the totalflux of magnetic induction (or, in air, magnetic force) from

224 MECHANICAL FORCE ON A CIBCUIT [9-6-

any such region is zero. Hence, if N, N', N" denote the fluxof force of the given field through S, 8' S" taken in the samesense, we have

N-N' + N" = O or 8N = N'-N = N".Hence if X denotes the resultant force on the circuit in

the direction 8*,X8x=-8W=j8N=jN".

But the contribution to N" arising from the element ofarea PQQ'P'

— area PQQ'P' x component of H perpendicular to this area= volume of parallelepiped of edges 8x, 8s, H= 8s.H sin d. 8a; cos e,

where 6 is the angle between 8s and H, e is the angle be-tween Sx and the perpendicular to 8s and H, and the positivedirection of this perpendicular is along the axis of a right-handed screw twisting from the direction of 8s to that of H.Therefore r

X8x=j8x\ Hsindcoseds

integrated round the circuit, or

X =j\ Hsin6 cos eds.

But if each element 8s of the circuit were acted upon bya f o r c e j8sHSin6 (2)

at right angles to 8s and H in the sense of the axis of a right-handed screw turning from mechanicalthe direction of j to that of y^forceH, the sum of all these forcesresolved in the direction 8a; current „would be the integral ex-pression for X. Since thedirection of 8a; is arbitrary, weconclude that the mechanical ^ - ^ magnetic forceeffect of the field on the circuit is the same as if every element8s was subject to a force j 8s H sin 8 in the direction andsense just indicated.

9-62] PLANE CIRCUIT IN A UNIFORM FIELD 225

The formula (2) with its accompanying rule of signs isknown as Ampere's Law for the force on an element of acircuit.

I t will be noticed that Ampere's Law does not give a repre-sentation of the resultant reaction of a magnetic field on acircuit, but leaves the resultant to be calculated by com-pounding the forces on the different elements. There are how-ever cases in which the resultant can be calculated directlyfrom the expression for the potential energy, as will be seenin the next article.

9*61. Couple acting on a plane circuit in a uniformmagnetic field. A plane circuit of area A carries a current jin a uniform magnetic field ofintensity H. The potentialenergy of the circuit is

W = -jN= -jAHsiad ...(1),

where 8 is the angle which theplane of the circuit makes withthe direction of the field H.

Any displacement which keeps the plane of the circuitparallel to itself does not alter N, so that there can be no forcetending to produce such a displacement. But there is ingeneral a couple tending to increase 9 measured by

(2).

If the circuit is free to rotate, it will set itself in equilibriumso that this couple vanishes; i.e. so that 8=\n, or the plane ofthe circuit is perpendicular to the direction of H and thecircuit then embraces as large a flux of force of the externalfield as possible.

9*62. Force between parallel straight currents. Let twoparallel wires carry currents,/, f flowing in the same directionand be at a distance a apart.

Let PP' be an element Ss of the wire carrying the current j .REM 15

226 FORCE BETWEEN PARALLEL CURRENTS [9-62-

H

The magnetic force H at P in the field produced by the currentj ' is given by H = 2j'/«, and it is at right angles to the plane ofthe parallel wires. Hence in accordancewith Ampere's Law (9*6) the elementPP' is acted upon by a force jSsH or2jj'8s/a directed towards the other wire.Therefore when the currents are in thesame sense there is a mutual attractionof amount 2jj'/a per unit length betweenthe wires.

If one of the currents is reversed, itis easy to see that the force is a repulsion.

9*63. Example. A current j flows round a circular wire of radius a,and a current j ' flows in a long straight wire in the same plane as thecircle and at a distance cfrom its centre. Determine the force between thecircuits.

Let PP' be a small arc of the circle of length 8s. At P the magneticforce due to the current j ' isH = 2j'/r, where r is the distance of P fromthe straight wire. The direction of H is at right angles to the plane of

the paper so that it is at right angles to the element j8s of current, andthe mechanical force on the element PP' of the wire is therefore j Ss H>or 2jj'hsjr, and in accordance with Ampere's Law it is directed alongthe radius outwards, provided that j is in the sense indicated in thefigure.

With </i denoting the angle between the radius CP and the perpen-dicular CO' to the straight wire, we have r=c—a cos ip and 8»=a$ip

9-7] CIRCULAR CURRENT 227

Also the resultant force on the circle can. easily be shewn to act in theline GO and its magnitude is

This force is an attraction when the currents are directed as in thefigure, i.e. so that the direction of the flux of magnetic force throughthe circle is related to the direction of the current round it in the right-handed screw sense j and we notice that, as in 9*61, the closed circuittends to move so as to increase the flux through it of magnetic force ofthe field in which it is located.

9-7. Magnetic field produced by a given circuit. Insteadof deducing the magnetic force from the potential function jw,we may often make use of Ampere's Law as given in 9'6. Takingthe formula jSs.Hsind for the mechanical force on theelement Ss of the circuit, suppose that the field H is due to asingle unit pole. Then the reactionof each element of the circuit onthe pole will be equal and oppositeto the reaction of the pole on theelement. Consider for examplethe magnetic force at a point onthe axis of a circular current (9*4).

Let 0 be the centre of the circle,a its radius and Q a point on theaxis at a distance z from 0. Aunit pole at Q would produce atP a force H=ljQP2 along QP.Therefore the mechanical forceon an element PP' of the circlecarrying a current j isj.PP'/QP2, at right angles to both QPand PP', i.e. along PO, in the figure. Hence the element j . PP'of the circuit would exert on the pole at Q a force of the samemagnitude in the opposite direction. To get the resultant forceon the pole we must resolve along the axis, i.e. multiply byain.OQP or OP/QP; so that the element j.PP' makes acontribution j . PP'. ajQP3 to the resultant force on the pole,and by summing for the whole circle we see that the force is

It must be observed however that there is something15-2

228 GALVANOMETERS [9-7-

unsatisfactory about this Hne of argument although it leadsto the correct result. We assume that the reaction betweena magnetic pole and an element of a current circuit consistsof equal and opposite forces acting at right angles to the Hnejoining them—rather a curious form of action at a distance;and we cannot demonstrate that the resultant force due tothe whole circuit is necessarily resolvable into componentsassigned to its elements in this way. It is as well to rememberthat an element of current, except in so far as it is a streamof electrons, has no independent existence, nor has a soHtarymagnetic pole. Both are convenient mathematical fictions.

9*8. The tangent galvanometer. Galvanometers are in-struments for measuring the strength of currents. The sim-plest—the tangent galvanometer—consists of a circular coilof wire through which the current canpass, placed with its plane vertical inthe plane of the magnetic meridian.There is a compass needle free to turnabout a vertical axis with its centre atthe centre of the coil.

If the coil is of mean radius a andconsists of n turns of wire, the passage _/of a current of strength j will producea magnetic force at the centre of magni-tude 2-irnJla at right angles to the planeof the coil (9-4).

When no current is passing the needle rests in the plane ofthe coil, but when a steady current j is passing the needle isdeflected and takes up an equiHbrium position in which thecouple due to the earth's magnetic field is balanced by thecouple due to the magnetic field produced by the current.Thus if H denotes the horizontal component of the earth'smagnetic field and 6 the angle of deflection, sometimes caUedthe 'throw of the needle', we have

(1).

Thus j can be determined by measurement of 6 when the

9-82] GALVANOMETERS 229

other constants are known. If we write j = G tan 6, G may becalled the constant of the instrument.

9-81. The sine galvanometer. If the coil of the instrumentdescribed in 9*8 can turn about a vertical axis through itscentre, and if, when a current passes, theplane of the coil is rotated until it over-takes the needle, then since the directionof the needle in equilibrium is that of theresultant of the two fields H and 2irnj/a,we must have

H sin 8=27rnjla, _/

or sin 6=2irnj/aH;

and an instrument of this kind is called \2Trnj/aa sine galvanometer.

9*82. Examples, (i) The poles of a battery (E.M.F. 2-9 volts andinternal resistance 4 ohms) are joined to those of a tangent galvanometerwhose coil has 20 turns of wire and is of mean radius 10 cm. Shew thatthe deflection of the galvanometer is approximately 45°. The horizontalintensity of the earth's magnetic force is 0-18 dyne and the resistance ofthe galvanometer is 16 ohms. [M. T. 1898]

The current j is determined by Ohm's law; thus

gives j = 0-145 amp.= 0-0145 absolute E.M.F. electromagnetic unit

of current (7*33).

™_ „ 2-nnj 2TT x 20 x 0-0145Then tanfl = = 1 Q x ( M 8

= 1-01,so that 8 is approximately 45°.

(ii) In a tangent galvanometer, the sensibility is measured by the ratioof the increment of deflection to the increment of current, estimated perunit current. Shew that the galvanometer will be most sensitive when thedeflection is Jw, and that in measuring the current given by a generatorwhose electromotive force is E, and internal resistance R, the galvanometerwill be most sensitive if there be placed across the terminals a shunt ofresistance HRr/{E—H(R + r)}, where r is the resistance of the galvano-meter, and H the constant of the instrument.

What is the meaning of the result if the denominator vanishes or isnegative? [M. T. 1899]

230 MOVING COIL GALVANOMETER [9-82-

Let j = H tan 8 give the current in terms of the deflection. Theincrement of current per unit current is Sj/j, so that the sensibility is

and this is greatest when 8 = \TT.Hence the galvanometer is most sensitive

when the current through it is H.Let x be the resistance of the shunt and C + H

C the current through it, so that the totalcurrent through the battery is C + H.

Then we haveE = R(C + H)+rH (1)

and xO = rH (2).

Therefore E =

and x=HRrl{E-H{B+r)}.If the denominator vanishes or is negative we see that E is only just

sufficient or insufficient to drive a current H through the galvanometerwithout a shunt.

9*83. The moving coil galvanometer. The disadvantagesof the tangent galvanometer are that it requires an accuratedetermination of the horizontal component of the earth'smagnetic field in an assigned position in thebuilding in which it is to be used and that theinstrument may be much affected by other localmagnetic fields such as those due to neighbouringelectric currents.

In the moving coil galvanometer the coilcarrying the current moves in a fixed strongmagnetic field. A coil, rectangular in shape, con-taining many turns of fine wire is suspendedbetween the poles of a horse-shoe magnet. Thecurrent enters the coil through a fine wire bywhich the coil is suspended; and it is the elasticproperties of this suspending wire, whose upper end is fixed,which resist the rotation of the coil. The current leaves thecoil through a fine wire attached to its lower end.

When a current passes through the coil it tends to set itselfat right angles to the magnetic field. For as was shewn in 9'5and 9*61, if A be its mean area and H the magnetic force sup-

9-84] MOVING OOIL GALVANOMETEB 231

posed uniform, the potential energy when a current j passesthrough the coil is w= _njAHsin0 ( 1 ) j

where 6 is the angle between H and the plane of the coil andn is the number of turns of wire.

It follows that there is a couple tending to increase 6 given by

- dW/d9 = njAH cos 6 (2).

This couple is resisted by the torsion of the suspending wire.The torsion in a straight wire one end of which is fixed whilethe other is twisted about the axis of the wire produces acouple proportional to the angle turned through.

Suppose that the coil is suspended in such a way that thereis no torsion in the wire when the plane of the coil is approxi-mately parallel to the magnetic field, and that we measure 6from the position in which the torsion is zero. Then when thecoil is turned through an angle 6 there will be a torsion coupleproportional to 6, say X9, opposing its rotation; and if theturning is due to the passage of a current j , the coil will takeup a position of equilibrium in which the couple given by (2)is balanced by the torsion couple, i.e.

njAH cos 0 = X6,

or j = \6lnAHcoa6 (3).

For small deflections cos 6 may be taken as unity, and then,; varies as 6, or . = kg ^

where k is the constant of the instrument.A good instrument will measure a millionth of an ampere

and a very sensitive one will measure a billionth or 10~12 amp.

9*84. The ballistic galvanometer. This is an instrument formeasuring the total quantity of electricity which passes in atransient current; for example, in the discharge of a condenser.

If we use a moving-coil galvanometer for this purpose andlet / denote the moment of inertia of the coil about its axis,its equation of motion during the passage of a current j is

IS=njAHcosd-X6 (1),

using the notation of 9*83.

232 BALLISTIC GALVANOMETEB [9-84-

For a transient current, the passage of the current is com-pleted before the coil has time to move, so that we may putcos 9— 1, in the preceding equation, and write

I6'=Gj-\6 (2),where G=nAH.

Integrating with regard to t through the duration of thecurrent, we have » ,.

I&=G\jdt-\\6dt,

were Mdtf = the total charge which passes = c, say; and \8dt

is negligible, so that I$ = Ge (3)

gives the initial velocity with which the coil begins to oscillateafter the passage of the current.

There is now no longer any current passing and the coiloscillates under the influence of the torsion couple alone, withan initial velocity given by (3).

The equation of motion is now

Id=-M (4),whence we get lfi=C-A02,

and d = Ge/1 when d = 0, so that C = G*ezjl, and

ifi=^-\ez (5).

Hence, if a be the amplitude of the oscillation, i.e. themaximum value of 6, Q2P2IT^\ 2.

or *=vWg (6)gives the quantity of electricity which passed through the coil.

Again, from (4), the period T of oscillation of the coil isgiven by T = 2ir\/(IIX), so that (6) may also be written

T A

Tor e = 2 ^ a

where k = X/G is the constant of the instrument.

9-85] BALLISTIC GALVANOMETER 233

9*85. To use a tangent galvanometer as a ballisticgalvanometer. Let M be the moment of the magnet, I its momentof inertia, H the horizontal component of the earth's magnetic field,and A the magnetic force at the centre of the coil when a unit currentpasses.

Let a transient current,;'pass through the coil. This produces a fieldjA perpendicular to the plane of the coil and while the magnet is stillin this plane its equation of motion is

I8=JAM (1).By integrating with regard to t through the duration of the current,

we get ,I6O = AM jdt = AMe (2),

where 60 is the angular velocity with which the needle begins to oscillateand e is the total charge which passed through the coil.

In the subsequent motion there being no longer any current theneedle oscillates in the earth's magnetic field and its equation ofm o t i o n i s Id=-HMsin6 (3),

which gives on integrationI&* = l60

i-2HM {l-coa6) (4),since 6 = (Jo when 8 = 0.

Hence if a be the amplitude, or maximum value of 6,

or, from (2), A*MH%\I = 4HMsin2 £a;

But, from (3), the time T of small oscillations is given by

so that (6) can also be writtenTH

e = —-. sin \a. (6).

EXAMPLES

1. Shew that the magnetic potential at P due to a current i in aclosed circuit is equal to iQp, where Op is the solid angle subtended atP by the circuit. Deduce the value of the magnetic potential andmagnetic force near a long straight wire, the rest of the circuit beingvery distant. [M. T. 1929]

2. Two equal magnetic poles repel one another with a force of 80dynes when a decimetre apart. A current through a coil of wire ofradius 3 cm. consisting of 4 turns of wire exerts a force of 5 dynes onone of the poles placed at the centre. Find the strength of the currentin amperes. [I. 1906]

234 EXAMPLES

3. Two infinitely long conducting wires A and B are parallel to theline ZOZ', and intersect the plane XOY (perpendicular to ZOZ') inx=a, y = 0, z = 0 and x= — a, j /=0,« = 0 respectively. Find the forceacting on a unit north magnetic pole situated at (i) (0, 0,0) and(ii) (3a, 0,0), when A carries a current I, and B a current 21 in theopposite direction. Draw a diagram to illustrate the distribution ofthe lines of magnetic force in the plane XOY around the wires whenthey carry equal currents (a) in the same direction, (6) in oppositedirections. [M. T. 1926]

4. Currents of strength i flow in opposite directions in two longparallel wires at distance 2a apart. A small magnet, of magneticmoment M and moment of inertia I, is mounted with its centre equi-distant from the two wires and distant b from the plane containingthem. If the magnet is free to swing in a plane perpendicular to thewires, shew that its period of oscillation is ir {I (a2 + b%)l(Mia)}*.

[M. T. 1935]

5. Four long vertical wires intersect a horizontal plane in A, B, Oand D which are the vertices of a rectangle such that AB = 2AD = 2a.The currents in the wires through A, B, C and D are — i, —j, +j and+ i respectively. Prove that the period of oscillation of a small magnetof moment p and moment of inertia I about its position of equilibriumwhen placed at the mid-point of AB and free to move in the planeABGD is 2 J 2 j

4 } [M.T.1933]

6. Three long parallel straight wires carrying currents of the samestrength/ meet a plane at right angles in the corners of an equilateraltriangle ABC and O is the centre of the triangle. Shew that themagnetic force at any point P in the plane is

6j.OP2/AP.BP.CP.Also shew that the wires are all acted upon by forces tending to

move them inwards, the force per unit length on each being 2j2j0A.[M. T. 1924]

7. A current I flows in a circular wire of radius a. A circular coil ofradius 2a coplanar and concentric with the wire is added. It haseight turns, and the same current I circulates in it in the sense oppositeto that in the wire. Shew that the magnetic field at the centre isnumerically three times as strong as it was before, and that the fieldat points on the axis in the neighbourhood of the centre is nearlyuniform. [M. T. 1932]

8. If the force normal to the plane of a circular coil of n turns andradius a which carries a current i is measured on the axis of the coilat a small distance h from the centre of the coil, shew that it is smallerthan at the centre by the fraction %hi\a%. [M. T. 1927]

EXAMPLES 235

9. A coil of wire is wound closely on a cylinder of n turns per unitlength. The diameter of the cylinder is equal to its length. Shew thatthe intensity at the centre of the cylinder is

[M. T. 1934]

10. Two circular coils A and B are set in planes at right angles toone another; the centres of the coils are coincident and their commondiameter is vertical, and the coil A is set with its plane in the magneticmeridian. Under the influence of a current in A a small horizontalmagnet at the centre is in equilibrium at an angle of 60° with the meri-dian. A current is then passed through B such that the deflection of themagnet is reduced to 45°. If H is the horizontal component of theearth's magnetic field and the intensities of the fields at the centreproduced by these currents separately are respectively H1 and J5Ta,find the values of the ratios H^.H^-.H. [M. T. 1911]

11. Two equal circular coils A and B^are mounted on a horizontalboard with their planes vertical and perpendicular to one another,their centres being in the same horizontal plane and equidistant fromthe line of intersection of their planes. A compass needle is pivoted atthe intersection of the axes of the coils. When the board is placed sothat the plane of A is in the magnetic meridian and currents GA, Cgare passed through A, B respectively, the deflection of the compassneedle is 6t. The board is now rotated through a right angle in thedirection from north to west, and the deflection is found to be #a.(Deflections of the needle are counted positive when its north polemoves toward the west.) Shew that

l + cotea IC» l - C O t 0 x j

In what circumstances does this fail as a method of comparison ofthe two currents? [M. T. 1923]

12. A battery of electromotive force E and internal resistance A,when connected to a tangent galvanometer of resistance R, gives adeflection a. A second battery of the same B.M.F. but of internal resist-ance B, when connected to the galvanometer gives a deflection /S.When the two batteries are connected in series to the galvanometer,the deflection is y. Shew that

A _ r, 2 cot y — cot j8"~ cota + cot|S —2coty'

and that the current through the galvanometer when the batteriesare connected to it in series is

E cot«+cotp-2cot y T_R COty L J

13. A given current sent through a tangent galvanometer deflectsthe magnet through an angle 8. The plane of the coil is slowly rotated

236 EXAMPLES

round the vertical axis through the centre of the magnet. Prove that,if 8 > \rt, the magnet will describe complete revolutions, but if 8 < JTT,the magnet will oscillate through an angle sin-1 (tan 0) on each sideof the meridian. [M. T. 1906]

14. A tangent galvanometer is to be constructed with a coil con-sisting of five turns of thick copper wire, so that the tangent of theangle of deflection may be equal to the number of amperes flowing inthe coil.

Shew that the radius of the coil must be about 17-45 cm., assumingthe earth's horizontal magnetic force to be 0-18 dyne. [I. 1895]

16. Shew that, if a slight error is made in reading the angle ofdeflection of a tangent galvanometer, the percentage error in thededuced value of the current is a minimum if the angle of deflectionis I*. [M. T. 1907]

16. The circumference of a sine galvanometer is 1 metre, the earth'shorizontal magnetic force is 0-18 C.G.S. unit. Shew that the greatestcurrent which can be measured by the galvanometer is 4-56 amperesapproximately. [M. T. 1896]

17. If there be an error a in the determination of the magneticmeridian, find the true strength of a current which is i as ascertainedby means of a sine galvanometer. [M. T. 1903]

18. Find the electromotive force of a Daniell cell, in C.G.S. units,from the following data. Five Daniell cells were connected in series toa tangent galvanometer, the coil of which had 10 turns of 11 cm. radius.The deflection produced was 45°. The total resistance of the circuitwas 16-9 x 10' C.G.S. units, and the horizontal component of the earth'smagnetic field at the point was 0-180 C.G.S. unit. [M. T. 1917]

19. A battery of 8 cells, placed 2 in series in 4 sets in parallel, isconnected with a galvanometer of 100 ohms resistance, which isshunted with a resistance of 100/9 ohms. Each cell has E.M.F. 2 voltsand internal resistance 12ohms. Find the deflection of the galvanometer,if a current 0-0005 ampere gives a deflection of one division.

[M. T. 1916]

20. Two galvanometers are wound respectively with 2 turns of wireof resistance 0-1 ohm per turn and 20 turns of wire of resistance 1 ohmper turn. The coils can be regarded as all having the same radius andtheir centres at the centre of the suspended magnet. Which galvanometerwill shew the greater deflection when connected to a cell whose internalresistance is 1 ohm? [M. T. 1908]

21. The coil of a tangent galvanometer consists of 2 turns of wire ofradius 20 cm. If the horizontal component of the earth's magnetic

EXAMPLES 237

field is equal to 0-17 dyne per unit pole, find the deflection due to acurrent of one-tenth of an electromagnetic unit.

Assuming that the magnetic field due to the earth is that due to asmall magnet placed at its centre, shew that the horizontal componentin magnetic latitude A varies as cos A. Hence find how the deflection ofthe galvanometer varies with latitude. [M. T. 1924]

22. If the coil of a tangent galvanometer consists of n turns of wireof radius a and cross-section A and specific resistance K, and a currentis sent through it from a cell of internal resistance B, and electromotiveforce E, find the deflection of the galvanometer.

Shewthat if the radius, specific resistance and the total cross-sectionnA are given, there will be definite values of A and n for which thedeflection is a maximum. [St John's Coll. 1913]

23. The plane of the coil of a tangent galvanometer made a smallangle with the magnetic meridian. A current was passed through thegalvanometer and the deflection of the needle from the magnetic meri-dian was 8. A second reading was made with an equal current but withhah0 as many turns of wire as were used before, and this time thedeflection was 8'.

Shew that the current passing through the galvanometer was

77-—TSI T~n approximately,O cotfl —cote " ^ J

where Q is the constant of the galvanometer when the full number ofturns are used, and H is the horizontal component of the earth'smagnetic force. [M. T. 1934]

24. A tangent galvanometer has two parallel coaxial coils of radiusa cm., each wound with N turns of fine wire. Their common axis pointsE. and W. They are distant a cm. apart and are connected in series.The magnet is suspended on the axis of the coils, midway betweenthem. Its length is small compared with a. Shew that if H is the hori-zontal component of the earth's magnetic field, in dynes per unit pole,

and if 8 is the deflection of the magnet, the current flowing is —=r=— tan 0

amperes. [M. T. 1925]

25. The terminals of a battery (B.M.F. E), a tangent galvanometer anda voltameter are connected in parallel arc, the resistance of each arcbeing B, G, and U respectively. The deflection of the galvanometer isobserved to be 8. After a time the battery arc is thrown out of thearrangement. Prove that the deflection of the galvanometer drops toi, where (E-KGt*xi8) U

and K is the reduction factor of the galvanometer. [I. 1902]

26. The coil of a moving-coil galvanometer is a square of side 1 cm-with 200 turns. The coil can rotate about a vertical axis and the

238 EXAMPLES

restoring couple is 10-* gm.-cm. per degree. The field in which the coilswings is 2 x 103

C.G.S. electromagnetic units. Find approximatelythe angular deflection due to a current of 10~4 amperes. [M. T. 1935]

27. If M is the moment of the magnet of a ballistic galvanometer,K its moment of inertia about its axis of rotation, H the intensity ofthe earth's magnetic field and A the magnetic intensity at the centreof the coil due to unit current, shew that neglecting frietional forcesthe quantity of electricity passing through the coil which causes themagnet to swing through a small angle 8 is Q, where

Find the relation connecting Q with 6, if there is a retarding frietionalcouple [j.6 on the magnet. [M. T. 1931]

28. Find (i) the couple exerted on a small circuit of m turns of areaA carrying a current i placed at the centre of a large circular coil ofradius a and n turns carrying a current,/, the planes of the coils beingat right angles;

(ii) the attraction between two circuits of areas A and B carryingcurrents i and,/ placed parallel to one another at a large distance apartand on a line perpendicular to them both;

(iii) the attraction between two long parallel straight wires carryinga flow and return current i. [M. T. 1934]

29. A thin bar magnet of moment M and length 21 lies along the axisof a circular wire of radius a in which a current i is flowing, the planeof the circle bisecting the bar. Prove that the reaction of the magneton the circuit causes a longitudinal stress in the wire of magnitudeMia/(a2 +1*)%. Determine in what cases the stress is a tension or a thrust.

[I. 1927]

ANSWERS

2. 0-00066 amp. 3. (i) 6I/a, (ii) 0. 10. V3:VS- 1:1.

17. i(l + xcot8), where 0 is the reading of the instrument.

18. 1-06 volts. 19. 60 divisions. 20. The former.

21. 20° 16'. 22. t&vr1{2TmAEjHa(BA + 2Trma)}.

26. 4°.

_„ n 6 2K //HK\ cosV(4HMK-ii.2)t a27. Q = je ^/{W), where VV ^ ^ = g28. (i) 2irrrmAijja. (ii) QABvj/d4; where d is the distance between the

circuits, (iii) — 2i2/d, where d is the distance between the wires.

29. Thrust or tension according as the sense of the current in relationto the axis of the magnet is that of a right-handed or a left-handedscrew.

Chapter X

MAGNETIC INDUCTION ANDINDUCED MAGNETISM

iO'l. Magnetic induction. In Chapter vni we limited ourconsiderations to the magnetic fields external to the magnetswhich produce them; and we saw in 8*12 that the magneticforce H was really the mechanical force acting upon a unit poleat the point under consideration and we have assumed thatin airHis the negative gradient of a potential function. In orderto define in this way the magnetic force at a point inside amagnet, it is necessary to imagine there to be at the pointunder consideration a small cavity inside the magnet, in whicha unit pole might be placed.

Let the cavity be cylindrical with its axis in the direction ofthe intensity of magnetization I. Let the length be 21, and letthe cross-section be such that its linear dimensions are smallcompared to I. Let a denote the 2Zarea of the cross-section. The re- /q() Q—Iamoval of the cylindrical portion of *" *the magnet will leave free polarity on the ends of the cylinder,equivalent to poles of strengths la. and — Ice at the ends. Theunit pole at the centre of the cavity may be regarded as in airand subject to a force H together with the force 2/a/Z2 in thedirection of I due to the polarity of the ends of the cylinder.But by hypothesis oc/Z2 is negligible, so that inside such a'needle-shaped' cavity the force per unit pole is independentof I and is simply the vector H. Hence we may define themagnetic force inside a magnet as the force per unit pole ina needle-shaped cavity whose axis is in the direction of themagnetization I.

If however we take the length of the cylindrical cavity to beshort compared to the linear dimensions of its cross-section,the free polarity on its ends produces a field of force on the axisof the cavity comparable with that produced by two infinite

240 MAGNETIC INDUCTION

plane sheets of surface density / and — / , viz. 4TT/.the total force per unit pole at the centre of thiscavity is the sum of two vectors H and 4TTI.

This vector is called the magnetic inductionand denoted by B. Thus

B = H + 4TTI (1);

and the magnetic induction is therefore the resultant force perunit pole in a disc-shaped cavity whose axis is in the direction ofthe magnetization.

We note that B and H are not necessarily in the same direc-tion, but that they will be so whenever I coincides in directionwith H.

We notice that outside the magnets, B = H since 1 = 0; i.e.in air magnetic induction is the same as magnetic force.

This is Clerk-Maxwell's method of defining magnetic in-duction. The vector B can also be introduced by an analyticalprocess and defined by the relations which it has to satisfy.

10*11. The total flux of magnetic induction out of anyclosed surface is zero. We have to prove that for any closedsurface in the magnetic field

JBndS =

where Bn denotes the outward normal component of themagnetic induction B.

Since B = H + 4TTI, therefore

(1).

Since the magnetic force H is related to magnetic polarity inexactly the same way as the electric intensity E is related toelectric charge, it follows that, by analogy from Gauss'sTheorem, .

\Hnd8 = ±TrM (2),

where M is the algebraical sum of the strengths of themagnetic poles inside S. But the sum of the strengths of thepoles of every elementary magnet is zero. Therefore the only

10-12] ELTTX OF INDUCTION 241

elementary magnets which contribute to the total M arethose which are intersected by the surface S in such a waythat one pole lies inside and one outside the surface.

Consider a small magnet whose section by the surface is anelement of area dS. The axis of the magnet is in the directionof the intensity of magnetization I; let this, for the sake ofprecision, make an acute angle e with the outward normal tod8. The negative pole is then inside S and the positive poleoutside. If I be the small length of the magnet, its volume isidS cos e and its moment is therefore IldS cos e, so that thestrength of its negative pole is — I coa ed8 or —IndS. Hence

the sum of the poles inside 8 is — IndS, so that, by sub-

stituting this expression for M in (2), we get

and therefore from (1) »)BndS=0 (3).

Since this relation is true for integration over every closedsurface that can be drawn in the field, it follows from thedefinition of divergence (1-52) that

divB = 0 (4)

at every point of the field; therefore magnetic induction is asolenoidal vector throughout the whole field (see 5*21).

10*12. Tubes of induction. Lines and tubes of inductionmay be defined like lines and tubes of force. Applying thetheorem of 10*11 to a tube of induction bounded by two cross-sections as in 2*31 (ii), we see that the induction multiplied bythe cross-section is constant along the tube. But whereas in anelectrostatic field tubes of force terminate on the surfaces ofcharged conductors, in a magnetic field a tube of inductioncannot have an end, for if such tubes could terminate therecould be closed surfaces for which the theorem of 10-ll wouldnot be true. Consequently all such tubes are re-entrant, alllines of induction forming closed curves. In air of course linesof induction are the same as lines of force, since in air B = H.

REM 16

242 FLUX OF INDUCTION [10-12-

In the case of a bar magnet whose positive and negativeends are P, Q, by experimenting with iron filings, it is easyto shew that the lines of force in air proceed from P to Q;but in air these are also lines of induction. It follows thattubes of induction pass from the end P in air round to theend Q and then enter the magnet and pass along it back to P,so that inside the magnet the induction B is directed fromQ towards P, though the magnetic force H inside the magnetis clearly directed from the positive end P towards the negativeend Q. Also, since H has a definite value at all points of thefield, it follows that the tubes of induction fill the whole field,and whilst all of them are crowded into the magnet, passingalong it from Q to P, they widen out considerably in air.

10*13. The amount of induction across any surface in agiven magnetic field depends only on the boundary curve.For if 8lt S2 denote any two surfaceshaving a common boundary curve s,then since 8lt 82 together form theboundary of a closed region, by 10*11

when the integral is taken over 81 and8Z and Bn is the outward normalcomponent over the whole surface.If therefore we reverse the sign over8Z so that induction is measured inthe same sense over both 8X and 82 in relation to the curve s, wehave .

Bnd8=\ Bnd8

so that the amount of induction is the same for all surfaceshaving the same boundary curve.

10*2. Continuity of normal induction. Let 8 be the surfaceof separation of two media in a magnetic field, such as theboundary surface of a magnet in air or in contact with a blockof iron. Let B l t B2 denote the magnetic induction at points

10-3] INDUCED MAGNETISM 243

close to one another and on opposite sides of the surface. Byapplying the theorem of 10-11 to a cylinder with ends parallelto the surface and so short that the induction through itssides is negligible, we get

(Bn)x = {Bn)2 (1).

or the normal component of induction is continuous.This result might also be inferred from the properties of

tubes of induction. For if the induction on opposite sides ofthe surface makes angles 91, 92

with the normal, and a tube ofinduction in the first mediumcuts the surface in an elementof area dS, the cross-section iscos didS in the first medium, andit is continued as a tube of cross-section cos 62dS in the second medium; but the inductionmultiplied by the cross-section is constant, so that

Bx cos 6xdS = B2 cos 02dS

or (-BJi = (-BJ2-

10*3. Induced magnetism. If a piece of soft iron is broughtnear to the positive pole of a permanent magnet it is found toacquire magnetic properties temporarily, exhibiting negativepolarity in its parts which are nearest to the positive pole ofthe permanent magnet and positive polarity in its remoterparts, so that if free to move it is attracted towards the per-manent magnet. The soft iron is then said to be magnetizedby induction and the magnetism which it acquires is 'in-duced magnetism'. Its tendency is clearly to move from theweak part of the field to the strong. Nickel and cobalt possessthe same property in this respect as iron, and such bodies arecalled Paramagnetic. There is another class of bodies calledDiamagnetic which have the opposite property of tending tomove from the strong part of the field to the weak. Bismuth isthe principal member of this class but its susceptibility isalmost infinitesimal as compared with that of iron.

In the elementary theory of induced magnetism it is16-2

244 INDTJOBD MAGNETISM [10-3--

assumed that the intensity of magnetization induced in apiece of metal by a given magnetic field is proportional to themagnetic force of the field; i.e. that

I = /cH (1),

where the constant K is called the susceptibility of the metal.If we put fj.= 1 + 4TT/<:, then /A is called the permeability of

the metal, and in the case in which the magnetism of a sub-stance is entirely induced magnetism, since

therefore B = (1 + 4TTK:) H,or B = /iH (2).

10-31. Retentiveness. The 'constants' K and /*, the sus-ceptibility and permeability of a piece of metal, are not con-stant in the strict sense of being independent of the magneticforce. If they were independent of H, then it might be ex-pected that the withdrawal of a field of magnetic force wouldinvolve the instant disappearance of induced magnetism fromany piece of metal which had been exposed to the action of thefield; since if I = KH, then 1 = 0 when H = 0. But this is foundnot to be the case. The induced magnetization does not whollydisappear. What remains is called residual magnetism, andsuch metals as retain residual magnetism when the externalmagnetic field is withdrawn are said to possess retentiveness.

Residual magnetism produces a magnetic field the tendencyof which is in general to reduce the residual magnetism. Forexample, an iron bar placed along the lines of force in amagnetic field becomes magnetized longitudinally, the directionof magnetization from the negative to the positive end beingthat of the field. But when the bar is removed from the fieldthe residual magnetism will produce a field in the iron from thepositive towards the negative end and this field will tend toreduce the residual magnetism. But the longer the bar the lesseffective will be the demagnetization since the central partswill be at a distance from the ends.

10*4. Equations for the potential. In magneto-static asin electrostatic fields we assume that the magnetic vector H

10-41] EQUATIONS VOB THE POTENTIAL 245

is the negative gradient of a potential function <f>, i.e. weassume that „ , . , , .

H = -grad<£ (1),

But in any case in which the magnetism is entirely inducedmagnetism, we have „ „ / o .

a=jujti \Z)and, from 10-11, divB = 0 (3),

or =c

By comparing with 5-3 we see that B, H and /x play the samepart in a magnetic field which contains pieces of soft iron, asdo D, E and K in an electric field containing dielectrics, andthe equations for the potential are similar in form.

At an interface between two media of permeabilities ^ andu2, since the normal induction is continuous, we have

or

where ^ , <£2 denote the potentials on opposite sides of thesurface, and the differentiations are along the normal in thesame sense on both sides. If either medium is air, the corre-sponding fi is unity.

10-41. When permanent magnetism is present we maydenote its intensity at any point of the field by Io, so that thetotal magnetization is given by

I = *H+I 0 (1),

where KH represents the induced magnetization. It is not tobe supposed that KH and Io both necessarily have non-zerovalues at the same point; H has a value at all points of the

246 EQUATIONS FOB. THE POTENTIAL [10-41-

field, K 4= 0 in such substances as iron and Io 4= 0 in permanentmagnets.

Again the induction is given by

.(2).

But divB = 0 (10-11),

therefore div ( iH) = — 4TT div I,,,

or, as in 10*4,

d

This is the characteristic equation for the potential in a fieldwhich contains permanent magnets; and 10-4 (4) is the specialcase in which Io = 0.

At the boundary of a permanent magnet, let the suffix 1denote the magnet and the suffix 2 the adjoining medium.Then, as in 10*4, we have

(£„)! = (Bn)2,where B = fiH. + 4TT-I0 , and I,, only exists in medium 1, so that

In terms of potential this gives

where the normal differentiation is outwards from the magnet.In every case the potential is continuous at the interface,

i.e. < x = <j>2 over the interface, so that the tangential componentof magnetic force is continuous. It is easy to shew, as in 5-26,that since normal induction and tangential forces are con-tinuous, at the interface between two media at which there isno permanent magnetism the lines of force undergo a refractionin accordance with a definite law.

10*5. Examples, (i) An infinite mass of soft iron has a plane J'ace.To find the magnetization induced by a magnetic particle outside the iron.

(a) When the axis of the magnet is normal to the plane face. Let m bethe moment of the magnet and A the position of its centre. Let B be

10-5] EXAMPLES OP INDUCED MAGNETISM 247

the point such that AB is bisected at right angles at 0 by the planeface. Let r1, ra denote distances of any point P from A, B and 8r, 8tthe angles made by AP, BP with BA.

There will be different potential functions fa, <f>t in air and in iron,and the conditions to be satisfied at the boundary are that

fa = fa (1)and that - = / * • •(2),

dn r dnwhere p is the permeability of the iron, and dn is normal to the planeface.

1Let us assume that the field in air is due to the magnet TO at A and

an image TO' at B with their axes in the same direction BA, and thatthe field in iron is due to a magnet TO" at A with its axis in the samedirection.

Also take an axis Ox along OA and axes Oy, Oz in the plane face,let OA = a and let P be the point (x, y, z).

Then A _TOCOS01 TO'COS98

and TO" COS 0J

91 =m(x — a) m'(x + a)

•(3)

and, m"(x-a)fa= .(4).

Substituting from (3), (4) in (1) and (2) and putting x= 0, we find that

TO — m'=m"and

so that

(compare 6*7).

TO + m' = f

m'=~—T+1 TO and m"= j

p+l(5)

248 EXAMPLES OF INDUCED MAGNETISM [10-5

Since all the necessary conditions are satisfied, this represents thesolution of the problem.

Hence the field inside the iron is such as would be produced by a

magnetic particle of moment — - in air; i.e. its potential is

2m

and the magnetic force has radial and transverse components

JJ 2m

It follows that the intensity of induced magnetization has components

_ _ 4 / c ? n C O S 0 J A T — ^Km s m ^ i

T 2/cm cos 8, . T Km sax 6,^ 1 d / i

where K is the susceptibility of the iron.

(j8) Wftew i^e M M o/ the magnet is parallel to the plane face. Take Ozparallel to the axis of m, and let AP, BP make angles ijix, ip2 with Oz.

Let the field ^x ' in air be due to m at A and a magnet m1 at B, and letthe field <f>2' in iron be due to a magnet m2 at 4 , all the axes pointing inthe same direction.

Then , , _ w c o a fxnen ^ _ —-^—+

and d>t

or (f,1/= ^ — — - | + 'J"1" ^ ^ | (6)

and &,' = % 1 (7);

and as before the boundary conditions are that when x = 0

ft' = #i' (8)

and "a == /A -2"•- (9).

Substituting (6) and (7) in (8) and (9), we get

1 = m,

and m—ml

so that mx = — —-^m, ma = —-—. (10);

whence the field in the iron and the components of induced magnetiza-tion can be written down as before.

10*5] EXAMPLES OF INDUCED MAGNETISM 249

(ii) To find the magnetization induced in an iron sphere placed in auniform field of force. This problem is the exact parallel of that of adielectric sphere in a uniform field of electric force (6'71).

If Ho denotes the uniform field of force, and we take an origin O atthe centre of the sphere and the axis Ox in the direction of the field,the potential of the field in the absence of the sphere is

^= -Hox= -H0roos9 (1).

Then arguing as in 6*71 when the sphere is present, we assume forthe potential in air M a

^ ^ (2)

and in iron ^2 = iVrcos0 (3).If a denotes the radius of the sphere, then <f>l, ^2 have to satisfy the

conditions a , P .§ = (4)

when r = a, where ju, is the permeability of the iron. These give

M = ^ H 0 a 3 and * = - ! ! ! (5).

The potentials are therefore

^=-Hor cos 0 + £ - j - ^ - ° _ i (6)

and ,A a =--^^rcos0 (7).

From (6) we see that the disturbance of the given field produced by thesphere is such as would be caused by a magnetic particle of moment

^ oH0a3 placed at its centre.

From (7) it follows that since <f>2 = -j^r the magnetic force H or

— d<j>s/dx is constant and equal to 3flr0/(/x. + 2) inside the iron. Theintensity of induced magnetization is given by

Since p = 1 + 4TTK, therefore

r_ SK TJ _ » rr3 + 4777C

and as the susceptibility K increases, I tends to a limit 3H0/4i7. Itfollows that no matter how great the susceptibility of the sphere theintensity of magnetization in a given field cannot exceed a fixed amount.

The induction inside the sphere is given by

and, as p or K increases, the induction tends to 3H0 and cannot exceedthis amount.

250 EXAMPLES [10-5

In air at a great distance from the sphere the uniform field is un-disturbed and B = H0. Suppose that dSt and dS0 denote the cross-sections of the same tube of induction inside the sphere and at a greatdistance from it, then since the product of the induction and cross-section is constant •>

so that

For large values of p it follows that dSt = idS0 so that the magneticinduction is gathered up and concentrated in the sphere, in the mannerindicated in the figure of 6-71.

In this connection we note that for paramagnetic substances suchas iron, cobalt and nickel the susceptibility K is positive so that /*> 1,and that for some samples of iron /t may be as great as 1000. But fordiamagnetic bodies K is negative and ^< 1. For all such bodies K isvery small, its largest value is for bismuth, for which K = 2-5 X 10~e.The direction of magnetization in a diamagnetic body is opposite tothat of the inducing field, and a diamagnetic sphere would take upfewer tubes of induction than would occupy the same space in anundisturbed field, its general effect being to push the tubes apartinstead of gathering them together.

EXAMPLES

1. Find the force with which a small magnet of moment m isattracted towards a large block of iron of permeability /t having aplane face at a distance a from the magnet, when the axis of themagnet is (i) perpendicular, (ii) parallel to the plane face.

2. Shew that, if the small magnet in Ex. 1 be free to turn about itscentre, either position in which its axis is normal to the plane face isone of stable equilibrium, and that the period of a small oscillation is

-, where K is the moment of inertia of the magnet.

3. Shew that, when the small magnet of Ex. 1 makes an angle awith the normal to the plane face, the mutual potential energy of themagnet and the block of iron is

- ] m1

4. A small magnet is placed at the centre 0 of a spherical cavity ina large mass of soft iron. Prove that at any point P in the iron thedirection of the induced magnetism lies in a plane through the axis ofthe magnet and makes with OP an angle tan~1(JtanS), where 6 isthe angle between OP and the axis of the magnet.

EXAMPLES 251

5. A small spherical cavity is cut in a permanent magnet. Shew fromthe expression for the potential that the force within it is H + ^irl,where H is the magnetic force and / the intensity of magnetization.

[I. 1923]

6. Shew that inside a sphere uniformly magnetized to intensity Ithere is a constant demagnetizing force fwJ. [I. 1922]

7. A long cylinder of circular section, of radius a and of given mag-netic permeability /*, is placed across a given uniform magnetic field.Find the ratio in which the intensity of the field inside the cylinder isreduced. [M. T. 1911]

8. Shew that the force at the centre of a long narrow circular cylin-drical cavity, whose axis is perpendicular to the direction of / , tends,as the cavity is indefinitely diminished, to half the vector sum of themagnetic force and the magnetic induction. [I. 1909]

9. A sphere of permeability /* is placed in a uniform magnetic field.A circle has its centre on the line of force which passes through thecentre of the sphere and its plane perpendicular to this line and liesoutside the sphere. Shew that the presence of the sphere increases theinduction through the circle in the ratio

where a is the greatest angle subtended by a radius of the sphere at apoint on the circle.

10. Prove that if P is the centre of a circular cylindrical cavitywhose diameter and length are equal, with generators parallel to theintensity of magnetization / in a magnet, the limit of the magneticforce at P as the cavity tends to vanishing is

11. A specimen of soft iron is placed in a uniform magnetic field, inthe form of (i) a sphere, (ii) a long rod with its axis parallel to theexternal field, (iii) a long rod of circular cross-section with its axis atright angles to the field. Compare the intensities of magnetization inthe three cases, estimating the values in cases (ii) and (iii) near themiddle of the rod. If the magnetic susceptibility be 20, shew that theintensities are approximately as

3:252:2. [M. T. 1915]

ANSWERS

Chapter XI

ELECTROMAGNETIC INDUCTION

11*1. Faraday's experiments. About the year 1832Michael Faraday was making experiments at the Royal In-stitution which led to the discovery of the law of electro-magnetic induction, which made possible all the modern appli-cations of electric power. He made two coils of insulated wire,each about two hundred feet in length, winding them onthe same wooden block, attaching the ends of one coil to theterminals of a battery and the ends of the other to a galvano-meter. Calling the circuit which contains the battery the'primary' circuit and the other the 'secondary', he observedthat whenever the primary circuit was completed the galvano-meter indicated the passage of a transient current in thesecondary circuit in the opposite direction to that in theprimary; also that whenever the primary circuit was broken atransient current passed through the secondary in the samedirection as the current in the primary. It was further observedthat so long as the current in the primary was steady there wasno current through the secondary. The currents in thesecondary circuit are called induced currents.

Other experiments were performed with primary andsecondary circuits separated from one another and situated inparallel planes. It was found that, when there was a steadycurrent in the primary, relative motion of the circuits resultedin an induced current in the secondary; and that the twocurrents were in the same sense when the motion was causingincreased separation of the circuits, and in opposite senseswhen the circuits were approaching one another.

Other experiments shewed that the relative motion of acircuit and a magnet resulted, in general, in the induction of acurrent in the circuit; that is to say, if a closed loop of wireis in motion in a magnetic field there is, in general, an electriccurrent induced in the wire.

11-2] LAW OF ELECTROMAGNETIC INDUCTION 253

11*2. Law of electromagnetic induction. Faraday attri-buted the phenomena described in 11*1 to what he calledthe electrotonic state of the medium. The precise formulationof the experimental law which accounts for the phenomenais due to the mathematician F. E. Neumann.* It is asfollows:

Whenever the number of tubes of magnetic induction whichpass through a circuit is varying, there is an electromotive forceproduced in the circuit equal to the rate of decrease of the numberof tubes of magnetic induction; and the positive direction of thiselectromotive force and of the tubes of induction have a right-handed screw relation.

As this law is based upon Faraday's experiments, it is com-monly called Faraday's circuital relation.

The reader will have no difficulty in following the way inwhich the law accounts for the results of the experiments. Inthe first case when the primary current is steady the magneticfield caused by it is also steady and there is no secondarycurrent; but when the primary circuit is first completed theprimary current has to increase from zero to its steady valueso that its magnetic field is also growing, i.e. the flux ofinduction through the secondary circuit is increasing and inconsequence there is an E.M.F. in the secondary in the left-handed screw sense in relation to the direction of the induction,and this produces a current in the secondary in the oppositesense to that in the primary. The break of the primary circuitinvolves a dying-away of the magnetic field through thesecondary and therefore an B.M.B1. in the secondary in theright-handed screw sense producing a current in the same senseas the original current in the primary.

In relation to the second experiment the separation of theparallel circuits decreases the amount of induction due to theprimary which passes through the secondary and the approachof the circuits increases it, and hence the currents are inducedas stated.

* F. E. Neumann (1798—1895), German mineralogist, physicist andmathematician. Berlin Akad. 1845 and 1847.

254 LENZ'S LAW [11-21-

11*21. Lenz's Law. An induced current of course producesa magnetic field of its own and this affects the total amount ofmagnetic induction threading the circuit. From what hasalready been said it will be observed that things happen asthough an effort were being made to preserve the status quo asregards the amount of induction which threads the circuit.For example, in Faraday's first experiment, before the pri-mary circuit is closed there is no current and no inductionthrough the secondary; but directly the circuit is closed themagnetic field begins to grow and so does the inductionthrough the secondary; and a current is induced in thesecondary in such a sense that its magnetic field gives inductionthrough the circuit in the opposite sense to that produced bythe primary, as though making an effort to keep the totalinduction through the secondary zero as long as possible.

This is in accord with a law formulated by Lenz,* viz./ / a current is induced in a circuit B by the relative motion ofcircuits A and B, the direction of the induced current will be suchthat by its electromagnetic action on A it tends to oppose therelative motion. Thus if the circuits are equal rings in parallelplanes and B is made to approach A, i.e. to increase the amountof induction due to A which passes through B, the currentinduced in B will be in the opposite sense to that in A so thatthey would repel one another, i.e. the relative motion wouldbe opposed.

We remark that, if Lenz's law were not true, stability ofequilibrium of any arrangement of circuits would be impossible.For relative displacement would start a current which wouldproduce a force tending to increase the displacement and ifunchecked the motion would increase indefinitely.

11*3. Self-induction and mutual induction. The law ofinduction of currents applies to all circuits whether theycontain batteries or condensers or are simply closed loops ofwire in a varying magnetic field. Thus in the case of a singlecircuit containing a battery, directly the circuit is closed thegrowth of the current means an increasing amount of magnetic

* Poggendorff, Annalen der Physik und Ghemie, xxxi, p. 483 (1834).

11*81] SELF AND MUTUAL INDUCTION 265

induction threading the circuit, and, by the law of induction,there must be an induced E.M.F. tending to produce a currentin the opposite sense to that produced by the battery. Theeffect will be more marked if the wire is coiled round a bar ofsoft iron, for this collects and concentrates the magneticinduction within the coil. It constitutes in fact an electro-magnet.

It is necessary to have a symbol to denote the amount ofinduction which passes through a circuit when a unit currentflows round it and there are no other currents in the field.This amount of induction is called the coefficient of self-induction or the inductance of the circuit and is denoted byL; so that if a current j is passing through the circuit Lj is theamount of induction passing through the circuit in the right-handed screw sense in relation to the current. The constantL depends of course on the form of the circuit and varies if theshape of the circuit is changed, and is in general larger for coilscontaining iron cores because they concentrate the induction.

In the same way if a unit current round a circuit A causesan amount of induction M through a circuit B, then M iscalled the coefficient of mutual induction, or the mutualinductance of the two circuits. It can be proved that aunit current round B produces the same amount M of induc-tion through A. Consequently if L' denotes the inductance ofB and currents,/, j ' flow round A and B, the total amounts ofinduction through A and B respectively are

Lj + Mj' and Mj + L'j'.

11*31. The phenomenon of self-induction is well illustrated by thefollowing experiment. An electric lamp Fof large resistance is in parallel with a coilG of small resistance surrounding an ironbar—i.e. an electromagnet. The terminalsP, Q are connected to a battery. Owingto the difference of the resistances most ofthe current passes along POQ and thereis not sufficient along PFQ to make thelamp incandescent. But when the batterycircuit is broken the lamp instantly flashes "up, because the tendency described in 11*21 to keep constant the flux

256 CIRCUIT WITH SELF-INDUCTION [ l l ' S l -

of induction through the circuit PFQG results in the passage of atransient current round this circuit much larger than the steady currentwhich passed along PFQ before the battery circuit was broken.

11*4. Single circuit with self-induction. When a singlecircuit of resistance R and inductance L contains a battery ofelectromotive force E, the current j in the circuit is determinedby the combination of Ohm's Law and Faraday's circuitalrelation. The magnetic induction through the circuit due to acurrent j is Lj, so that in addition to the electromotive force

of the battery there is an induced electromotive force — -=- (Lj),dt

or — L-jf: and therefore by Ohm's Lawdt

Thus we have a differential equation for the current

L^ + Rj = E (2).

When E is constant, the solution of this equation is

! » (3),

where Q is a constant of integration.From (2) we see that Ldjjdt is finite when j is zero. We infer

that a finite current is not started instantaneously but by agradual growth, though it may be that a finite value isattained in a very short time. In (3) therefore we put j = 0 whent = 0, giving C=- EjR, so that

| (4).

The final steady value of the current, after a sufficient lapseof time, is therefore EjR, the value given by Ohm's Law whenthe circuit possesses no self-induction. Whether this value isattained quickly or slowly depends on whether LjR is small orgreat, for in LjR seconds the current attains the definitefraction (1 — e-1) of its final value.

When the circuit is first closed the current is at first very

11-5] CIRCUIT WITH SELF-INDUCTION 257

small and the battery is doing work in creating the magneticfield of the circuit, but after the permanent state is reacheddj/dt is negligible and the work done by the battery is entirelyspent in heating the circuit.

The inductance L would be increased by inserting some ironin the circuit and this would retard the attainment of thepermanent state.

It appears therefore that in a single circuit with a battery ofconstant B.M.F. the self-induction of the circuit is only effectiveduring the interval of time that is occupied in setting up thepermanent current as given by Ohm's Law.

11*41. Flow of electricity caused by the sudden creation of auniform magnetic field. Consider a plane circuit of wire of area Aand suppose that a uniform magnetic field is brought into existenceand that B denotes the induction normal to the plane of the circuit.The amount of induction of this field which passes through the circuitis AB and the amount due to self-induction when a current,;' is passingis Lj, so that the total amount of induction through the circuit isAB + Lj. Therefore by Ohm's Law and the law of induction

Rj= -^(AB + Lj) (1),

where JB denotes the resistance of the circuit. If we integrate thisequation with regard to t from t = 0 to t = oo, we have

f°°But / jdt=Q, the total quantity of electricity which crosses any

section of the wire while the magnetic field is being created. The initialvalue of B is zero and its final value is the strength of the induction ofthe ultimate uniform field, which we may still denote by B; the initialvalue of j is zero and its final value when the permanent state of thefield is attained is zero also. Hence it follows that

BQ=-AB (2).

11-5. A single circuit with a periodic electromotive force.We can easily see that, if a wire ring were to rotate steadily ina uniform magnetic field, the amount of magnetic inductionpassing through the ring would be continually changing, sothat, in accordance with the law of induction, there would bean electromotive force induced in the wire, which would beperiodic in the sense that its direction along the wire would

RIM 17

258 PERIODIC ELECTROMOTIVE FORCE [11-5-

alternate with every half revolution and its magnitude wouldvary between zero and a definite maximum.

Let us therefore investigate the current which would beproduced in the circuit of 11*4 if, instead of a constant electro-motive force E, the circuit is subject to a periodic electro-motive force E oospt.

Instead of 11*4 (2) we now have the equation

^ (1).dt

This equation has an integrating factor eie'/i, and the solution is

jeBilL = C + % L

o r j = Oe-m +o r 3 ve +where G is a constant of integration. After the lapse of suffi-cient time the term e~miL will disappear and the current willcontain periodic terms only, viz.

. E (B cos pt +Lpein.pt)3~ ( '

. Econ(pt-oL)

where tan a = LpjB.

The current has the same frequency (number of alternationsper second) p/2n as the electromotive force, but its phasept — a. lags behind that of the electromotive force by an amounttan - 1 (Lp/B), and this will be approximately \n when Lp islarge compared to B, i.e. if the alternations of E.M.F. are ofhigh frequency. In this case we have, approximately,

shewing that the current is independent of the resistance, inthis case, and depends only on the inductance and thefrequency.

In the general case the maximum value of theE.M.F. is E andthe maximum value of the current is E[^/(B2+L2p2), and the

11-51] CIRCUIT ROTATING IN UNIFORM FIELD 259

ratio of these, viz. \/(Bz + L2p2), is called the impedance ofthe circuit.

11*51. Plane circuit rotating uniformly in a uniformmagnetic field of force H. Let A be the area of the circuit and attime t let 9 be the angle between the plane of the circuit and thedirection of the magnetic field H. The amount of induction threadingthe circuit is then AH sin 9. By the law of induction of currents thereis therefore an E.M.F. in the circuit of amount — Aff cos 9.6; and if weput 9= wt, where eo is the angular velocity, the induced E.M.F. is— AHa> cos tat.

Hence, as in 11*5, the equation for the current,?' is

L-l. + Rj = -AHwcoswt (1),dt

where L, B denote the inductance and resistance of the circuit. As in11*5 the solution is A „ „ „ , , •>

. AH (o COS ((at — a.)3~ VlB*+ &<»*) ( >'where tan a = L<a/R.

As in 9*5, the potential energy of the circuit in the given field H isW= -jAH sin 9,

so that the field exerts on the circuit a couple tending to increase 9 ofmoment JTIT-

^ (3).

Substituting the value of? from (2) and taking note of the minus signwe see that there is a couple opposing the rotation of moment

A2H*co COS <at COS ((at —a.)

This may be written1 A2H*(a {cos (2wt- a) + cos a}2 -y/(R* + L*w2)

and since the mean value of cos (2<ot — a.) is zero, the mean value of themoment of the couple is

1

When the rotation is so rapid that Lu> is large compared to R, a isapproximately \n and the current is approximately

AH sin wt

so that Lj + AH sin 9 = 0,

or the total induction passing through the circuit is zero.17-2

260 DISCHARGE OP A CONDENSER [11-6

11*6. Circuit containing a condenser. Discharge of acondenser. Let a circuit contain a condenser of capacity Cand a battery of electromotive force E;let R denote the total resistance and Lthe inductance of the circuit. At time tlet Q and — Q denote the charges onthe plates of the condenser and <j> thedifference of their potentials. Also letj denote the current in the circuit fromthe positive towards the negative plateof the condenser and suppose that theelectromotive force is in the same sense.By the law of induction there is aninduced E.M.F. equal to — Ldjjdt, therefore by Ohm's Law, asin 7-3(5), ..

dt'

or •(I)-

= —dQ/dt, and Q = C<f>, so that (1) is equivalent to

.(2).dt* ' ~ dt ' c "

When E is constant, this equation admits of simple solution,for the left-hand side retains the same form if we writeQ= — CE + Q'. But the only case of special interest is whenE = 0, i.e. when there is no battery in the circuit and we aresimply concerned with the discharge of a condenser througha wire of given resistance and inductance. The equation is then

LW+Ii~dt+c=0 (3)>

and, if we substitute for Q in terms of either^' or <f>, we find thatj and (f> satisfy the same differential equation. The ordinaryprocess of solution gives

where — Als — A2 are the roots of

i-o (.).

11-6] DISCHARGE OF A CONDENSER 261

When B2 > ±LjC, the roots of this quadratic are real andnegative, i.e. A1( A2 are positive numbers and the solution (4)shews that the charge dies away continuously.

For the determination of the constants Ax, A2, we note thatif Qo is the initial charge, since the initial current — dQ/dt iszero, we have by putting t — 0

and 0=-X1A1—X2A2)

so tha t Ax = A2 Q0l(X2 - Ax) and A2 = - \ QO/(XZ - XJ,

and Q = _^o_(A2e-Ai '-A1e-A^) (6).A2 —Ax

The current j is then given by

' - - * - £ * & « ' * - ' * > <"•From this we find that j increases from zero to a maximum,which it attains at time t given by Xre~^l = X2e~^, and thendies away continuously.

When B2 — 4:LjG, the quadratic (5) has equal roots — RJ2L and the

solution of (4) is Q = e-W2L(Al + Ait) (8),

and the initial values of Q and dQ/dt give

Qo — Aj,, 0 = — ^^Al i

so that Q = e-MI2LQ0(i + Rtj2L) (9)

and y = _ * « = e-JW»«og* (10);

and j attains a maximum value after a time t = 2L/R and then diesaway continuously.

The more interesting case is when R2 < 1LJC; the quadratic(5) then has imaginary roots and the solution of (3) has the form

(11),

representing harmonic oscillations, Q changing sign regularly

and vanishing at regular intervals of v J 11 ^TF ~ Tfl) a s * n e

charge is transferred from one plate of the condenser and back

262 DISCHARGE OP A CONDENSER [11-6-

again, but all the time dying away because of the dampingfactor erm^.

As a special case, if the condenser be 'short-circuited', e.g.if its plates were allowed to come into contact, this is equi-valent to making its capacity infinite, so that (3) becomes

Hf+Jy-O;

with a solution j=Joe~BllL

shewing that the current would decrease exponentially withthe time.

I t frequently happens that Rz is small compared to 4.L/C.The damping is then small and the period of the oscillationsis approximately 2TT-\/(CL). This is known as Thomson'sFormula, as the theory of the discharge was first given bySir William Thomson (Lord Kelvin) in 1853. The actualoscillations were first detected by Feddersen in 1857.

11*61. It should be remarked that currents only flow in closedcircuits and that in the case of the discharge of a condenser thecircuit is to be regarded as completed by a displacement current, i.e.a time rate of change of the displacement in the dielectric mediumbetween the plates. But this does not necessitate any changes in theforegoing results.

I t should be noted that while the inductance of a wire bent into asingle loop may be small, it may be greatly increased by being coiledin spiral form and it is still more increased if an iron bar is inserted inthe coil.

11-62. Unit of inductance. The practical unit of induct-ance is the henry,* which is equal to 109 electromagneticabsolute c.G.s. units of inductance.

11*7. Examples, (i) A Wheatstone's bridge is adjusted so that thereis no permanent current through the galvanometer. One of the arms whoseresistance is P contains a coil of inductance L. If the resistance of thegalvanometer is negligible, prove that when the battery circuit is broken aquantity

of electricity will flow through the galvanometer.

* Joseph Henry (1797—1878), American physicist.

11-7] EXAMPLES 263

[P, Q, R, S are the resistances of the arms, R being opposite to Q; E isthe E.M.F. of the battery which con-nects the junction of P and Q with thejunction of R and S; and B is theresistance of the battery.] [M. T. 1923]

Since in the permanent state thereis no flow through the galvanometer,therefore PS=QB , ( 1 ) .

Let the steady currents before thebattery circuit is broken be x, yand x — y as indicated in the diagram.Then from the two circuits con-taining the battery we have

(2);

so that (P+Q + R + S)y = (or, substituting for 8 from (1),

(P + Q)y=Qx.Hencefrom(2) y{(P+Q) B + {P + R) Q} = EQ (3).

After the battery circuit is broken, the change in the amount ofinduction through the coil will cause a current z, say, in the arm KMand since MN is of negligible resistance the whole current will flowalong MN and return along NK. Hence for the current z

and z has an initial value y.The total flow through the galvanometer is then

Lj o dtdt

and from (3) the required result follows.

(ii) Between A and B there is a coil of wire of resistance R and induc-tance L, and a condenser of capacity G has its plates connected to B and toD by wires the sum of whose resistances is r, and a potential difference

L R B D

E oospt is maintained between A and D. If Ex and Et are the amplitudesof the potential differences between A and B and between B and D respec-tively and if LjO= Rr, shew that E1

i + Ei2 = Ei. 1.1911]

264 EXAMPLES [11-7

lietQ denote the charge on the plate of the condenser nearer to B,then Q denotes the current.

The difference of potential of the condenser plates is Q/C, anddenoting the other potential differences by <J>A~<I>B

a n ( i <f>B — <t>D> w e

have equations for the current in AB and in BD

and rQ = 4>B-^J)-QIO.It is clear that if a potential difference E cos pt is maintained

between A and D all the variables must have the same period 2ir/p; wemay therefore suppose that Qoce^'so that Q = ipQ and Q = — p2Q and

4>A-<I>B =

Hence

But a complex number x+iy may be written r (cos 8 + i sin 8) orreie and its amplitude is r or \/(a;2 + 2/2). Hence from (1)

E* E2 El

and

if

or if Lp*IC=p*Rr, i.e. L/C = Rr.

(iii) A condenser is formed of a pair of parallel circular plates of1 metre radius 1 centimetre apart. If it is charged up to 100 volts anddischarged through a circuit of induction -fa henry and resistance 100 ohms,find the period of the oscillations and the maximum current.

Neglecting edge effects the capacity C of the condenser is equal to'area/4w x thickness'= 10*7r/47r=104/4 absolute electrostatic units= 104/4-9-1011 Farads; i.e. C= 1/36.10' Farads. Also L = henry andB= 100 ohms.

As in 11*6 (3) we can shew that the differential equation for thepotential difference + is i (£ + i ^ + (£/G = o (1)

and if this has a periodic solution with <f> = <f>o where t = 0, it is of theform

whereR/2L = 500, and ^ / { ^ - ^ } = V{36.108- 25.104} = 6.104approx.,

so that (£ = 0oe-5OO'cos(6.104*).The period of the oscillations is therefore 2w/6.104, or approximately

11-7] EXAMPLES 265

wisrs second, and the damping factor drops to e~1 in ^ 5 second orabout 19 oscillations..

Again the current is given byj--C^>= (7<£0e-500'{500cos(6.104«) + 6.104sin(6.104*)}

and the first term is negligible compared with the second, and, since^0= 100, J_

with a maximum value approximately 6. l O ' C ^ ^ amp.It can be shewn that electromagnetic effects are propagated with

the velocity of light, i.e. 3.1010 cm. per sec. Hence the length of thewaves in the above oscillations being 'velocity x period' is

or about 31-5 kilometres.In reference to the units in this example, the capacity of the con-

denser was actually calculated as so many centimetres but the capacityso found was essentially an electrostatic measure, and in electrostaticunits the Farad is 9.1011 absolute units, and as the other measures inthe data were all in practical units so the capacity was the onlymeasure which needed conversion.

EXAMPLES

1. A coil of resistance R and self-induction L is joined to a batteryof electromotive force E. Shew that the current after a time * is(1 — x) E/B, and that if contact be then broken the current after afurther time t is (x-x*) E/B, where Llogx + Bt = 0. [I. 1898]

2. A coil is rotated with constant angular velocity ai about an axisin its plane in a uniform field of force perpendicular to the axis ofrotation. Find the current at any time, and shew that it is greatestwhen the plane of the coil makes an angle tan-1 (Lw/B) with the linesof magnetic force. [M. T. 1897]

3. A metal ring rotates uniformly round a horizontal diameterwhich is at right angles to the magnetic meridian. If the effects ofself-induction be neglected, in what parts of the revolution is theinduced current strongest and when does it vanish ?

[Trinity CoU. 1895]

4. A condenser of capacity C is discharged through a resistance Bso high that inductance is inappreciable; prove that the charge fallsaccording to the law e~'°lB.

If the two plates of the condenser are each circular of radius 10 cm.,and are a millimetre apart, and are discharged through a damp threadof resistance 109 ohms, shew that the charge will fall in the ratio of 1 toe in about a quarter of a second. Would the same mode of calculationapply if the resistance were only 106 ohms? [M. T. 1911]

266 EXAMPLES

5. A circuit of resistance R and of self-induction L has a condenserof capacity 0 inserted in it, and is acted on by an electromotive forceE sin pt. Prove that the current at any time is

TPr COB (pt —a),

\tp-^)where tana = J ? / ^ ^ — L p \ . [1.1903]

6. A circular wire of radius a and resistance R is spun in a magneticfield of strength H with angular velocity w about an axis in the planeof the wire and at right angles to the lines of force. Shew that theaverage rate of dissipation of energy is 7r2o4o)2i3ra/2iJ approximately.

Shew that if the speed of rotation becomes larger the rate of dissipa-tion of energy would be less than that given by the above formula.

[M. T. 1919]

7. The resistance and self-induction of a coil are R and L and itsends A and B are connected with the electrodes of a condenser ofcapacity C by wires of negligible resistance. There is a current I cos ptin a circuit connecting A and B, and the charge of the condenser is inthe same phase as this current; shew that the charge at any time is

~-j-cospt and that C(R2+p2Li) = L. Obtain also the current in the

coil. [M. T. 1897]

8. Shew that, when a coil of inductance L and resistance R isattached to two terminals at which an electromotive force E COB pt ismaintained, the average rate of consumption of energy is

Shew also that , if p — 500, L = 0-2 henry and R = 75 ohms and energyis consumed at the rate of 24 watts, E must be 100 volts.

9. Four points A, B, C, D are connected up as follows: A, B arejoined through a coil of self-induction L and resistance P ; A, Dthrough a resistance Q; B, C through a resistance R; G, D through aresistance $ and through a condenser of capacity K, the resistance andthe condenser being in parallel; B, D through a galvanometer; A, Cthrough a source of current of period 2w/p. Shew that , if no currentpasses through the galvanometer,

PS = QR and L = QRK.(The resistances of the connecting wires may be neglected.)

[M. T. 1908]

10. Two insulated conductors whose coefficients of potential arePu > Pit • Pt2 have charges Qo, Qo' given to them, and are then con-nected by a coil of resistance R and self-induction L; shew that the

EXAMPLES 267

current will be oscillatory if L ( p u — 2pti+pi%) — £B2 is positive; and,denoting this expression by K1, shew that, in this case, the current attime t will be

1 _5? KtK{Qo(Pn-Pia)~Qo'(Pn-Pit)}e 2 isin- j- ,

and that the difference of phase of charge and current will be

t a n - i ^ . [M. T. 1904]

11. The arms AB, AD of a Wheatstone's bridge are of inductancesLlt Lt and of resistances Blt i?2; the arms DO, BO contain con-densers of capacities Gx, G2 and are of resistances B3, B4. BD containsthe galvanometer and AC an alternating source of E.M.F. Shew thatthe bridge will not be balanced for all frequencies unlessLtB3 = LiBl, Oji?a = CaB1 and either BtB3 = BsBt or Lx = Ct JR2BA .

[I. 1926]

12. A thin bar magnet of moment M and length 2c lies along theaxis of a circular ring of radius a and has a small longitudinal vibration/S sin pt about its mean position in which the centre of the magnetcoincides with the centre of the ring. Shew that there is an inducedcurrent in the ring of strength

ain(2pt—a.)

where B, L are the resistance and inductance of the ring and[M. T. 1924]

ANSWERS2. See 11-51.

3. Strongest when the plane of the ring contains the lines of force andzero when at right angles.

7. (B cos pt + Lp sin pt) I/B.

CAMBRIDGE: PBINTED BY WALTER LEWIS, M.A., At THE UNIVERSITY PRESS