ph501set1

5/21/2018 ph501set1

1/23

Princeton University

Ph501

ElectrodynamicsProblem Set 1

Kirk T. McDonald

(1998)

[email protected]

http://puhep1.princeton.edu/~mcdonald/examples/

References:

R. Becker,Electromagnetic Fields and Interactions(Dover Publications, New York, 1982).D.J. Griffiths, Introductions to Electrodynamics, 3rd ed. (Prentice Hall, Upper Saddle

River, NJ, 1999).J.D. Jackson, Classical Electrodynamics, 3rd ed. (Wiley, New York, 1999).

The classic is, of course:

J.C. Maxwell, A Treatise on Electricity and Magnetism(Dover, New York, 1954).

For greater detail:

L.D. Landau and E.M. Lifshitz, Classical Theory of Fields, 4th ed. (Butterworth-Heineman, Oxford, 1975); Electrodynamics of Continuous Media, 2nd ed. (Butterworth-Heineman, Oxford, 1984).

N.N. Lebedev, I.P. Skalskaya and Y.S. Ulfand, Worked Problems in Applied Mathematics(Dover, New York, 1979).

W.R. Smythe, Static and Dynamic Electricity, 3rd ed. (McGraw-Hill, New York, 1968).J.A. Stratton, Electromagnetic Theory(McGraw-Hill, New York, 1941).

Excellent introductions:

R.P. Feynman, R.B. Leighton and M. Sands, The Feynman Lectures on Physics, Vol. 2(Addison-Wesely, Reading, MA, 1964).

E.M. Purcell, Electricity and Magnetism, 2nd ed. (McGraw-Hill, New York, 1984).History:

B.J. Hunt, The Maxwellians(Cornell U Press, Ithaca, 1991).E. Whittaker, A History of the Theories of Aether and Electricity (Dover, New York,

1989).

Online E&M Courses:

http://www.ece.rutgers.edu/~orfanidi/ewa/

http://farside.ph.utexas.edu/teaching/jk1/jk1.html

5/21/2018 ph501set1

2/23

Princeton University 1998 Ph501 Set 1, Problem 1 1

1. (a) Show that the mean value of the potential over a spherical surface is equal to thepotential at the center, provided that no charge is contained within the sphere.

(A related result is that the mean value of the electric field over the volume of acharge-free sphere is equal to the value of the field at its center.)

(b) Demonstrate Earnshaws theorem: A charge cannot be held at equilibrium solelyby an electrostatic field.1

(c) Demonstrate that an electrostatic field E cannot have a local maximum of E2,using the mean value theorem mentioned in part (a) or any other technique.

Remark: An interesting example of nonelectrostatic equilibrium is laser trappingof atoms. Briefly, an atom of polarizability takes on an induced dipole momentp = E in an electric field. The force on this dipole is then (Notes, p. 26),F= (p E) =E2. Since an electrostatic field cannot have a local maximumof E2, it cannot trap a polarizable atom. But consider an oscillatory field, inparticular a focused light wave. The time-average force,F = E2 drawsthe atom into the laser focus where the electric field is a maximum. See,http://puhep1.princeton.edu/~mcdonald/examples/tweezers.pdf

1http://puhep1.princeton.edu/~mcdonald/examples/EM/earnshaw_tcps_7_97_39.pdf

5/21/2018 ph501set1

3/23


2. Calculate the potential(z) along the axis of a disk of radius R in two cases:

(a) The disk is a uniform layer of charge density, and

(b) The disk is a uniform dipole layer of dipole moment densityp = pzper unit area.

5/21/2018 ph501set1

4/23


3. Suppose the electric field of point chargeqwereE = qr/r2+ where 1, rather thenE= qr/r2.

(a) Calculate Eand Efor r = 0. Find the electric potential for such a pointcharge.

(b) Two concentric spherical conducting shells of radiia and b are joined by a thinconducting wire. Show that if chargeQaresides on the outer shell, then the chargeon the inner shell is

Qb Qa2(a b)[2b ln 2a (a+ b)ln(a+ b) + (a b)ln(a b)] (1)

5/21/2018 ph501set1

5/23


4. (a) Starting from the dipole potential = p r/r3 explicitly show that

E=3(p r)r p

r3 4p

3 3(r). (2)

Hint: to show the need for the 3

(r) term, consider the volume integral ofEovera small sphere about the dipole. You may need a variation of Gauss theorem:V dVol =

S

n dS, (3)

wherenis the outward normal to the surface.

(b) The geometric definition of the lines of force is that this family of curves obeysthe differential equation:

dx

Ex=

dy

Ey=

dz

Ez. (4)

For a dipole p= pz, find the equation of the lines of force in the x-zplane. It iseasiest to work in spherical coordinates. Compare with the figure on the cover ofthe book by Becker.

5/21/2018 ph501set1

6/23


5. Find the two lowest-order nonvanishing terms in the multipole expansion of the po-tential due to a uniformly charged ring of radius a carrying total charge Q. Take theorigin at the center of the ring, and neglect the thickness of the ring.

5/21/2018 ph501set1

7/23


6. (a) A long, very thin rod of dielectric constant is oriented parallel to a uniformelectric field Eext. What are E and Dinside the rod?

(b) What are E and D inside a very thin disc of dielectric constant if the disc isperpendicular toEext?

(c) Find E and D everywhere due to a sphere of fixed uniform polarization densityP. Then calculate

E DdVol for the two volumes inside and outside the spheres

surface.

Hint: this problem is equivalent to two oppositely charged spheres slightly dis-placed.

(d) Show that for any finite electret, a material with fixed polarization P,

all space

E DdVol = 0. (5)

5/21/2018 ph501set1

8/23


7. A spherical capacitor consists of two concentric conducting shells of radii a and b.The gap is half filled with a (non-conducting) dielectric liquid of constant . You mayassume the fields are radial. The inner shell carries charge +Q, the outer shell Q.Calculate E and D in the gap, and the charge distribution in the inner shell. Alsocalculate the capacitance, defined as C = Q/V, where V is the potential differencebetween the inner and outer shells.

5/21/2018 ph501set1

9/23


8. (a) As a classical model for atomic polarization, consider an atom consisting of a fixednucleus of charge +ewith an electron of chargeein a circular orbit of radius aabout the nucleus. An electric field is applied at right angles to the plane of theorbit. Show that the polarizability is approximately a3. (This happens to bethe result of Beckers (26-6), but the model is quite different!)

Assuming that radius a is the Bohr radius, 5.3 109 cm, use the modelto estimate the dielectric constant of hydrogen gas at S.T.P. Empirically, 1 + 2.5 104.

(b) Another popular classical model of an atom is that the electron is bound to aneutral nucleus by a spring whose natural frequency of vibration is that of somecharacteristic spectral line. For hydrogen, a plausible choice is the Lyman line at1225 Angstroms. In this model, show that = e2/m2, and estimate . Recallthat e= 4.8 1010 esu, and m= 9.1 1028 g.

5/21/2018 ph501set1

10/23

Princeton University 1998 Ph501 Set 1, Solution 1 9

Solutions

1. a) We offer two solutions: the first begins by showing the result holds for small spheres,and then shows the result is independent of the size of the (charge-free) sphere; thesecond applies immediately for spheres of any size, but is more abstract.

We consider a charge-free sphere of radiusR centered on the origin.

In a charge-free region, the potential (r) satisfies Laplaces equation:

2= 0. (6)

First, we simply expand the potential in a Taylor series about the origin:

(r) = (0) +i

(0)

xixi+

1

2

i,j

2(0)

xixjxixj+ ... (7)

We integrate (7) over the surface of the sphere:S

(r)dS= 4R2(0) +i

(0)

xi

S

xidS+1

2

i,j

2(0)

xixj

S

xixjdS+ ... (8)

For a very small sphere, we can ignore all terms except the first, In this case, eq. (8)becomes

1

4R2

S

(r)dS=(0), [Rsmall], (9)

which was to be shown.

Does the result still hold at larger radii? One might expect that since small is not

well defined, (9) holds for arbitraryR, so long as the sphere is charge free.

Progress can be made staying with the Taylor expansion. By spherical symmetry, theintegral of the product of an odd number ofxi vanishes. Hence, only the terms witheven derivatives of the potential survive. And of these, only some terms survive. Inparticular, for the 2nd derivative, only the integrals ofx21,x

22, andx

23survive, and these

3 are all equal. Thus, the 2nd derivative term consists of

1

2

2(0)

x21+

2(0)

x22+

2(0)

x23

S

x21dS, (10)

which vanishes, since 2

(0) = 0.It is less evident that the terms with 4rth and higher even derivatives vanish, althoughthis can be shown via a systematic multipole expansion in spherical coordinates, whichemphasizes the spherical harmonics Yml .

But by a different approach, we can show that the mean value of the potential over acharge-free sphere is independent of the radius of the sphere. That is, consider

M(r) = 1

4r2

S

dS= 1

4

d cos

d(r,,), (11)

5/21/2018 ph501set1

11/23


in spherical coordinates (r,,). Then

dM(r)

dr =

1

4

d cos

d

r =

1

4

d cos

d r

= 1

4r2S

dS= 1

4r2V

2dVol = 0, (12)

for a charge-free volume. Hence, the mean value of the potential over a charge-freesphere of finite radius is the same as that over a tiny sphere about the center of thelarger sphere. But, as shown in the argument leading up to (9), this is just the valueof the potential at the center of the sphere.

A second solution is based on one of Greens theorems (sec. 1.8 Of Jackson). Namely,for two reasonable functions(r) and (r),

V

2 2

dVol =S

n

n

dS, (13)

wherenis coordinate normally outward from the closed surface Ssurrounding a volumeV.

With as the potential satisfying Laplaces equation (6), the second term on the l.h.s.of (13) vanishes. We seek an auxiliary function such that2 = 3(0), so the l.h.s.is just (0). Further, it will be helpful if vanishes on the surface of the sphere ofradiusR, so the second term on the r.h.s. vanishes also.

These conditions are arranged with the choice

= 1

4

1

R1

r , (14)

recalling pp. 8-9 of the Notes. On the surface of the sphere, coordinaten is just theradial coordinate r, so

n =

1

4r2. (15)

Thus, we can evaluate the expression (13), and get:

(0) = 1

4R2

S

dS, (16)

which means that the value of the potential at the center of a charge-free sphere of

any size is the average of the potential on the surface of the sphere.

b) The potential energy of a charge qat pointr, due to interaction with an electrostaticfield derivable from a potential (r), is:

U=q(r). (17)

For the point r0 to be the equilibrium point for a particle, the potential should havea minimum there. But we can infer from part (a) that:

5/21/2018 ph501set1

12/23


Harmonic functions do not have minima,

harmonic functions being a name for solutions of Laplaces equation (6).

Indeed, a minimum ofUat point r0 would imply that there is a small sphere centeredon r0 such that(r0) is less than at any point on that sphere. This would contradict

what we have shown in part (a): (r0) is the average of over the sphere.

We continue with an example of Earnshaws theorem. Consider 8 unit charges locatedat the corners of a cube of edge length 2, i.e., the charges are at the locations (xi, yi, zi)= (1,1,1), (1,1,-1), (1,-1,1), (1,-1, -1), (-1,1,1), (-1,1,-1), (-1,-1,1), (-1,-1,1). It is sug-gestive, but not true, that the electric field near the origin points inwards and couldtrap a positive charge.

The symmetry of the problem is such that a series expansion of the electric potentialnear the origin will have terms with only even powers, and we must go to 4rth order tosee that the potential does not have a maximum at the origin. To simplify the series

expansion, we consider the electric field, for which we need expand only to third order.The electric potential is given by

=8

i=1

1(xi x)2 + (yi y)2 + (zi z)2

. (18)

Thex component of the electric field is

Ex = x

= 8

i=1

xi x[(xi x)2 + (yi y)2 + (zi z)2]3/2

= 1

33/2

8i=1

xi

x

[1 + (2xix 2yiy 2ziz+ x2 + y2 + z2)/3]3/2

133/2

8i=1

xi

1 + yiy+ ziz+

2x2 + y2 + z2 + 5yiyziz3

133/2

8i=1

xi

11yiy3 + 11ziz3

54 x

2yiy+ x2ziz

6 +

3y2ziz+ 3z2yiy

2

133/2

8i=1

2xyiy+ 2xziz

3 7x

3

54 +

7xy2 + 7xz2

6 +

20xyiyziz

9

= 28

813(x3

9xy2

9xz2), (19)

noting that x2i =y2i =z

2i = 1 and that

xi= 0 =

xiyi, etc. Similarly,

Ey 2881

3

(y3 9yx2 9yz2) and Ez 2881

3

(z3 9zx2 9zy 2). (20)

The radial component of the electric field is therefore

Er = E r

r =

xEx+ yEy+ zEzr

2881

3r

[x4 + y4 + z4 18(x2y2 + y2z2 + z2x2)]. (21)

5/21/2018 ph501set1

13/23


Along the x axis, r = x and the radial field varies like

Er 2881

3

r3 >0, (22)

but along the diagonal x= y = z= r/3 it varies likeEr 476

243

3r3

5/21/2018 ph501set1

14/23


2. a) The potential(z) along the axis of a disk of radius R of charge density (per unitarea) is given by the integral:

(z) =

R

r=0

2rdr r2 + z2

= 2

R2 + z2 |z| . (26)

Notice that (z) behaves as R2/|z| for|z| R, which is consistent with the ob-servation that in this limit the disk may be considered as a point charge q = R2.Notice also, that at z = 0 the potential is continuous, but its first derivative (E)

jumps from2 atz= 0+ to 2 at z= 0. This reflects the fact that at small|z|(|z| R) the potential may be calculated, in first approximation, as the potential forthe infinite plane with charge density .

b) A disk of dipole-moment density p= pz can be thought of as composed of a layerof charge density + separated in z from a layer of charge density by a distanced = p/. Say, the + layer is at z= d/2, and the layer is as z=d/2. Then, thepotential b at distance zalong the axis could be written in terms ofa(z) found in(26) as

b(z) =a(z d/2) a(z+ d/2) da(z)

z = 2p

sign(z) z

R2 + z2

. (27)

In this, we have taken the limit as d 0 while , but p= d is held constant.At large zwe get the potential of a dipole P = pR2 on its axis. But near the plate(|z| R), the potential has a discontinuity:

b(0+) b(0) = 4p. (28)We may explain this in our model of the dipole layer as a system of two close plateswith charge density and distancedbetween them, where p = d. The field betweenthe plates isEz = 4, and the potential difference potential between two plates is

= Ezd= 4p, (29)

as found in (28).

5/21/2018 ph501set1

15/23


3. a) For a chargeqat the origin, the proposed electric field is

E= q r

r2+ =q

r

r3+. (30)

This still has spherical symmetry, so we can easily evaluate the divergence and curl in

spherical coordinates: E= q r

r3+ + qr 1

r3+ =

3q

r3+ q3 +

r3+ = q

r3+; (31)

E= q rr3+

+ 1

r3+ qr= 0 (3 + ) r

r5+ qr= 0. (32)

Since E= 0, the field can be derived from a potential. Indeed,

E= , where = r

Erdr= qr

dr

r2+ =

1

+ 1

q

r+1. (33)

b) Let us first compute the potential due to a the spherical shell of radius athat carries

charge Q, as observed at a distance r from the center of the sphere. We use (33) andintegrate in spherical coordinates (r,,) to find

(r) = Q

4a21

1 +

11

2ar2 d cos

(a2 + r2 2ar cos ) 1+2=

Q

2ar

(a+ r)1 |a r|11 2 . (34)

Now consider the addition of a sphere of radiusb < athat carries chargeQb. The totalpotential at r = a is

a= Qa2a2

(2a)1

1 2 + Qb2ab

(a + b)1 (a b)11 2 , (35)

while that at r = b is

b= Qa2ab

(a+ b)1 (a b)11 2 +

Qb2b2

(2b)1

1 2. (36)

We require Qb such that a = b, (as guaranteed by the wire connecting the twospheres). However, we neglect terms ofO(2), assuming to be small. Then, (35-36)lead to the relation

Qb= Qa (b/a)(2a)1 (a + b)1 + (a b)1

(a/b)(2b)1 + (a + b)1 (a b)1 (37)

What about the factors of form x, which are approximately 1 for small ? Let x 1 + . Taking logarithms, ln x

ln(1 + )

. Thus,

x 1 + ln x, so x1 x1 + ln x

x(1 ln x). (38)Using (38) in (37), we find

Qb= Qa2(a b)[2b ln 2a (a + b)ln(a+ b) (a b)ln(a b)] . (39)

Measurement of the ratio Qb/Qa provides a stringent test of the accuracy of the 1/r2

law for electrostatics.

5/21/2018 ph501set1

16/23


4. a) The first terms in E can be obtained by explicit differentiation, perhaps best doneusing vector components. For r >0,

Ei= xi

pjxjr3

=3(pjxj)xi

r5 pi

r3. (40)

To justify the-term, consider the integral over a small sphere surrounding the (point)dipole:

VEdVol =

VdVol =

S

ndS=S

p rr3

rdS, (41)

according to the form of Gauss law (3) given in the hint. Evaluating the last integral ina spherical coordinate system withzaxis along p, we find that only the pcomponentis nonzero:

EdVol =p

11

2d cos cos2 =4p

3 . (42)

No matter how small the sphere, the integral (42) remains the same.

On the other hand, if we insert the field E from (40) in the volume integral, only thezcomponent (along p) does not immediately vanish, but then

V

EzdVol =

r0

2rdr

11

d cos p(3cos2 1)

r3 = 0, (43)

if we adopt the convention that the angular integral is performed first.

To reconcile the results (42) and (43), we write that the dipole field has a spike near

the origin symbolized by(4p/3)3

(r).Another qualitative reason for the -term is as follows. Notice, that without this termwe would conclude that the electric field on the axis of the dipole (z-axis) would alwaysalong +z. This would imply that if we moved some distribution of charge from largenegative z to large positive z, then that charge would gain energy from the dipolefield. But this cannot be true: after all, the dipole may be thought as a system of twocharges, separated by a small distance, and for such a configuration the potential atlarge distances is certainly extremely small.

We can also say that the-term, which points in the pdirection, represents the largefield in the small region between two charges that make up the dipole.

b) In spherical coordinates withzalong p, the dipole potential in the x-zplane is

(r, ) =p cos

r2 (44)

Then,

Er = r

=2p cos

r3 , and E =

r =

p sin

r3 . (45)

5/21/2018 ph501set1

17/23


Thus, the differential equation of the field lines

dr

Er=

rd

E, implies

dr

r = 2

d sin

sin , (46)

which integrates to

r= Csin2 = Cx2

r2, (47)

etc.

5/21/2018 ph501set1

18/23


5. The multipole expansion of the potential about the origin due to a localized chargedistribution(r) is

(r) =Q

r +

P rr2

+1

2

Qij rirjr3

+ . . . (48)

where

Q=

(r)dVol, Pi =

ridVol, Qij =

(3rirj ijr2)dVol, . . . (49)

For the ring, Q is just the total charge, while the dipole moment P is zero because ofthe symmetry.

Let us find the quadrupole moment, Qij. Take thezaxis to be along that of the ring.Then, the quadrupole tensor is diagonal, and due to the rotational invariance,

Qxx= Qyy = 12

Qzz . (50)

We compute Qxx in spherical coordinates (r,,):

Qxx =

(3x2 r2)dVol = 2

0

Q

2aa2(3cos2 1)ad= Qa

2

2 . (51)

Thus, the third term in the expansion (48) is:

1

2r3

Qxxsin

2 cos2 + Qyysin2 sin2 + Qzzcos

2

= Qa2

4r3(3cos2 1). (52)

5/21/2018 ph501set1

19/23


6. a) Remember thatE is defined as a mean electric field, due to both the external fieldand the microscopic charges.

In the presence of external field Eext, polarization P is induced in the rod. Since E = 0, the tangential electric field is continuous across the surface of the rod.This suggests that inside the rod, which is parallel to Eext, we have E= Eext.

To check for consistency, note that in this case, P is constant apart from the ends.There is no net polarization charge in the bulk of the rod, and no change in the electricfield from Eext. But at the ends there are chargesQ, where Q = P A, and A isthe cross-sectional area of the rod. Since A is very small for the thin rod, and theends of a long rod are far away from most of the rod, these charges do not contributesignificantly to the electric field. Hence our hypothesis is satisfactory.

The electric displacement can then be deduced as D= E= Eext inside the rod.

b) For a dielectric disk perpendicular to Eext, the normal component of the fields isnaturally emphasized. Recall that D implies that the normal component of thedisplacementD is continuous across the dielectric boundary.Outside the disk, where = 1,D = E = Eext is normal to the surface. (That E= Eextmay be justified by noting that the induced charges on two surfaces of the thin diskhave opposite signs and do not contribute to the field outside of the disk.) Thus, insidethe disk, D = Eext. Lastly, inside the disk E = D/= Eext/.

c) The problem of a dielectric sphere of uniform polarization density P is equivalentto two homogeneous spheres with charge densities and, displaced by distance d,such that in the limit d 0, but d= P.Recall that for a sphere of uniform charge density , the interior electric field is

E= 43

r. (53)

Thus, insidethe polarized sphere we have

E= lim

4

3(r d) 4

3r

= 4

3 lim d= 4P

3 . (54)

The displacement is

D= E + 4P=8P

3 . (55)

Then, inside

E DdVol = 4P3

8P3

4r33

= 1282r3P227

. (56)

Outside the sphere, the electric field is effectively due to two point charges q= 4r3/3separated by small distance d, where d = P. The external field is simply that of apoint dipole at the origin of strength

p=4r3P

3 . (57)

5/21/2018 ph501set1

20/23


Then,

E=3r(p r) p

r3 , (58)

and D= E. Thus,

outside

E DdVol =

p2

+ 3(p r)2

)r6

dVol (59)

= p2r

4r2dr

r6 + 3p2

r

2r2dr

r6

11

cos2 d cos =8p2

3r3 =

1283r3P2

27 ,

and the sum of the inside and outside integrals vanishes.

d) Let us show, on general grounds, that for an electret the integral ofE D over thewhole space is zero. First, note that D = 0 everywhere for an electret, and thatthe electric field can be derived from a potential, . Then

all space

E DdVol = V DdVol =

V (D)dVol =

S

D dS. (60)

But if electret occupies finite volume, then 1/r at large r, as seen from multipoleexpansion; in the same limit,D 1/r2, whiledS r2. So the integral over the surfaceat infinity is zero.

What would be different in this argument if the material were not an electret? Ingeneral, we would then have D= 4free, and the 3rd step in (60) would have theadditional term

V D= 4

Vfree= 8U. Thus, the usual electrostatic energy

is contained inV

E

D/8, as expected.

This argument emphasizes that the work done in putting a field on an ordinary di-electric can be accounted for using only the free charges (which establish D). Oneneed not explicitly calculate the energy stored in the polarization charge distribution,which energy is accounted for via the modification to the potential in the presenceof the dielectric. But the whole argument fails for an electret, which remains polarized(energized) even in the absence of a free charge distribution.

5/21/2018 ph501set1

21/23


7. In the upper half of the capacitor (where there is no dielectric) we have via Gauss law:

Eup(r) =Dup(r) =4Qup

2r2 =

2Qupr2

, (61)

whereQup is the charge on the upper half of the inner sphere, assuming the fields are

radial.In the lower part we have:

Edown(r) =Eup(r), (62)

which follows from the continuity of the tangential component ofEacross the boundarybetween dielectric and vacuum. Also,

Ddown(r) =Edown(r), (63)

and

Ddown(r) =2Qdown

r2 , (64)

as follows from D = 4free, where Qdown is the charge on the lower part of theinner sphere.

Combining (61-64),Qdown = Qup, (65)

holds for the free charge on the inner shell.

What about the total charge distribution, which include polarization charges in thedielectric? The polarization vector P obeys 4P= ( 1)E. Thus, the charge density,= 4P n, which appears microscopically on the boundary of the dielectric adjacentto the inner shell, equals 1 times the charge density on the upper shell (sincen =r). In other words, the total microscopic charge densities on the lower andupper parts of the shell are equal. This ensures that the electric field is the same inthe upper and lower part of the capacitor.

Of course,Qdown= Q Qup. (66)

From (65) and (67),Q Qup= Qup, (67)

and hence,

Qup= Q

1 + . (68)

This implies that the potential difference Vbetween the shells is

V = b

a

Edr = 2Q1 +

ba

dr

r2 =

2Q

1 +

1

b 1

a

. (69)

So, the capacitance is

C=Q

V =

1 +

2

ab

a b . (70)

5/21/2018 ph501set1

22/23


8. a) In the presence of an electric field along thezdirection, which is perpendicular tothe plane of the orbit of our model atom, the plane is displaced by a distance z. Wecalculate this displacement from the equilibrium condition: the axial component of theCoulomb force between the electron and the proton should be equal to the force eEon the electron of charge e due to the electric field. Namely,

Fz =e2

r2z

r =eE, (71)

where r =

a2 + z2, and a is the radius of the orbit of the electron. The inducedatomic dipole momentp is

p= ez= r3E a3E, (72)where the approximation holds for small displacements, i.e., small electric fields. Since

p= Ein terms of the atomic polarizability , we estimate that

a3, (73)

wherea is the radius of the atom.

The dielectric constant is related to the atomic polarizability via

1 = 4N, (74)whereNis the number of atoms per cm3. For hydrogen, there are 2 atoms per molecule,and 6 1023 molecules in 22.4 liters, at S.T.P. Hence N= 2(6 1023)/(22.4 103) =5.4 1019 atoms/cm3. Estimating the radiusaas the Bohr radius, 5.3 109 cm, wefind

1 4(5.4 1019)(5.3 109)3 1.0 104. (75)b) If an electric field E is applied to the springlike atom, then the displacement d ofthe electron relative to the fixed (neutral) nucleus is related by F = kd =eE, wherek = m2 is the spring constant in terms of characteristic frequency . The induceddipole momentp is given by

p= ed =e2E

k =

e2E

m2. (76)

Thus the polarizability is given by

= e2

m2

. (77)

The frequency that corresponds to the Lyman line at 1225

A is

=2c

=

2(3 1010)1225 108 = 1.54 10

16 Hz. (78)

From (77),

= (4.8 1010)2

(9.1 1028)(1.54 1016)2 = 1.07 1024 cm3. (79)

5/21/2018 ph501set1

23/23


Finally, 1 = 4N = 4(5.4 1019)(1.07 1024) = 7.3 104 (80)

Thus, our two models span the low and high side of the empirical result. Of course,we have neglected the fact that the hydrogen atoms are actually paired in molecules.

ph501set1

Documents

new york

oscillatory field

electrostatic field

mean value theorem

e2 drawsthe atom

local maximum of e2

electricity dover

classical electrodynamics