Chapter 1

Basic Electrostatics

1.1 Coulombs law, electrostatic eld

1.1.1 Coulombs law

The suppose we have two small charges fq1; q2g centered at positions fx1;x2g. The force on1 due to 2 is then

F12 = kq1q2x12jx12j3 (1.1)

where x12 = x1 x2. See the diagram. This is known as Coulombs law. Notice that the

O

= 1 2

12

1 2 12-

xx

x x x

force is (1) proportional to the strength of the charges; (2) inversely proportional to thesquare of the separation; (3) directed along the line connecting the charges; (4) repulsive forlike charges and attractive for opposite charges.

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1.1.2 Units

We need to discuss units at this point. In SI units, the force is in newtons (N) (1 N = 1 kg ms2), and charge is in Coulombs (C) (an electron carries a charge e equal to 1:6 1019 C).The constant k is

k =1

40 9 109 N m2 C2; (1.2)

so that two charges of 1 C each at a distance of 1 m repel each other with the very large forceof 9 109 N. In the SI system, current is the dened unit. The unit of current is the Ampere(A); it is the current, which if maintained in two innitely long parallel wires of negligiblecross section, placed 1 m apart in vacuum, produces an attractive force per unit length of2 107 N m1. The Coulomb is then a derived unit; it is the quantity of electrical chargetransported in one second by a current of 1 A.

In Gaussian units we chose k = 1; the unit of charge is then the statcoulomb. In thissystem, two charges of 1 statcoulomb each, separated by a distance of 1 cm, produce a forceof 1 dyne (recall that 1 dyne = 1 g cm s2 =105 N). Then

1 statcoulomb = 3:3356 1010 C; e = 4:803 1010 statcoulomb: (1.3)

1.1.3 Electric eld

We dene the electric eld E of a charged object as follows. Take a very small test charge q(small so that it does not disturb the charge distribution whose eld were measuring), andmeasure the force on the test charge as a function of position x. Then the electric eld is

E(x) = limq!0 F(x)=q: (1.4)

Q

2

Of course, physically the smallest q would be the electron charge e. For a point chargeof charge Q centered at the origin the electric eld is

E(x) =Q

40

x

jxj3 =Q

40

err2

; (1.5)

with er the outwardly directed unit vector and r the radial distance from the origin. Now itis an experimental fact that electrostatics is linear, so that the electric elds produced by acollection of point charges fqig at positions fxig simply add:

E(x) =1

40

NXi=1

qix xijx xij3 : (1.6)

This can be written in a dierent form using the Dirac delta function:

E(x) =1

40

Zd3x0

"NX

i=1

qi(3)(x0 xi)

#x x0jx x0j3

=1

40

Zd3x0(x0)

x x0jx x0j3 ; (1.7)

where we have introduced the charge density (x),

(x) =NX

i=1

qi(3)(x xi): (1.8)

1.1.4 The Dirac delta function

The Dirac delta function is a mathematically convenient way of representing singularitiessuch as point charges. While we call it a function, real mathematicians bristle at this termi-nology; mathematically the delta function is dened in terms of the theory of distributions,but well ignore this and abuse the mathematics liberally. One way of dening the deltafunction in one dimension is

(x) =

(1=w if w=2 < x < w=20 otherwise,

(1.9)

and then take the limit that w ! 0 (see gure).We see that the delta function is a spike of unit area. Other representations are

possible. The delta function has the following properties:

1.R11 (x) dx = 1.

2.R11 (x a)f(x) dx = f(a).

3.R11

0(x a)f(x) dx = f 0(a) [integrate by parts and use (2)].

3

x

d (x)

-w/2 w/2

1/w

4. Let f(x) have simple zeros at fxig (i.e., f(x) f 0(xi)(x xi) for x near xi). Then

[f(x)] =X

i

1

jf 0(xi)j(x xi): (1.10)

5. In three dimensions, (3)(x) = (x)(y)(z). In other coordinate systems things are abit more complicated (see homework).

6. Note that in d dimensions, (d)(x) has dimensions of Ld, so that the three dimensionaldelta function has the dimensions of 1/volume.

1.1.5 An application: charged rod

Calculate the electric eld of a rod of length L on which a charge Q is uniformly distributed.Assume that the rod is centered on the z axis. The problem has cylindrical symmetry so weshould use cylindrical coordinates. Then the charge density can be written as

(x) =Q

L

()

2[(L=2)2 z2]; (1.11)

where I have introduced the Heavyside function (x), dened as

(x) =

(0 if x < 01 if x > 0.

(1.12)

To check this, integrate the charge density:

Z L=2L=2

dzZ

2 dQ

L

()

2= Q; (1.13)

which is correct. In cylindrical coordinates (; ; z), x = e + z ez, and

jx x0j =q

2 + 02 20 cos( 0) + (z z0)2: (1.14)

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r,f ,z)(

f

rx

y

z

-L/2

L/2

Then substituting into Eq. (1.7), we have for the electric eld

E(x) =1

40

Z 11

dz0Z 20

d0Z 10

0 d0Q

L

(0)20

[(L=2)2 z02]

e 0e0 + (z z0) ez

[2 + 02 20 cos( 0) + (z z0)2]3=2 : (1.15)

The 0 integration is easily performed because of the delta function; the integrand is thenindependent of 0, so the 0 integral is just 2; and the theta function restricts the z0 integralto be between L=2 and L=2. We then have for the components of the electric eld

E(; z) =Q

40L

Z L=2L=2

dz0

[2 + (z z0)2]3=2 ; (1.16)

Ez(; z) =Q

40L

Z L=2L=2

dz0z z0

[2 + (z z0)2]3=2 : (1.17)

The integrals are complicated, so it is instructive to consider limiting cases. Specializeto z = 0, so that x lies in the plane which bisects the line charge. The integrand of theintegral for Ez is then odd in z

0, so the integral vanishes, which is to be expected due to thesymmetry. For E we have

E(; 0) =1

40

Q

L

Z L=2L=2

dz01

[z02 + 2]3=2

=(Q=L)

20

1q1 + (2=L)2

: (1.18)

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First, suppose that L; then E Q=402, which is the electric eld of a point charge.On the other hand, if L, then E (Q=L)=20, which is the result for an innitelylong line of charge. This latter result is more easily obtained using Gausss law.

Another special case which is easily checked is when the observation point is on thez-axis (i.e., = 0). You should check for yourself that when jzj L=2, the electric eld hasonly a z-component given by Ez Q=40z2 (for z > 0), which is again the eld due to apoint charge Q.

1.2 Gausss law

q

n

E

da

S

r

d Wq

This is proved by noting that for a single charge q

E n = q40

cos

r2; (1.19)

so that the flux of the electric eld through the area element da is

E n da = q40

da cos

r2=

q

40d; (1.20)

where d is the solid angle subtended by da (i.e., r2 d = cos da). Then, integrating overthe entire surface, and noting that

HS d = 4, we haveI

SE n da = q=0: (1.21)

For a collection of charges fqig inside the surface, this becomesI

SE n da = 1

0

Xi

qi = qenc=0

=1

0

ZV

(x) d3x; (1.22)

where the last line holds for a continuous distribution of charge.

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1.2.1 Dierential form of Gausss law

We can convert Gausss law into dierential form by appealing to the divergence theorem,which states that for a vector eld A,

ISA n da =

ZVr A d3x; (1.23)

so that a surface integral can be related to a volume integral. Applying this to Eq. (1.22),we have I

SE n da =

ZVr E d3x

=Z

V

1

0(x) d3x: (1.24)

Then equating the integrands, we have

r E = =0; (1.25)

which is the dierential form of Gausss law.

1.2.2 Applications of Gausss law

These are left as homework problems, and can be found in more elementary books. Thereare three basic types:

Problems with spherical symmetry, which are solved by noting that on a sphere ofradius r the eld is radial and given in magnitude by

4r2E = q=0 (1.26)

where q is the total charge inside the sphere. To the outside world, it is as if all thecharge were at the center of the sphere. The analogous result for the gravitational eldis due to Newton.

Problems with cylindrical symmetry, which similarly give

2E = =0; (1.27)

where is the total charge per unit length inside the cylinder of radius .

\Pillbox" arguments (see Jacksons Fig. 1.4). The best known of these leads to theconclusion that the eld at the surface of a conductor (which is normal to the surface)is related to the surface charge density by E = =0.

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1.3 The electrostatic potential

By using the identity r(1=r) = er=r2, we can write the electric eld of a point charge q asE = r(q=40r). The eld of a set of charges can also be written as a gradient:

E(x) = 140

Xi

qir 1jx xij = r(x); (1.28)

where is the electrostatic potential

(x) =1

40

Xi

qi1

jx xij =1

40

Z (x0)jx x0j d

3x0: (1.29)

For any smooth function , r (r) = 0. Hence

r E = 0: (1.30)

The physical meaning of (x) is as follows: if there are no charges at innity, so that(1) = 0, then q(x) is the work required to bring a charge q from 1 to x (the othercharges being held xed). More generally, the work required to bring a charge q from pointA to point B is

W = Z B

AqE dl = q [(B) (A)] (1.31)

The work done depends only on the end points, not on the path; hence the net workin going around a closed path is zero|the electric eld is conservative. These propertiesare a consequence of r E = 0, as can also be seen from Stokess theorem:

IC

E dl =Z

Sr E n da = 0; (1.32)

where S is any surface bounded by the closed contour C.In the SI system the unit of potential is the Volt (V); 1 V = 1 J/C. The Gaussian

unit is the statvolt; 1 statvolt = 1 statcoulomb/cm. Therefore,

1 statvolt 300 V: (1.33)

1.3.1 Equipotentials and lines of force

The lines of force (also called the eld lines) provide a method for graphing the electric eld.These lines have the following properties:

They are everywhere tangent to the electric eld E. They begin on positive charges and terminate on negative charges. The local density of the eld lines is proportional to the strength of the electric eld.

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The electric eld lines do not cross (otherwise the eld would not be unique at thatpoint).

The lines of force are tangent to the electric eld and therefore for a point charge aretangent to the force exerted by the eld on the particle.

The lines of force are not particle trajectories! The particle trajectories are obtainedby solving F = ma with F = qE.

The equipotentials are contours of constant electrostatic potential. They are analogousto the contours on a topographic map. They are perpendicular to the lines of force.

1.3.2 Fields of a point charge, point dipole, line charge, and sur-face charge

Point charge (x) = q(3)(x) = q(x)(y)(z):

E =q

40

err2

; (1.34)

=q

40r: (1.35)

Dipole. Start with two point charges q and q at positions x0 and x00, separated by adisplacement d = x0x00. The observation point P is at position x. Then the potential

O

x

P

x"

x

dq-q

at P is

(x) =1

40

"q

jx x0j q

jx x0 + dj#: (1.36)

Expand for small d:

(x) 140

p (x x0)jx x0j3 ; (1.37)

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where the dipole moment p = qd. This can also be written as

(x) =1

40

p cos

r2; (1.38)

where is the angle between the dipole moment and the observation point. The electriceld is

E(x) =1

40

3n(p n) pjx x0j3 ; (1.39)

where n is a unit vector directed from x0 to x.

Straight line charge (x) = (x)(y) :

E =

20

e

; (1.40)

= (=20) ln(0=): (1.41)

Plane charge (x) = (x):Ex = =20 sign(x); (1.42)

= jxj =20: (1.43)That is, the potential has a cusp and the eld is discontinuous. For a non-planar chargelayer, there is the same discontinuity, but is not symmetric for x:

1.4 Surface Distributions of Charges and Dipoles

E

n

E1

2

s (x)

Suppose that we have a surface with a surface charge distribution (charge per unitarea) (x). Then by applying Gausss Law in integral form to the surface (see gure), wehave

(E2 E1) n = (x)=0: (1.44)

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For electrostatic elds we also haveHC E dl = 0, which when applied to the surface (take a

loop which cuts through the surface) tells us that the component of E tangent to the surfaceis continuous; i.e., that

E1;t = E2;t: (1.45)

It turns out that this is also true for time dependent elds. Equations (1.44) and (1.45) arethe fundamental boundary conditions used in solving boundary value problems in electro-statics.

Another problem of interest is a surface distribution of dipoles. This is constructed bytaking two surfaces with surface charge densities (x) and (x) and letting them approacheach other. If they are separated by a distance d(x), then we can dene a surface dipolestrength D(x) as

D(x) = limd(x)!0

(x)d(x): (1.46)

We can think of the surface dipole density as a collection of point dipoles of dipole momentp = nD(x0) da0. For a single point dipole, recall that the potential is

(x) =1

40

p (x x0)jx x0j3 ; (1.47)

so that for a surface dipole density we sum (integrate) over the surface,

(x) =1

40

ZS

D(x0)n (x x0)jx x0j3 da

0: (1.48)

This result has a simple interpretation. First,

n

da

dW

|x-x|

q

n (x x0)jx x0j3 da

0 = cos da0

jx x0j2 = d; (1.49)

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where d is the element of solid angle subtended at the observation point by the area elementda0.

Therefore, the potential due to the surface dipole layer is

(x) = 140

ZS

D(x0) d: (1.50)

A surface with a constant surface dipole density D then produces a potential

= D40

; (1.51)

with the solid angle subtended by the surface at the observation point. The surface dipolelayer produces a discontinuity in the electrostatic potential.

Surface dipole layers are important in discussions of the physics of metal surfaces. Ina simple model, the positive ionic charge is assumed to be uniformly distributed throughoutthe metal (\jellium"); the lighter, negatively charged electrons have a nonuniform density,especially near the surface, where they leak out into the vacuum via quantum mechanicaltunneling. The result is a surface dipole layer which serves to keep the electrons within thesample, and which makes a contribution to the work function of the metal surface.

electron charge density

ionic charge density

net charge density

z0

1.5 Conductors

Next, a refresher on conductors.

A conductor is a material which has mobile electric charges. Due to the mobile charges, an electrostatic eld cannot penetrate into a conductor.

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The electric eld at the surface of a conductor is normal to the surface|if it had acomponent parallel to the surface, the mobile charges would move until an equilibriumwith zero parallel component is achieved.

Using Gausss law, the normal component of the electric eld at the surface is E n ==0, where n is a normal directed outward from the conductor and is the surfacecharge density.

The entire conductor must be at the same potential (since E = r = 0); in partic-ular, the surface of a conductor is an equipotential surface.

1.6 Poissons equation

Starting from Coulombs law we have derived the two dierential eld equations of electro-statics :

r E = 0; (1.52)r E = =0: (1.53)

The most general solution of the rst equation can be written E = r, where at this point is unspecied. Inserting in the second equation, we nd that must satisfy Poissonsequation:

r2 = =0: (1.54)In a region of space with no sources ( = 0), this reduces to Laplaces equation,

r2 = 0: (1.55)

A good part of this course is devoted to nding solutions of these two equations.Conversely, one can show directly that the only solution of the Poisson equation that

vanishes at innity is the familiar

(x) =1

40

Z (x0)jx x0 j d

3x0: (1.56)

To prove that (1.56) is a solution of (1.54), we make use of

r2 1jx x0j = 4(3)(x x0); (1.57)

which is really nothing but Poissons equation when is a point charge. If this equation isgranted, then

r2(x) = 140

Z(x0)r2

1

jx x0j!

d3x0 = 10

Z(x0)(3)(x x0) d3x0 (1.58)

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gives immediately r2(x) = (x)=0. Then, putting r = x x0, all we need to show isthat

r2

1

r

= 4(3)(r): (1.59)

The easiest way to proceed is to note that for r 6= 0,

r2

1

r

=

1

r2@

@rr2

@

@r

1

r= 0; (1.60)

but for r = 0 we have a singularity with integrated strength

Zr2

1

r

d3x =

ZS

r1

r

n da (1.61)

by the divergence theorem. The integral is conveniently evaluated on a sphere of nite radiusR, and it gives 4R2 (1=R2) = 4.

Another way to handle the singular behavior is to replace 1/r with a \regularized"function, such as

1pr2 + a2

; (1.62)

and then let a ! 0. This is done in Jackson and may appeal to the rigor-minded.

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