Chapter 1

Electrostatics I. Potential due toPrescribed Charge Distribution,Dielectric Properties, Electric Energyand Force

1.1 Introduction

In electrostatics, charges are assumed to be stationary. Electric charges exert force on other charges

through Coulomb�s law which is the generic law in electrostatics. For a given charge distribution,

the electric �eld and scalar potential can be calculated by applying the principle of superposition.

Dielectric properties of matter can be analyzed as a collection of electric dipoles. Atoms having

no permanent dipole moment can be polarized if placed in an electric �eld. Most molecules have

permanent dipole moments. In the absence of electric �eld, dipole moments are oriented randomly

through thermal agitation. In an electric �eld, permanent dipoles tend to align themselves in the

direction of the applied �eld and weaken the �eld. Some crystals exhibit anisotropic polarization and

the permittivity becomes a tensor. The well known double di¤raction phenomenon occurs through

deviation of the group velocity from the phase velocity in direction as well as in magnitude.

1.2 Coulomb�s Law

Normally matter is charge-neutral macroscopically. However, charge neutrality can be broken

relatively easily by such means as mechanical friction, bombardment of cosmic rays, heat (e.g.,

candle �ame is weakly ionized), etc. The �rst systematic study of electric force among charged

bodies was made by Cavendish and Coulomb in the 18th century. (Cavendish�s work preceded

Coulomb�s. However, Coulomb�s work was published earlier. Lesson: it is important to publish

your work as early as possible.) The force to act between two charges, q1 at r1 and q2 at r2; follows

1

the well known Coulomb�s law,

Figure 1-1: Repelling Coulomb force between like charges.

F = const.q1q2r2

= const.q1q2

jr1 � r2j2; (N) (1.1)

where r = jr1 � r2j is the relative distance between the charges. In CGS-ESU (cm-gram-second

electrostatic unit) system, the constant is chosen to be unity. Namely, if two equal charges q1 =

q2 = q separated by r = 1 cm exert a force of 1 dyne = 10�5 N on each other, the charge is de�ned

as 1 ESU ' 13 � 10

�9 C. The electronic charge in ESU is e = 4:8 � 10�10 ESU. In MKS-Ampere(or SI) unit system, 1 Coulomb of electric charge is de�ned from

1 Coulomb = 1 Ampere � 1 sec,

where 1 Ampere of electric current is de�ned as follows. If two in�nitely long parallel currents of

equal amount separated by 1 meter exert a force per unit length of 2 � 10�7 N/m on each other

(attracting if the currents are parallel and repelling if antiparallel), the current is de�ned to be 1

Ampere. Since the force per unit length to act on two parallel currents I1 and I2 separated by a

distance d is given byF

l= B1I2 =

�02�

I1I2d; (N/m) (1.2)

the magnetic permeability �0 = 4��10�7 H/m in MKS-Ampere unit system is an assigned constantintroduced to de�ne 1 Ampere current. (The permittivity "0 is a measured constant that should

be determined experimentally.)

In MKS-Ampere unit system, the measured proportional constant in Coulomb�s law is approx-

imately 9:0� 109 N m2 C�2: It is customary to write the constant in the form

const. =1

4�"0;

2

Figure 1-2: In MKS unit system, I = 1 Ampere current is de�ned if the force per unit length betweenin�nite parallel currents 1 m apart is 2 � 10�7 N/m. The magnetic permeability �0 = 4� � 10�7H/m is an assigned constant to de�ne 1 Ampere current.

where

"0 = 8:85� 10�12C2

N �m2

�=Faradm

�;

is the vacuum permittivity. Since

c2 =1

"0�0; m2s�2;

the permittivity "0 can be deduced from the speed of light in vacuum that can be measured

experimentally as well.

1.3 Coulomb Electric Field and Scalar Potential

The Coulomb�s law may be interpreted as a force to act on a charge placed in an electric �eld

produced by other charges, since the Coulomb force can be rewritten as

F =1

4�"0

q1(r2 � r1)jr1 � r2j3

� q2

= q2Eq1 ; (N) (1.3)

where

Eq1 =1

4�"0

q1(r2 � r1)jr1 � r2j3

;

�NC=Vm

�(1.4)

is the electric �eld produced by the charge q1 at a distance r1� r2: The �eld is associated with thecharge q1 regardless of the presence or absence of the second charge q2: (The factor 4� is the solid

3

angle pertinent to spherical coordinates. If it is ignored as in the CGS-ESU system, it pops up in

planar coordinates. Corresponding Maxwell�s equation is

r �E = �

"0; in MKS-Ampere,

or

r �E = 4��; in CGS-ESU.

Which unit system to choose is a matter of conveneince and it is nonsense to argue one is superior

to other.)

Since two charges q1 and q2 exert the Coulomb force on each other, assembling a two-charge

system requires work. A di¤erential work needed to move the charge q2 against the Coulomb force

is

dW = �F � dr2 = �1

4�"0

q1q2(r2 � r1)jr1 � r2j3

� dr2:

Noting

rr21

jr1 � r2j= +

r1 � r2jr1 � r2j3

;

integration from r2 =1 to r2 can be readily carried out,

W =1

4�"0

q1q2jr1 � r2j

; (J) (1.5)

This is the potential energy of two-charge system. Note that it can be either positive or negative. If

W is positive, the charge system has stored a potential energy that can be released if the system is

disassembled. Energy released through nuclear �ssion process is essentially of electrostatic nature

associated with a system of protons closely packed in a small volume. On the other hand, if W

is negative, an energy jW j must be given to disassemble the system. For example, to ionize ahydrogen atom at the ground state, an energy of 13:6 eV is required. In a hydrogen atom, the total

system energy is given by

W =1

2mv2 � e2

4�"0rB= � e2

8�"0rB< 0;

where rB = 5:3� 10�11 m is the Bohr radius and

1

2mv2 =

e2

8�"0rB= 13:6 eV,

is the electron kinetic energy. The electric potential energy is

� e2

4�"0rB= �27:2 eV.

4

The potential energy of two-charge system may be written as

W =1

4�"0

q1jr1 � r2j

� q2; (J)

which de�nes the potential associated with a point charge q1 at a distance r;

� =1

4�"0

q

r; (J C�1 = V); (1.6)

and the electric �eld is related to the potential through

E = �r�; (V m�1): (1.7)

Both potential � and electric �eld E are created by a charge q regardless of the presence or absence

of a second charge.

1.4 Maxwell�s Equations in Electrostatics

In electrostatics, electric charges are stationary being held by forces other than of electric origin

such as molecular binding force. Since charges are stationary, no electric currents and thus no

magnetic �elds are present, B = 0:

For a distributed charge density �(r), the electric �eld can be calculated from

E(r) =1

4�"0

ZV

�(r0)(r� r0)jr� r0j3

dV 0 = � 1

4�"0rZ

�(r0)

jr� r0jdV0; (1.8)

since the di¤erential electric �eld due to a point charge dq = �(r0)dV 0 located at r0 is

dE =1

4�"0

�(r0)dV 0

jr� r0j3(r� r0):

Note that

r 1

jr� r0j = �r0 1

jr� r0j = �r� r0

jr� r0j3;

where r0 is the gradient with respect to r0. Eq. (1.8) allows one to calculate the electric �eld for agiven charge density distribution �(r):

In order to �nd a di¤erential equation to be satis�ed by the electric �led, we �rst note that the

surface integral of the electric �eld IE � dS;

can be converted to a volume integral of the divergence of the �eld (Gauss�mathematical theorem,

5

not to be confused with Gauss�law),ISE � dS =

ZVr �EdV

= � 1

4�"0

ZVdVr2

ZV 0

�(r0)

jr� r0jdV0:

However, the function 1= jr� r0j satis�es a singular Poisson�s equation,

r2 1

jr� r0j = �4��(r� r0): (1.9)

Therefore, ISE � dS = 1

"0

ZVdV

ZV 0�(r� r0)�(r0)dV 0 = 1

"0

Z�(r)dV: (1.10)

This is known as Gauss� law for the electric �eld. Note that Gauss� law is a consequence of the

Coulomb�s inverse square law. FromZr �EdV = 1

"0

Z�(r)dV; (1.11)

it also follows that

r �E = �

"0: (1.12)

This is the di¤erential form of the Gauss�law for the longitudinal electric �eld and constitutes one

of Maxwell�s equations.

For the electric �eld due to a static charge distribution,

E(r) =1

4�"0

ZV

�(r0)(r� r0)jr� r0j3

dV 0 = � 1

4�"0rZV

�(r0)

jr� r0jdV0;

its curl identically vanishes because

r�E = � 1

4�"0r�r

Z�(r0)

jr� r0jdV0 � 0:

(Note that for any scalar function F; r � rF � 0:) Therefore, the second Maxwell�s equation inelectrostatics is

r�E = 0: (1.13)

This is a special case of the more general Maxwell�s equation

r�E = �@B@t; (1.14)

which determines the transverse component of the electric �eld. Evidently, if all charges and �elds

are stationary, there can be no magnetic �eld. It should be remarked that a vector �eld can be

uniquely determined only if both its divergence and curl are speci�ed. (This is known as Helmholtz�s

6

theorem.) In electrodynamics, two vector �elds, the electric �eld E and magnetic �eld B, are to befound for given charge and current distributions �(r; t) and J(r; t): Therefore it is not accidental

that four Maxwell�s equations emerge specifying the four functions r �E; r�E; r �B; and r�Bwhich determine the longitudinal and transverse components of E and B.Digression: A vector A can be decomposed into the longitudinal and transverse components Aland At which are, by de�nition, characterized by

r�Al = 0; r �At = 0:

The longitudinal component can be calculated from

Al (r) = �1

4�rZ r0 �A (r0)

jr� r0j dV0;

since

r2 1

jr� r0j = �4���r� r0

�:

The transverse component is given by

At (r) = A (r)�Al (r)

=1

4�r�r�

ZA (r0)

jr� r0jdV0;

where the identity

r�r�A =rr �A�r2A;

is exploited. It is known that for a vector to be uniquely de�ned, its divergence and curl have to

be speci�ed. Since r �A =r �Al; divergence of a vector spci�es the divergence of the longitudinalcomponent. Likewise, r�A =r�At specifes the curl of the tarnverse component. As a concreteexample, let us consider the electric �eld. Its divergence is

r �E = �

"0;

and the solution to this di¤erential equation is

El (r) = �1

4�"0rZV

�(r0)

jr� r0jdV0:

The tranverse component is speci�ed by the magnetic �eld as

r�E = �@B@t:

The vanishing curl of the static electric �eld allows us to write the electric �eld in terms of a

7

gradient of a scalar potential �,

E = �r�; (1.15)

for the curl of a gradient of a scalar function identically vanishes,

r�r� � 0:

For a given charge distribution, the potential has already been formulated in Eq. (1.8),

�(r) =1

4�"0

ZV

�(r0)

jr� r0jdV0: (1.16)

In terms of the scalar function �(r); Eq. (1.12) can be rewritten as

r2� = � �"0; (1.17)

which is known as Poisson�s equation. In general, solving the scalar di¤erential equation for � is

easier than solving vector di¤erential equations for E.

The physical meaning of the scalar potential is the amount of work required to move adiabati-

cally a unit charge from one position to another. Consider a charge q placed in an electric �eld E.The force to act on the charge is F = qE and if the charge moves over a distance dl; the amount

of energy gained (or lost, depending on the sign of qE � dl) by the charge is

qE � dl = �qr� � dl:

Therefore, the work to be done by an external agent against the electric force to move the charge

from position r1 to r2 is

W = �qZ r2

r1

E � dl = qZ r2

r1

r� � dl = q [�(r2)� �(r1)] ; (1.18)

where �(r2) � �(r1) is the potential di¤erence between the positions r1 and r2: The work isindependent of the choice of the path of integration. The potential is a relative scalar quantity and

a constant potential can be added or subtracted without a¤ecting the electric �eld.

In a conductor, an electric �eld drives a current �ow according to the Ohm�s law,

J = �E; (A m�2) (1.19)

where � (S m�1) is the electrical conductivity. In static conditions (no time variation and no �ow

of charges), the electric �eld in a conductor must therefore vanish. A steady current �ow and

constant electric �eld can exist in a conductor if the conductor is a part of closed electric circuit.

However, such a circuit is not static because of the presence �ow of charge. For the same reason,

a charge given to an isolated conductor must entirely reside on the outer surface of the conductor.

Since E (static) = 0 in a conductor, the volume charge density � must also vanish according to

8

� = "0r � E = 0: A charge given to a conductor can only appear as a surface charge � (C m�2)

on the outer surface. The time scale for a conductor to establish such electrostatic state may be

estimated from the charge conservation equation

@�

@t+r � J = 0: (1.20)

The current density can be estimated from the equation of motion for electrons

m@v

@t= �eE�m�ev; (1.21)

where �c (s�1) is the electron collision frequency with the lattice ions. Noting J = �nev with nthe density of conduction electrons, we obtain�

@

@t+ �c

�J =

ne2

mE: (1.22)

Therefore, the equation for the excess charge density in a conductor becomes

@

@t

�@

@t+ �c

��+ !2p� = 0; !2p =

ne2

"0m; (1.23)

which describes a damped plasma oscillation �0e� t�i!pt with an exponential damping factor =

�c=2: The typical electron collision frequency in metals is of order �c ' 1013= sec and the time

constant to establish electrostatic state is indeed very short. (If the low frequency Ohm�s law is

used, an unrealistically short transient time emerges,

@�

@t+�

�0� = 0; � = �0e

�t=� ;

where � = �0=� ' 10�19 sec.)The principle that a charge given to a conductor must reside entirely on its outer surface is also

a consequence of Coulomb�s 1=r2 law. For a conductor of an arbitrary shape, the surface charge

distribution is so arranged that the electric �eld inside the conductor vanishes everywhere. In Fig.

1-3, one conducting spherical shell surrounds a smaller one. The spheres are connected through

a thin wire. A charge originally given to the inner sphere will end up at the outer surface of the

larger sphere and the electric �eld inside should vanish if Coulomb�s law is correct. Experiments

have been conducted to measure residual electric �elds inside and it has been established that the

power � in the Coulomb�s law F _ 1=r� is indeed very close to 2 within one part in 1016: � is

probably exactly equal to 2.0. If not, there would be a grave consequence that a photon should

have a �nite mass. This is because if a photon has a mass mp; corresponding Compton wavenumber

kC = mpc=~ will modify the wave equation for all potentials, including the Coulomb potential, inthe form �

r2 � k2C �1

c2@2

@t2

�� = � �

"0: (1.24)

9

Figure 1-3: Charge initially given to the inner conductor will all end up at the outer surface of theouter conductor. The absence of the electric �eld inside the sphere is a consequence of Coulomb�slaw.

In static case @=@t = 0; this reduces to

�r2 � k2C

�� = � �

"0; (1.25)

and for a point charge �(r) = q�(r); yields a Debye or Yukawa type potential,

�(r) =1

4�"0

q

rexp (�kCr) ; (1.26)

and electric �eld

Er =q

4�"0

�1

r2+kCr

�e�kCr: (1.27)

The power � in the Coulomb�s law Er _ q=ra experimentally established is � � 2 = O(10�16);and this corresponds to an upper limit of photon mass of mp < 10�51 kg. The lower limit of

photon Compton wavelength is �C > 4� 105 km at which distance a signi�cant deviation from the

Coulomb�s law, if any, is expected. (Incidentally, in a plasma, the static scalar potential does have

a Debye form,

�(r) =1

4�"0

q

rexp (�kDr) ; (1.28)

where

kD =

sne2

"0T; (1.29)

is the inverse Debye shielding distance, n is the plasma density and T (in Joules) is the plasma

10

temperature. Likewise, in a superconductor, static magnetic �eld obeys

�r2 � k2L

�B = 0; (1.30)

where kL = !p=c is the London skin depth with !p =pne2="0me the electron plasma frequency.)

The potential energy of a two-charge system is

W =1

4�"0

q1q2jr1 � r2j

= 2� 12

1

4�"0

q1q2jr1 � r2j

; (1.31)

where1

2

1

4�"0

q1q2jr1 � r2j

; (1.32)

is the potential energy associated with each charge. For many charge system, this can be generalized

in the form

W =1

2

Xj

�jqj =1

2

1

4�"0

Xi6=j

qiqjjri � rj j

; (1.33)

where

�j =1

4�"0

Xi6=j

qijri � rj j

; (1.34)

is the potential at the location of charge qj due to all other charges. For distributed charge with

a local charge density �(r) (C/m3); the potential energy of a di¤erential charge dq = �dV , which

can be regarded as a point charge if dV is su¢ ciently small, is

dW =1

2�dq =

1

2��dV:

Since � = "0r �E;dW =

1

2"0�r �EdV: (1.35)

Integrating over the entire volume, we �nd

W =1

2"0

ZV�r �EdV

=1

2"0

ZV[r � (�E)� (r � �)E] dV

=1

2"0

IS�E�dS+

ZV

1

2"0E

2dV: (1.36)

where use is made of Gauss�theorem,ZVr � FdV =

ISF�dS; (1.37)

for an arbitrary well de�ned vector �eld F. The surface integral vanishes because at in�nity, bothpotential and electric �eld vanish. Therefore, the potential energy associated with a distributed

11

charge system is

W =

Z1

2"0E

2dV; (1.38)

which is positive de�nite. In the expression for the potential energy of discrete charge system,

the potential energy due to a point charge itself is excluded while in the integral form, spatial

distribution of charge is assumed even for point charges and the so-called self energy is included.

The quantity

ue =1

2"0E

2; (J/m3) (1.39)

is the electric energy density associated with an electric �eld. This expression for the energy density

holds regardless of the origin of the electric �eld, and can be used for �elds induced by time varying

magnetic �eld as well.

The self-energy of an ideal point charge evidently diverges because the electric �eld proportional

to 1=r2 does in the limit of r ! 0. However, if the electron is assumed to have a �nite radius re;

the self-energy turns out to be of the order of

We '1

4�"0

e2

re: (1.40)

This should not exceed the rest energy of the electron mec2: Equating these two, the following

estimate for the electron radius emerges,

re '1

4�"0

e2

mc2= 2:85� 10�15 m. (1.41)

Although this result should not be taken seriously, the scattering cross section of free electron

placed in an electromagnetic wave (the process known as Thomson scattering) does turn out to be

� =8

3�r2e ;

and the concept of electron radius is not totally absurd. For proton whose radius is also of order

10�15 m, the analogy obviously breaks down.

The Poisson�s equation in Eq. (1.17) is the Euler�s equation to make the energy-like integral

stationary

U =

Z �1

2�0E

2 � ���dV =

Z �1

2�0(r�)2 � ��

�dV: (1.42)

Indeed, the variation of this integral

�U =

Z(�0r� � r��� ���) dV

= �Z �

�0r2�+ ����dV;

12

becomes stationary (�U = 0) when the Poisson�s equation holds,

r2�+ �

"0= 0:

In other words, electrostatic �elds are realized in such a way that the total electric energy becomes

minimum. In general, electric force act so as to reduce the energy in a closed (isolated) system.

Macroscopically, electric force tends to increase the capacitance as we will see in Chapter 2.

If the charge density distribution is known, the potential �(r) can be readily found as a solution

of the Poisson�s equation. However, if the charge density � is unknown a priori, as in most potential

boundary value problems, the potential and electric �eld must be found �rst. Then the charge

density is to be found from � = "0r �E; or in the case of surface charge on a conductor surface,

� = "0En; (C m�2);

where En is the electric �eld normal to the conductor surface.

1.5 Formal Solution to the Poisson�s Equation

As shown in the preceding section, if the spatial distribution of the charge density �(r) is given,

the solution for the Poisson�s equation

r2�(r) = ��(r)�0; (1.43)

can be written down as

�(r) =1

4��0

ZV

�(r0)

jr� r0jdV0: (1.44)

This is understandable because the di¤erential potential due to a point charge dq = �dV is

d� =1

4�"0

�(r0)dV 0

jr� r0j ;

and the solution in Eq. (1.44) is a result of superposition.

It is noted that the Poisson�s equation for the scalar potential still holds even for time varying

charge density,

r2�C(r; t) = ��(r; t)

"0; (1.45)

if one employs the Coulomb gauge characterized by the absence of the longitudinal vector potential

r �A = 0: (1.46)

13

Figure 1-4: Di¤erential potential d� due to a �point charge�dq = �dV 0.

In the Coulomb gauge, the transverse vector potential satis�es the wave equation,�r2 � 1

c2@2

@t2

�At = ��0Jt; (1.47)

where Jt is the transverse current. The solution for the scalar potential in the Coulomb gauge is

non-retarded,

�C(r; t) =1

4��0

ZV

�(r0; t)

jr� r0jdV0; (1.48)

and so is the resultant longitudinal electric �eld,

El(r; t) = �r�C =1

4�"0

ZV

r� r0

jr� r0j3�(r0; t)dV 0: (1.49)

Such non-retarded (instantaneous) propagation of electromagnetic disturbance is clearly unphysical

and should not exist. In fact, the non-retarded Coulomb electric �eld is exactly cancelled by a term

contained in the retarded transverse electric �eld

Et(r; t) = �@At@t

= ��04�

@

@t

ZJt(t� �)jr� r0j dV

0; � =jr� r0jc

;

as will be shown in Chapter 6.

The potential in Eq. (1.44) is in the form of convolution between the function

G(r; r0) = G(r� r0) = 1

4�jr� r0j ; (1.50)

14

and the source function �(r0)=�0: The function G(r; r0) is called the Green�s function and introduced

as a solution to the following singular Poisson�s equation

r2G = ��(r� r0); (1.51)

subject to the boundary condition that G = 0 at r = 1: (G subject to this boundary condition

is called the Green�s function in free space. It is a particular solution to the singular Poisson�s

equation. Later, we will generalize the Green�s function so that it vanishes on a given closed

surface by adding general solutions satisfying r2G = 0:) It is evident that the Green�s function

is essentially the potential due to a point charge, for the charge density of an ideal point charge

(without spatial extension) q located at r0 can be written as

�c = q�(r� r0): (1.52)

Here �(r� r0) is an abbreviation for the three dimensional delta function. For example, in thecartesian coordinates,

�(r� r0) = �(x� x0)�(y � y0)�(z � z0); (1.53)

in the spherical coordinates (r; �; �),

�(r� r0) = �(r � r0)�[r(� � �0)]�[r sin �(�� �0)] (1.54)

=�(r � r0)rr0 sin �

�(� � �0)�(�� �0) (1.55)

=�(r � r0)rr0

�(cos � � cos �0)�(�� �0); (1.56)

and in the cylindrical coordinates (�; �; z),

�(r� r0) = �(�� �0)�[�(�� �0)]�(z � z0) (1.57)

=�(�� �0)

��(�� �0)�(z � z0): (1.58)

The singular Poisson�s equation

r2G = ��(r� r0); (1.59)

can be solved formally as follows. Let the Fourier transform of G(r� r0) be g(k); namely,

g(k) =

ZG(r� r0)e�ik�(r�r

0)dr: (1.60)

Its inverse transform is

G(r� r0) = 1

(2�)3

Zg(k)eik�(r�r

0)d3k: (1.61)

Noting that the operator r is Fourier transformed as ik and �(r� r0) as unity, we readily �nd from

15

Eq. (1.59),

g(k) =1

k2: (1.62)

Substitution into Eq. (1.61) yields

G(r� r0) =1

(2�)3

Z1

k2eik�(r�r

0)d3k

=1

(2�)3

Z 1

0dk

Z �

0eikjr�r

0j cos � sin �d�

Z 2�

0d�

=1

(2�)2�

Z �

0�(jr� r0j cos �) sin �d�

=1

4�jr� r0j ; (1.63)

where the polar angle � in the k-space is measured from the direction of the relative position vector

r� r0; andd3k = k2dk sin �d�d�; (1.64)Z 1

�1eikxdk = 2��(x); (1.65)

�(ax) =1

jaj�(x); (1.66)

are noted. (The same technique will be used in �nding a Green�s function for the less trivial wave

equation, �r2 � 1

c2@2

@t2

�G(r; r0; t; t0) = ��(r� r0)�(t� t0); (1.67)

in Chapter 6.)

If the charge density distribution is spatially con�ned within a small volume such that r � r0;

the inverse distance function 1=jr� r0j may be Taylor expanded as follows:

1

jr� r0j =1

r�r

�1

r

�� r0 + 1

2!rr

�1

r

�: r0r0 � � � �

=1

r+r � r0r3

+1

2r5

Xi;j

(3rirj � r2�ij)r0ir0j + � � � (1.68)

and so is the potential

�(r) =1

4��0

ZV

�(r0)

jr� r0jdV0 =

1

4��0

�1

r

ZV�(r0)dV 0 +

r

r3�ZVr0�(r0)dV 0 +

3rr� r212r5

: Q+ � � ��;

(1.69)

where 1 is the unit dyadic. The quantity q =R�(r0)dV 0 is the total charge contained in the (small)

volume V and the corresponding lowest order monopole potential is

�monopole =1

4��0

q

r: (1.70)

16

The vector quantity

p =

Zr�(r)dV; (1.71)

is the dipole moment and the corresponding dipole potential is

�dipole =1

4��0

p � rr3: (1.72)

The tensor

Q = Qij =

Zrirj�(r)dV; (1.73)

de�nes the quadrupole moment and the corresponding quadrupole potential is

�quadrupole =1

4��0

3rr� r212r5

: Q =1

4��0

1

2r5

Xi;j

(3rirj � r2�ij)Qij : (1.74)

Example 1 Electric Field Lines of a Dipole

The potential due to a dipole moment directed in z direction p = pez at the origin is given by

�(r; �) =1

4��0

p

r2cos �:

The equipotential surfaces are described by

cos �

r2= const.

The electric �eld can be found from

E = �r� = 1

4��0

�2p cos �

r3er +

p sin �

r3e�

�: (1.75)

The equation to describe an electric �eld line is, by de�nition,

dr

Er=rd�

E�; (1.76)

which givesdr

r= 2 cot �d�;

and thus

r = const. sin2 �: (1.77)

Evidently, the electric �eld lines are normal to the equipotential surfaces. In Fig.1-5, one equipo-

tential surface and three electric �eld lines are shown.

Example 2 Linear quadrupole

17

10.50­0.5­1

1

0.5

0

­0.5

­1

Figure 1-5: Equipotential surface and electric �eld lines of an electric dipole pz: The z-axis isvertical.

A linear quadrupole consists of charges q at z = �a and �2q at z = 0 as shown in Fig.1-6. Thecharge density is

�(r) = q[�(z � a)� 2�(z) + �(z + a)]�(x)�(y):

Therefore, only Qzz is nonvanishing,

Qzz =

Zz2�(r)dV = 2a2q;

and the quadrupole potential at r � a is given by

�(r) =1

4��0

1

2r5(3z2 � r2)Qzz =

1

4��0

2qa2

r33 cos2 � � 1

2=

1

4��0

2qa2

r3P2(cos �);

where

P2(cos �) =3 cos2 � � 1

2;

is the Legendre function of order l = 2.

Alternatively, the potential can be found from the direct superposition of the potentials pro-

duced by each charge,

�(r; �) =q

4��0

�1p

r2 + a2 � 2ar cos �� 2r+

1pr2 + a2 + 2ar cos �

�:

For r > a; the functions 1=pr2 + a2 � 2ar cos � can be expanded in terms of the Legendre functions

18

Figure 1-6: Linear quadrupole.

Pl(cos �) as1p

r2 + a2 � 2ar cos �=1

r

1Xl=0

�ar

�l(�1)lPl(cos �); r > a:

Retaining terms up to l = 2 (quadrupole), we recover

�(r; �) ' 1

4��0

2qa2

r3P2(cos �); r � a:

At r � a; the leading order potential is of quadrupole. The total charge is zero, thus no monopole

potential at r � a. Also, the charge system consists of two dipoles of equal magnitude oriented in

opposite directions. Therefore, the dipole potential vanishes at r � a as well. Equipotential pro�le

is shown in Fig. ??.

1.6 Potential due to a Ring Charge: Several Methods

A given potential problem can be solved in di¤erent coordinates systems. Of course, the solution

is unique, and answers found by di¤erent methods should all agree. For the purpose of becoming

familiar with several coordinates systems and some useful mathematical techniques, here we �nd

the potential due to a ring charge having radius a and total charge q uniformly distributed using

several independent methods.

Method 1: Direct integrationThe potential is symmetric about the z-axis because of the uniform charge distribution. There-

fore, we may evaluate the potential at arbitrary azimuthal angle � and here we choose � = �=2,

that is, observing point in the y � z plane. The di¤erential potential due to a �point charge�

19

­1

­0.5

0

0.5

1

­1 ­0.5 0.5 1

Figure 1-7: Equipotential surface of a linear quadrupole. The z axis (� = 0) is in the verticaldirection. The thick line is at +1 unit potential, and the thin line is at �1:

dq = qd�0=2� located at angle �0 is

d� =1

4��0

dq

jr� r0j =1

4��0

q

2�pr2 + a2 � 2ar cos

d�0; (1.78)

where is the angle between the vectors r = (r; �; � = �=2) and r0 = (a; �0 = �=2; �0): cos reduces

to

cos = cos � cos �0 + sin � sin �0 cos(�� �0) = sin � sin�0: (1.79)

Integrating over �0 from 0 to 2�; we �nd

�(r; �) =1

4��0

q

2�

Z 2�

0

d�0pr2 + a2 � 2ar sin � sin�0

: (1.80)

Changing the variable from �0 to � through

2� = �0 +�

2;

the integral can be reduced to

4pr2 + a2 + 2ar sin �

Z �=2

0

1p1� k2 sin2 �

d� =4p

r2 + a2 + 2ar sin �K(k2); (1.81)

20

Figure 1-8: Uniform ring charge.

where

k2 =4ar sin �

r2 + a2 + 2ar sin �; (1.82)

21

is the argument of the complete elliptic integral of the �rst kind de�ned by

K(k2) �Z �=2

0

1p1� k2 sin2 �

d�: (1.83)

The �nal form of the potential is

�(r; �) =q

2�2�0

1pr2 + a2 + 2ar sin �

K(k2): (1.84)

f (x) =

Z �=2

0

1p1� x sin2 �

d�

2

2.5

3

3.5

4

4.5

0 0.2 0.4 0.6 0.8 1

Figure 1-9: K(x) the complete elliptic integral of the �rst kind. It diverges logarithmically at x . 1;K(x) ' ln

�4=p1� x

�:

The function K(k2) is shown in 1-9. It diverges as x = k2 approaches unity, that is, near the

ring itself, as expected. However, the divergence is only logarithmic,

limk2!1��

= ln

�4p�

�:

Example 3 Capacitance of a Thin Conductor Ring

Let us assume a thin conducting ring with ring radius a and wire radius � (� a) shown in

Fig.1-10. The potential on the ring surface can be found by letting r = a��; � = �=2: In this limit,

22

the argument k2 approaches unity,

k2 =4a(a� �)

(a� �)2 + a2 + 2a(a� �) ' 1�� �2a

�2;

and the ring potential becomes

�ring 'q

4�2�0aln

�8a

�

�: (1.85)

Then the self-capacitance of the ring can be found from

C =q

�ring=

4�2�0a

ln

�8a

�

� ; a� �: (1.86)

This formula is fairly accurate even for a not-so-thin ring because of the mere logarithmic depen-

dence on the aspect ratio, a=�: For example, when a=� = 5 (which is a fat torus rather than a thin

ring); Eq. (1.86) still gives a capacitance within 2 % of the correct value. An exact formula for the

capacitance of a conducting torus will be worked out later in terms of the toroidal coordinates.

Figure 1-10: A thin conductor ring with major radius a and minor radius �; a� �:

Method 2: Multipole Expansion in the Spherical CoordinatesAs the second method, we directly solve the Poisson�s equation in the spherical coordinates,

r2� = ��(r)�0

= � 1�0

q

2�a2�(r � a)�(cos �): (1.87)

Except at the ring itself, the charge density vanishes. Therefore, in most part of space, the potential

satis�es Laplace equation,

r2� = 0; r 6= a; � 6= �

2;

and we seek a potential in terms of elementary solutions to the Laplace equation. Since @=@� = 0;

23

the Laplace equation reduces to

@2�

@r2+2

r

@�

@r+1

r21

sin �

@

@�

�sin �

@�

@�

�= 0: (1.88)

Assuming that the potential is separable in the form �(r; �) = R(r)F (�), we �nd

r2

R

�d2R

dr2+2

r

dR

dr

�+1

F

1

sin �

d

d�

�sin �

dF

d�

�= 0: (1.89)

Since the �rst term is a function of r only and the second term is a function of � only, each term

must be constant cancelling each other. Introducing a separation constant l(l + 1); we obtain two

ordinary di¤erential equations for R and F;

d2R

dr2+2

r

dR

dr� l(l + 1)

r2R = 0; (1.90)

1

sin �

d

d�

�sin �

dF

d�

�+ l(l + 1)F = 0; (1.91)

which can be written asd

d�

�(1� �2)dF

d�

�+ l(l + 1)F = 0; (1.92)

where � = cos �: Eq. (1.92) is known as the Legendre�s di¤erential equation. The solutions for

R(r) are

R(r) = rl and1

rl+1; (1.93)

and the solutions for F (�) are the familiar Legendre functions,

F (�) = Pl(cos �) and Ql(cos �); (1.94)

where Ql(cos �) is the Legendre function of the second kind. Some low order Legendre functions

for real x (jxj � 1) are listed below.

P0(x) = 1; P1(x) = x; P2(x) =3x2 � 12

; P3(x) =5x3 � 3x

2; � �� (1.95)

Ql(x) =1

2Pl(x) ln

1 + x

1� x �Wl�1(x); W�1(x) = 0; W0(x) = 1; W1(x) =3

2x; � �� (1.96)

For a general complex variable z; the de�nition for Ql(z) is modi�ed as follows:

Ql(z) =1

2Pl(z) ln

z + 1

z � 1 �Wl�1(z); W�1(z) = 0; W0(z) = 1; W1(z) =3

2z; � �� (1.97)

In the present problem, the potential should remain �nite everywhere except at the ring. Since

Ql(cos �) diverges at � = 0 and �; it should be discarded and we assume the following series solutions

24

for the potential separately for interior (r < a) and exterior (r > a);

�(r; �) =

8>>><>>>:PlAl

�ra

�lPl(cos �); r < a

PlBl

�ar

�l+1Pl(cos �); r > a

(1.98)

where Al and Bl are constants to be determined. The potential should be continuous at r = a

because there are no double layers to create a potential jump. Therefore, Al = Bl: To determine

Al; we substitute the assumed potential into the Poisson�s equation,

@2�

@r2+2

r

@�

@r+1

r21

sin �

@

@�

�sin �

@�

@�

�= � 1

�0

q

2�a2�(r � a)�(cos �);

to obtain Xl

�d2

dr2+2

r

d

dr� l(l + 1)

r2

�AlRl(r)Pl(cos �) = �

1

�0

q

2�a2�(r � a)�(cos �); (1.99)

where

Rl(r) =

8>>><>>>:�ra

�l; r < a

�ar

�l+1; r > a

(1.100)

If we multiply Eq. (1.99) by Pl0(cos �) sin � and integrate over � from � = 0 to �; the summation

over l disappears because of the orthogonality of the Legendre functions,Z �

0Pl(cos �)Pl0(cos �) sin �d� =

Z 1

�1Pl(�)Pl0(�)d� =

2

2l + 1�ll0 : (1.101)

We thus obtain

Al

�d2

dr2+2

r

d

dr� l(l + 1)

r2

�Rl(r) = �

1

�0

q

2�a2�(r � a)2l + 1

2Pl(0): (1.102)

The LHS should contain a delta function �(r� a) to be compatible with that in the RHS. This canbe seen as follows. The radial function Rl(r) has a discontinuity in its derivative at r = a;

dRldr

����r=a+0

= � l + 1a;dRldr

����r=a�0

= +l

a: (1.103)

Therefore, the second order derivative d2Rl=dr2 yields a delta function,

d2Rldr2

= �2l + 1a

�(r � a): (1.104)

25

We thus �nally �nd the expansion coe¢ cient Al;

Al =q

4��0aPl(0) =

q

4��0a

(�1)l=2l!2l[(l=2)!]2

; (l = 0; 2; 4; 6; � � �); (1.105)

and the potential,

�(r; �) =q

4��0a

Xl even

(�1)l=2l!2l[(l=2)!]2

Rl(r)Pl(cos �): (1.106)

Figure 1-11: The derivative of the radial function R(r) is discontinuous at r = r0 and thus itssecond order derivative yields a delta function.

The disappearance of the odd harmonics is understandable because of the up-down symmetry

of the problem. At r � a; the leading order term is the monopole potential,

�l=0(r) =q

4��0a

a

r;

followed by the quadrupole potential,

�l=2(r; �) = �q

4��0a

1

2

�ar

�3P2(cos �);

and so on. The dipole potential (l = 1) vanishes because of the symmetric charge distribution,

�(�r) = �(r);p =

Zr�(r)dV = 0:

For problems with axial symmetry (@=@� = 0) as in this example, knowing the potential along

the axis �(z) is su¢ cient to �nd the potential at arbitrary point �(r; �) in the spherical coordinates.

This is because for all of the Legendre functions, Pl(1) = 1; Pl(�1) = (�1)l: In the case of the ring

26

charge, the axial potential can be readily found,

�(z) =q

4��0

1pz2 + a2

: (1.107)

For jzj > a; this can be expanded as

�(z) =q

4��0

1

jzj

�1� 1

2

�az

�2+3

8

�az

�4� � � �

�: (1.108)

Therefore, at arbitrary point (r > a; �); the potential is

�(r; �) =q

4��0

1

r

�1� 1

2

�ar

�2P2(cos �) +

3

8

�ar

�4P4(cos �)� � � �

�; r > a: (1.109)

For interior region (r < a); the potential becomes

�(r; �) =q

4��0

1

a

�1� 1

2

�ra

�2P2(cos �) +

3

8

�ra

�4P4(cos �)� � � �

�; r < a: (1.110)

They agree with the potential given in Eq. (1.106).

27

Method 3: Cylindrical CoordinatesIn the cylindrical coordinates (�; �; z); the Poisson�s equation for the potential due to a ring

charge becomes �@2

@�2+1

�

@

@�+@2

@z2

��(�; z) = � q

2��0

�(�� a)a

�(z): (1.111)

Since the z-coordinate extends from �1 to1; we seek a solution in the form of Fourier transform,

�(�; z) =1

2�

Z 1

�1�(�; k)eikzdk; (1.112)

where �(�; k) is the one-dimensional Fourier transform of the potential having dimensions of V�m.Since

@

@z�(�; z);

is Fourier transformed as

ik�(�; k);

and �(z) as unity, the equation for �(�; k) becomes an ordinary di¤erential equation,�d2

d�2+1

�

d

d�� k2

��(�; k) = � q

2��0

�(�� a)a

: (1.113)

Elementary solutions of the di¤erential equation at � 6= a;�d2

d�2+1

�

d

d�� k2

��(�; k) = 0; (1.114)

are the zero-th order modi�ed Bessel functions,

�(�; k) = I0(k�); K0(k�); (1.115)

shown in Fig.1-12.

The Fourier potential �(�; k) which is continuous at � = a and remains bounded everywhere

may be constructed in the following form,

�(�; k) =

8><>:AI0(k�)K0(ka); � < a

AI0(ka)K0(k�); � > a

(1.116)

The coe¢ cient A can be determined readily from the discontinuity in the derivative of �(�; k) at

� = a;d�

d�

�����=a+0

= AkI0(ka)K00(ka);

d�

d�

�����=a�0

= AkI 00(ka)K0(ka); (1.117)

28

x 21.510.50

5

4

3

2

1

Figure 1-12: Modi�ed Bessel functions I0(x) (starting at 1) and K0(x) (diverging at x = 0):

from which it follows that

d2�

d�2

�����=a

= Ak�I0(ka)K

00(ka)� I 00(ka)K0(ka)

��(�� a) = �A

a�(�� a); (1.118)

where the Wronskian of the modi�ed Bessel functions

I0(ka)K00(ka)� I 00(ka)K0(ka) = �

1

ak; (1.119)

has been exploited. From

d2�

d�2

�����=a

= �Aa�(�� a) = � q

2��0

�(�� a)a

; (1.120)

we �nd

A =q

2��0; (1.121)

and the �nal form of the potential is

�(�; z) =q

4�2�0

Z 1

�1

8><>:I0(k�)K0(ka)

I0(ka)K0(k�)

9>=>; eikzdk;� < a

� > a

(1.122)

Noting that the modi�ed Bessel functions are even with respect to the argument, this may be

further rewritten as

�(�; z) =q

2�2�0

Z 1

0

8><>:I0(k�)K0(ka)

I0(ka)K0(k�)

9>=>; cos(kz)dk;� < a

� > a

(1.123)

29

The convergence of the integral is rather poor since the function I0K0 decreases with k only

algebraically. An alternative solution for the potential can be found in terms of Laplace transform

rather than Fourier transform,

�(�; z) =

Z 1

0�(�; k)e�kjzjdk: (1.124)

The Laplace�s equation thus reduces to�d2

d�2+1

�

d

d�+ k2

��(�; k) = 0; (1.125)

whose solution is the ordinary Bessel function J0(k�): We thus assume for �(�; k)

�(�; k) = A(k)J0(k�); (1.126)

where A(k) may still be a function of k: The Poisson�s equation becomes�@2

@�2+1

�

@

@�+@2

@z2

�Z 1

0A(k)J0(k�)e

�kjzjdk = � q

2��0

�(�� a)a

�(z): (1.127)

Notingd2

dz2e�kjzj = k2e�kjzj � 2k�(z) (1.128)

we thus �nd Z 1

0kA(k)J0(k�)dk =

q

4��0

�(�� a)a

: (1.129)

However, the Bessel function forms an orthogonal set according toZ 1

0kJ0(ka)J0(k�)dk =

�(�� a)a

; (1.130)

which uniquely determines the function A(k);

A(k) =q

4��0J0(ka): (1.131)

Therefore, the �nal form of the potential is

�(�; z) =q

4��0

Z 1

0J0(ka)J0(k�)e

�kjzjdk: (1.132)

The convergence of this solution is much faster than the solution in Eq. (1.123) and thus more

suitable for numerical evaluation.

30

Method 4: Toroidal CoordinatesThe toroidal coordinates (�; �; �) are related to the cartesian coordinates (x; y; z) through the

following transformation, 8>>>>>>>>>><>>>>>>>>>>:

x =R sinh � cos�

cosh � � cos � ;

y =R sinh � sin�

cosh � � cos � ;

z =R sin �

cosh � � cos � :

(1.133)

This coordinate system is one example in which the Laplace equation is not separable, that is, the

potential cannot be written as a product of independent functions, �(�; �; �) 6= F1(�)F2(�)F3(�):However, the potential is partially separable and can be sought in the form

�(�; �; �) =pcosh � � cos �F1(�)F2(�)F3(�): (1.134)

Figure 1-13: Cross-section of the toroidal coordinates (�; �; �): � ! 0 corresponds to a thin ring ofradius R:

In the toroidal coordinates, the � = constant surfaces are toroids having a major radius R coth �

and minor radius R= sinh �: � !1 degenerates to a thin ring of radius R: In the other limit, � ! 0

describes a thin rod on the z-axis. The � = constant surfaces are spherical bowls as illustrated in

Fig.1-13.

Assuming the partial separation for the potential in Eq. (1.134) leads to the following ordinary

31

equations for the functions F1(�); F2(�) and F3(�);

1

sinh �

d

d�

�sinh �

dF1d�

�+

�1

4� l2 � m2

sinh2 �

�F1 = 0; (1.135)

�d2

d�2+ l2

�F2 = 0;

�d2

d�2+m2

�F3 = 0:

Comparing Eq. (1.135) with the standard form of the di¤erential equation for the associated

Legendre function Pml (cos �); Qml (cos �);�

1

sin �

d

d�

�sin �

d

d�

�+ l(l + 1)� m2

sin2 �

�(Pml ; Q

ml ) = 0; (1.136)

we see that solutions for F1(�) are

F1(�) = Pml� 1

2

(cosh �); Qml� 1

2

(cosh �): (1.137)

The functions F2 and F3 are elementary,

F2(�) = eil�; F3(�) = e

im�; (1.138)

and the general solution for the potential may be written as

�(�; �; �) =pcosh � � cos �

Xl;m

�AlmPl� 1

2(cosh �) +BlmQl� 1

2(cosh �)

�eil�+im�: (1.139)

Of course, the potential is a real function. The solution above is an abbreviated form of a more

cumbersome expression,

�(�; �; �) =pcosh � � cos �

1Xl=0

�AlPl� 1

2(cosh �) +BlQl� 1

2(cosh �)

�(Cl cos l� +Dl sin l�)

�1Xm=0

(El cosm�+ Fl sinm�) :

We now consider a conducting toroid with a major radius a and minor radius b: Its surface is

described by �0 = const. where

a = R coth �0; b =R

sinh �0: (1.140)

Because of axial symmetry, only the m = 0 term is present. If the toroid is at a potential V; the

potential o¤ the toroid can be written as

�(�; �) =pcosh � � cos �

1Xl=0

AlPl� 12(cosh �) cos l�; (1.141)

32

where the Legendre function of the second kind Ql� 12(cosh �) has been discarded because it diverges

at � ! 0 which corresponds to the z�axis. (The potential along the z�axis should be bounded.)The expansion coe¢ cients Al can be determined from the boundary condition, � = V at � = �0;

V =pcosh �0 � cos �

1Xl=0

AlPl� 12(cosh �0) cos l�; (1.142)

or 1Xl=0

AlPl� 12(cosh �0) cos l� =

Vpcosh �0 � cos �

:

Multiplying both sides by cos l0� and integrating over � from 0 to �; we obtain

Al = VQl� 1

2(cosh �0)

Pl� 12(cosh �0)

p2

��l; (1.143)

where

�0 = 1; �l = 2 (l � 1);

and the following integral representation has been exploited,Z �

0

cos l�px� cos �

d� =p2Ql� 1

2(x): (1.144)

Then, the �nal form of the potential is

�(�; �) =

p2V

�

pcosh � � cos �

1Xl=0

Ql� 12(cosh �0)

Pl� 12(cosh �0)

Pl� 12(cosh �) cos(l�)�l: (1.145)

The capacitance of a conducting torus can be found from the behavior of the potential at a

large distance from the torus which should be in the form of monopole potential,

�(r !1) = q

4��0r: (1.146)

At a large distance from the torus r � R; both � and � approach zero,

r2 = x2 + y2 + z2 =R2(sinh2 � + sin2 �)

(cosh � � cos �)2 ! 4R2

�2 + �2; (1.147)

pcosh � � cos � '

p�2 + �2p2

'p2R

r; (1.148)

Pl� 12(cosh �) ' 1: (1.149)

33

Therefore, the asymptotic potential is

�(r � R) =2V R

r

1Xl=0

Ql� 12(cosh �0)

Pl� 12(cosh �0)

=q

4��0r; (1.150)

from which the capacitance of the torus can be found,

C � q

V= 8�0R

1Xl=0

Ql� 12(cosh �0)

Pl� 12(cosh �0)

�l = 8�0a tanh �0

1Xl=0

Ql� 12(cosh �0)

Pl� 12(cosh �0)

�l; (1.151)

where a is the major radius of the torus. The function Ql� 12(cosh �0) can be evaluated from Eq.

(1.144) and Pl� 12(cosh �0) from

Pl� 12(cosh �0) =

1

�

Z �

0

d�

(cosh �0 + sinh �0 cos �)l+ 1

2

: (1.152)

Example 4 Capacitance of a Fat Torus

We wish to �nd the capacitance of a conducting torus having a major radius of a = 50 cm and

minor radius of b = 10 cm. The torus is de�ned by cosh �0 = 5:0 and R = a tanh �0 = 49 cm. For

cosh �0 = 5:0; the Legendre functions numerically evaluated are:

l Ql� 12(5) Pl� 1

2(5)

Ql� 12(5)

Pl� 12(5)

0 1:00108 0:74575 1:34234

1 0:05063 2:03557 0:02487

2 0:00384 13:32184 0:00029

The ratio Ql� 12(5)=Pl� 1

2(5) rapidly converges as l increases and it su¢ ces to truncate the series at

l = 2: The capacitance is

C = 8"0R(1:3423 + 2� 0:0249 + 2� 0:0003 + � � �) ' 8�0R� 1:393 = 8�0a� 1:36:

The approximate formula worked out earlier gives

C ' 4�2�0a

ln

�8a

b

� = 8�0a� 1:34;which is in reasonable agreement with the numerical result.

34

1.7 Spherical Multipole Expansion of the Scalar Potential

The potential due to a prescribed charge distribution,

�(r) =1

4�"0

Z�(r0)

jr� r0jdV0;

can be expanded in terms of the general spherical harmonics as follows. For this purpose, it is

su¢ cient to expand the Green�s function,

G(r; r0) =1

4�

1

jr� r0j ; (1.153)

which satis�es the singular Poisson�s equation,

r2G = ��(r� r0): (1.154)

The Green�s function G satis�es Laplace equation except at r = r0;

r2G = 0; r 6= r0; (1.155)

and we �rst seek elementary solutions for Laplace equation. Assuming that the function G(r; �; �)

is separable in the form

G(r; �; �) = R(r)F1(�)F2(�); (1.156)

and substituting this into Laplace equation�@2

@r2+1

r

@

@r+1

r21

sin �

@

@�

�sin �

@

@�

�+

1

r2 sin2 �

@2

@�2

�G(r; �; �) = 0; (1.157)

we obtain three ordinary equations,�d2

dr2+1

r

d

dr� l(l + 1)

r2

�R(r) = 0; (1.158)

�1

sin �

d

d�

�sin �

d

d�

�+ l(l + 1)� m2

sin2 �

�F1(�) = 0; (1.159)�

d2

d�2+m2

�F2(�) = 0; (1.160)

where l(l+1) and m2 are separation constants. Solutions of the radial function R(r) are as before,

R(r) = rl;1

rl+1:

In free space, the Green�s function must be periodic with respect to the azimuthal angle �: Therefore,

F2(�) = eim� with m being an integer.

35

Equation (1.159) is known as the modi�ed Legendre equation and its solutions are

F1(�) = Pml (cos �); Q

ml (cos �); (1.161)

where

Pml (x) = (1� x2)m=2dm

dxmPl(x); (1.162)

Qml (x) = (1� x2)m=2dm

dxmQl(x): (1.163)

That Pml (cos �) given in Eq. (1.162) satis�es the modi�ed Legendre equation can be seen as follows.

The ordinary Legendre function Pl(cos �) = Pl(�) satis�es

d

d�

�(1� �2)dPl(�)

d�

�+ l(l + 1)Pl(�) = 0:

Di¤erentiating m times yields

(1� �2)dm+2Pld�m+2

� 2�(m+ 1)dm+1Pld�m+1

+ [l(l + 1)�m(m+ 1)] dmPld�m

= 0:

Let us assume

F1(�) = (1� �2)m=2f(�);

and substitute this into Eq. (1.159). f(�) satis�es the same equation as dmPld�m ;

(1� �2)d2f

d�2� 2�(m+ 1) df

d�+ [l(l + 1)�m(m+ 1)] f = 0:

Therefore, f(�) = dmPld�m :

Since the Legendre function Pl(x) is a polynomial of order l; the azimuthal mode number m

is limited in the range 0 � m � l: For later use, we combine the functions F1(�) and F2(�) and

introduce the spherical harmonic function de�ned by

Ylm(�; �) = const. Pml (cos �)e

im�; (1.164)

where the constant is chosen from the following normalization,ZYlm(�; �)Y

�lm(�; �)d = 1: (1.165)

Noting Z �

0[Pml (cos �)]

2 sin �d� =2

2l + 1

(l +m)!

(l �m)! ; (1.166)

we �nd

const. =

s2l + 1

4�

(l �m)!(l +m)!

;

36

Ylm =

s2l + 1

4�

(l �m)!(l +m)!

Pml (cos �)eim�: (1.167)

For historical reasons, it is customary to write Ylm(�; �) in the form

Ylm(�; �) = (�1)ms2l + 1

4�

(l �m)!(l +m)!

Pml (cos �)eim�; for 0 � m � l;

Yl;�m(�; �) = (�1)mY �lm(�; �) ; for � l � m � 0;

or for arbitrary m; positive or negative,

Ylm(�; �) = (�1)(m+jmj)=2s2l + 1

4�

(l � jmj)!(l + jmj)!P

jmjl (cos �)eim�: (1.168)

Some low order forms of Ylm(�; �) are:

l = 0 m = 0 Y00 =1p4�

l = 1 m = 0 Y10 =

r3

4�cos �

m = �1 Y1;�1 = �r3

8�sin �e�i�

l = 2 m = 0 Y20 =

r5

4�

3 cos2 � � 12

m = �1 Y2;�1 = �r15

8�sin � cos �e�i�

m = �2 Y2;�2 =1

4

r15

2�sin2 �e�2i�

Having found the general solutions of the Green�s function, we now assume the following ex-

pansion for G;

G(r; r0) =Xlm

Almgl(r; r0)Ylm(�; �);

where the radial function gl(r; r0) is

gl(r; r0) =

8>>>><>>>>:rl

r0l+1; r < r0;

r0l

rl+1; r > r0:

37

In this form, the function gl(r; r0) remains bounded everywhere. The expansion coe¢ cient Alm can

be determined by multiplying the both sides of

r2G = ��(r� r0) = ��(r � r0)

rr0�(cos � � cos �0)�(�� �0); (1.169)

by Y �l0m0(�; �) and integrating the result over the entire solid angle,

Almd2

dr2gl(r; r

0) = ��(r � r0)

rr0Y �lm(�

0; �0);

where orthogonality of Ylm(�; �);ZYlm(�; �)Y

�l0m0(�; �)d = �ll0�mm0 ; (1.170)

is exploited to remove summation over l andm: Since the radial function gl(r; r0) has a discontinuity

in its derivative at r = r0; it follows that

d2

dr2gl(r; r

0) = �(2l + 1)�(r � r0)

r2: (1.171)

Therefore,

Alm =1

2l + 1Y �lm(�

0; �0); (1.172)

and the desired spherical harmonic expansion of the Green�s function is given by

G(r; r0) =1

4� jr� r0j =Xlm

1

2l + 1gl(r; r

0)Ylm(�; �)Y�lm(�

0; �0): (1.173)

For a given charge distribution �(r); this allows evaluation of the potential in the form of spherical

harmonics,

�(r) =1

4�"0

Z�(r0)

jr� r0jdV0

=1

4�"0

Xlm

4�

2l + 1

1

rl+1Ylm(�; �)

Zr0lY �lm(�

0; �0)�(r0; �0; �0)dV 0

=1

4�"0

Xlm

4�

2l + 1

1

rl+1Ylm(�; �)qlm; in the region r > r

0; (1.174)

where

qlm �Zr0lY �lm(�

0; �0)�(r0; �0; �0)dV 0; (1.175)

is the electric multipole moment of order (l; m):

Example 5 Planar Quadrupole

A planar quadrupole consists of charges +q and �q alternatively placed at each corner of a

38

square of side a as shown in Fig.1-14. The charge density may be written as

�(r) = q

�(r)� �(r � a)

a2�(cos �)� (�) +

�(r �p2a)

2a2�(cos �)�

��� �

4

�� �(r � a)

a2�(cos �)�

��� �

2

�!:

The potential in the far �eld region r � a is of quadrupole nature and given by

�(r; �; �) =1

"0

1

5r3(Y2;2q2;2 + Y2;�2q2;�2) ;

where

q2;�2 =

Zr2Y �2;�2(�; �)�(r)dV =

1

2

r15

2�a2qe�i�=2:

The potential reduces to

�(r; �; �) ' 3

8�"0

a2

r3sin2 � sin(2�); r � a:

The reader should check that this is consistent with the direct sum of four potentials in the limit

r � a:

Figure 1-14: Planar quadrupole in the x� y plane.

1.8 Collection of Dipoles, Dielectric Properties

Dielectric properties of material media originate from dipole moments carried by atoms and mole-

cules. Some molecules carry permanent dipole moments. For example, the water molecule has a

dipole moment of about 6 � 10�30 C�m due to deviation of the center of electron cloud from the

center of proton charges. When water is placed in an external electric �eld, the dipoles tend to be

39

aligned in the direction of the electric �eld and a resultant electric �eld in water becomes smaller

than the unperturbed external �eld by a factor "="0; where " is the permittivity of water. Even a

material composed of molecules having no permanent dipole moment exhibits dielectric property.

For example, a dipole moment is induced in a hydrogen atom placed in an electric �eld through

perturbation in the electron cloud distribution which is spatially symmetric in the absence of ex-

ternal electric �eld. In this section, the potential and electric �eld due to a collection of dipole

moments will be analyzed.

The potential due to a single dipole p located at r0 is

�(r) =1

4��0

(r� r0) � pjr� r0j3 : (1.176)

Consider a continuous distribution of many dipoles. It is convenient to introduce a dipole moment

density P =np (C�m/m3 = C/m2) where n is the number density of dipoles. An incremental

potential due to an incremental �point�dipole dp = P(r0)dV 0 is

d� =1

4��0

(r� r0) �P(r0)jr� r0j3 dV 0: (1.177)

Integration of this yields

�(r) =1

4��0

Z(r� r0) �P(r0)jr� r0j3 dV 0: (1.178)

Sincer� r0jr � r0j3 = r

0�

1

jr� r0j

�; (1.179)

where r0 means di¤erentiation with respect to r0; the integral in Eq. (1.178) can be rewritten asZ(r� r0) �P(r0)jr� r0j3 dV 0 =

Zr0 �

�P(r0)

jr� r0j

�dV 0 �

Z r0 � P (r0)jr� r0j dV

0:

The �rst term in the right can be converted into a surface integral through the Gauss theorem,ZVr0 �

�P(r0)

jr� r0j

�dV 0 =

IS

P(r0)

jr� r0j � dS0;

which vanishes on a closed surface with in�nite extent on which all sources should be absent.

Therefore, the potential due to distributed dipole moments is

�dipole(r) = �1

4��0

Z r0 �P(r0)jr� r0j dV

0: (1.180)

Comparing with the standard form of the potential,

�(r) =1

4�"0

Z�(r0)

jr� r0jdV0;

40

we see that

�e¤ = �r �P;

can be regarded as an e¤ective charge density. To distinguish it from the free charge density,

�e¤ de�ned above is called bound charge density. In general, �e¤ cannot be controlled by external

means. In a dielectric body, the bound charge appears as a surface charge density. Adding the

monopole potential due to a free charge density �free(r); we �nd the total potential

�(r) =1

4��0

Z�free(r

0)

jr� r0j dV0 � 1

4��0

Z r0 �P(r0)jr� r0j dV

0: (1.181)

Noting

r2 1

jr� r0j = �4��(r� r0);

we �nd

r �E = �r2� = �free�0

� r �P�0

: (1.182)

In this form too, it is evident that �r � P can be regarded as an e¤ective charge density. Eq.

(1.181) can be rearranged as

r � (�0E+P) = �free: (1.183)

Introducing a new vector D (displacement vector) by

D = �0E+P; (1.184)

we write Eq. (1.183) as

r �D = �free; (1.185)

which is equivalent to the original Maxwell�s equation

r �E =�all�0; (1.186)

provided that the total charge density �all consists of free charges and dipole charges and that

higher order charge distributions (quadrupole, octupole, etc.) are ignorable. The vector D was

named the displacement vector by Maxwell. Its fundamental importance in electrodynamics will be

appreciated later in time varying �elds because it was with this displacement vector that Maxwell

was able to predict propagation of electromagnetic waves in vacuum. As brie�y discussed in Sec-

tion 2, Maxwell�s equations would not be consistent with the charge conservation principle if the

displacement current,@("E)

@t=@D

@t;

were absent.

In usual linear insulators, both D and P are proportional to the local electric �eld E which

41

allows us to introduce an e¤ective permittivity ";

D = "0E+P ="E: (1.187)

In some solid and liquid crystals, the permittivity takes a tensor form,

D = � �E or Di = �ijEj ; (1.188)

because of anisotropic polarizability. Double refraction phenomenon in optics already known in

the 17th century is due to the tensorial nature of permittivity in some crystals as we will study in

Chapter 5.

Figure 1-15: In an electric �eld, the hydrogen atom exhibits a dipole moment due to the displace-ment of the center of electron cloud from the proton.

The origin of �eld induced dipole moment may be seen qualitatively for the case of hydrogen

atom as follows. In an electric �eld, the distribution of the electron cloud around the proton

becomes asymmetric because of the electric force acting on the electron. The center of electron

cloud is displaced from the proton by x given by

m!20x = eE;

where !0 is the frequency of bound harmonic motion of the electron and m is the electron mass.

(~!0 is of the order of the ionization potential energy.) Then, the dipole moment induced by theelectric �eld is

p = ex =e2

m!20E:

If the atomic number density is n; the dipole moment density P is

P = np =ne2

m!20E; (1.189)

42

and the displacement vector is given by

D = "0E+P

= "0

�1 +

ne2

"0m!20

�E = "E: (1.190)

The permittivity can therefore be de�ned by

" = "0

1 +

!2p!20

!; (1.191)

where

!p =

sne2

"0m; (1.192)

is an e¤ective plasma frequency. (The plasma frequency pertains to free electrons in a plasma. !pintroduced here is so called only for dimensional convenience.)

In the incremental contribution to the permittivity normalized by "0,

!2p!20= n

e2

m"0!20;

the quantitye2

4�"0m!20;

is of the order of r30 where r0 is the Bohr radius. This can be seen from the force balance for the

electron,

mr0!20 =

e2

4�"0r20;

which indeed givese2

4�"0m!20= r30:

The quantity

� = 4�"0r30 =

e2

m!20; (1.193)

is called the atomic polarizability. A more rigorous quantum mechanical calculation yields for this

quantity

� =9

2� 4�"0r30; (1.194)

indicating a substantial correction (by a factor 4.5) to the classical picture based on the assumption

of rigid shift of the electron cloud. For molecualr hydrogen H2; the polarizability at T = 273 K is

approximately given by

� ' 5:4� 4�"0r30:

43

Satisfactory quantum mechanical calculation was performed by Kolos and Wolniewicz relatively

recently (1967). This is in a fair agreement with the permittivity experimentally measured in the

standard conidition 20 C, 1 atmospheric pressure,

" '�1 + 2:7� 10�4

�"0:

The atomic polarizability � and the macroscopic relative permittivity "r are related through a

simple relationship known as the Clausius-Mossoti equation,

n�

"0= 3

"r � 1"r + 2

; (1.195)

where n is the number density of atoms. This may qualitatively be seen as follows. A dipole

moment induced by an atom,

p = ex =e2

m!20E = �Eext; (1.196)

produces an electric �eld

E0 = � p

4�"0r3; (1.197)

at a distance r: If the number density of atoms is n; the average inter-atom distance R can be found

from4�

3R3 =

1

n: (1.198)

Then,

E0 = � �

4�"0R3Eext; (1.199)

and the total electric �eld is

E =

�1� �

4�"0R3

�Eext (1.200)

=

�1� n�

3"0

�Eext: (1.201)

Therefore, using this total �eld in calculation of the polarization �eld P; we �nd

P =n�

1� n�

3"0

E =n�="0

1� n�

3"0

"0E; (1.202)

which de�nes the relative permittivity "r as

"r � 1 =n�="0

1� n�

3"0

: (1.203)

Solving for n�="0; we obtain Eq.(1.195). Note that the relative permittivity is an easily measurable

quantity and it re�ects, somewhat surprisingly, microscopic atomic polarizability through a simple

44

relationship.

In an oscillating electric �eld, the displacement x is to be found from the equation of motion,

m

�@2

@t2+ !20

�x = �eE0e�i!t; (1.204)

which yields

x(t) =eE0e

�i!t

!2 � !20: (1.205)

A resultant permittivity is

"(!) = "0

1�

!2p!2 � !20

!: (1.206)

This exhibits a resonance at the frequency !0: The resonance plays an important role in calculation

of energy loss of charged particles moving in a dielectric medium as will be shown in Chapter 8.

Most molecules have permanent electric dipole moments. In the absence of electric �elds,

dipoles are randomly oriented due to thermal agitation. When an electric �eld is present, each

dipole acquires a potential energy,

U = �p �E = �pE cos �; (1.207)

where � is the angle between the dipole moment p and electric �eld E: Dipoles are randomly

oriented but should obey the Boltzmann distribution in thermal equilibrium,

n(cos �) = n0 exp

�� U

kBT

�= n0 exp

�pE cos �

kBT

�; (1.208)

where n(cos �)=n0 is the probability of dipole orientation in � direction, n0 is the number density

of the dipoles and kB = 1:38� 10�23 J/�K is the Boltzmann constant. The average dipole numberdensity oriented in the direction of the electric �eld can be calculated from

�n =

Rn(cos �) cos �dRexp

�pE cos �kBT

�d: (1.209)

In practice, the potential energy is much smaller than the thermal energy pE � kBT: Therefore,

the exponential function can be approximated by

exp

�pE cos �

kBT

�' 1 + pE cos �

kBT; (1.210)

and we obtain

�n ' n0p

3kBTE: (1.211)

45

Resultant dipole moment density is

P = �np =n0p

2

3kBTE: (1.212)

This de�nes a permittivity in a medium consisting of molecules having a permanent dipole moment

p;

" = "0

�1 +

n0p2

3"0kBT

�: (1.213)

Note that the correction term

�" =n0p

2

3kBT; (1.214)

is inversely proportional to the temperature. This property can be exploited in separating con-

tributions from atomic and molecular polarizability in a given medium. In liquids and solids, the

additional permittivity given above can be comparable with "0 primarily because of the large num-

ber density n0: For example, a water molecule has a dipole moment of p = 6 � 10�30 C m. At

room temperature, the additional permittivity due to molecular polarizability of water is approxi-

mately �" ' 11"0: The measured permittivity of water at room temperature is 81"0: The atomic

polarizability is thus dominant.

1.9 Boundary Conditions for E and D

In electrostatics, the electric �eld E obeys the Maxwell�s equations,

r �E = �

"0;

and

r�E = 0:

In problems involving dielectrics, it is more convenient to introduce the displacement vector D,

r �D = �free; (1.215)

where �free is the free charge density that can be controlled by external means.

Let us consider a boundary of two dielectrics having permittivities "1 and "2; respectively.

Integration of r �D = �free over the volume of a pancake dV = dS � dn (dn is the thickness of thepancake in the direction perpendicular to the surface dS) on the boundary yieldsZ

r �DdV =ID � dS =

Z�freedV: (1.216)

In the limit of in�nitesimally thin pancake, this reduces to

(Dn1 �Dn2)dS = �freedndS = �freedS; (1.217)

46

Figure 1-16: Gauss�law applied to a thin pancake at a boundary between two dielectrics.

where �free = �freedn is the free surface charge density residing on the boundary. Note that the

volume charge density in the presence of surface charge can be written as

� = ��(n� n0): (1.218)

General boundary condition for the normal component of the displacement vector is thus given by

D1n �D2n = �free: (1.219)

In the absence of free surface charge, the normal component of the displacement vector should be

continuous,

Dn1 = Dn2; no free charge. (1.220)

Figure 1-17:HE � dl = 0 applied to a rectangle at the boundary of two dielectrics. The tangential

component of the electric �eld Et is continuous. This holds in general for time varying �elds aswell.

47

The equation r�E = 0 demands that the tangential component of electric �eld be continuousacross a boundary of dielectrics. This can be seen by integrating r�E = 0 over a small rectangulararea on the boundary, Z

r�E � dS =IE � dl = 0; (1.221)

which yields

(E1 �E2) � dl = 0; E1t = E2t: (1.222)

In fact, the continuity of the tangential component of the electric �eld holds for general time varying

case,

r�E = �@B@t; (1.223)

for the area integral ZSB � dS;

vanishes in the limit of in�nitesimally thin rectangle, S ! 0: In contrast to charge and current

densities, a singular magnetic �eld involving a delta function is not physically realizable because

then the magnetic energy simply diverges. Note that the square of a delta function is not integrable,Z�2(x)dx = �(0) =1:

In a conductor, there can be no static electric �elds and E = 0 must hold inside conductors:

Therefore, at a conductor surface, the tangential component of the electric �eld should vanish and

the electric �eld lines fall normal to the surface. The normal component of the electric �eld and

the surface charge density � are related through

En =�

"0; at conductor surface. (1.224)

The potential of a conducting body is constant. The surface electric �eld depends on the amount

of charge carried by the body and also curvature of the surface. A trivial case is the electric �eld

at the surface of charged conducting sphere of radius a;

E =1

4�"0

q

a2:

In general, the �eld increases as the curvature radius decreases. The electric �eld at the tip of

needle can be strong enough to allow emission of electrons as in tunneling electron microscopes. In

power engineering, conductor surfaces should be made smooth and round as much as possible to

avoid breakdown.

Example 6 Dielectric Sphere in an Electric Field

Let us consider an uncharged dielectric sphere having a uniform permittivity " placed in an

uniform external electric �eld E0 = E0ez: The sphere perturbs the external �eld because a bound

48

surface charge � = �r �P will be induced on the sphere. The interior and exterior potentials maybe expanded in the spherical harmonics,

�(r; �) = �0 +Xl

Al

�ra

�lPl(cos �); r < a (interior); (1.225)

�(r; �) = �0 +Xl

Bl

�ar

�l+1Pl(cos �); r > a (exterior); (1.226)

where

�0(r; �) = �E0z = �E0r cos �; (1.227)

is the potential associated with the external electric �eld E0 = E0ez. Since there is no double layer

to cause potential jump at the sphere surface, the potential is continuous at the surface r = a;

from which it follows Al = Bl: This also follows from the continuity of the tangential component

(� component) of the electric �eld,

E� = �1

r

@�

@�: (1.228)

The continuity of the normal (radial) component of the displacement vector requires

"

�E0 cos � +

Xl

All

aPl(cos �)

!= �"0

E0 cos � +

Xl

All + 1

aPl(cos �)

!: (1.229)

Since cos � = P1(cos �); only the l = 1 terms are non-vanishing and we readily �nd

A1 ="� "0"+ 2"0

aE0: (1.230)

Then the solutions for the potentials are

�(r; �) =

8>>>><>>>>:�0(r; �) +

"� "0"+ 2"0

E0r cos �; r < a

�0(r; �) +"� "0"+ 2"0

E0a3

r2cos �; r > a

(1.231)

The electric �eld in the dielectric sphere is uniform and given by

Eiz =3"0

"+ 2"0E0; r < a: (1.232)

This is smaller than the external �eld E0 since " > "0: The interior displacement vector is

Diz = "Eiz =3"

"+ 2"0"0E0; (1.233)

49

which is larger than D0 = "0E0: The perturbed exterior potential,

"� "0"+ 2"0

a3

r2E0 cos �; (1.234)

is of dipole form with an e¤ective dipole moment

pz = 4�"0"� "0"+ 2"0

a3E0; (1.235)

located at the center of the sphere.

In the limit of "� "0; the potential reduces to

�(r; �) =

8>><>>:0; r < a

�0(r; �) +a3

r2E0 cos �; r > a

(1.236)

which describes the potential when a conducting sphere is placed in an external electric �eld. A

sphere of in�nite permittivity is mathematically identical to a conducting sphere.

The polarization vector P can be found from

Dz = "0Ez + Pz; (1.237)

Pz =3("� "0)"+ 2"0

"0E0; r < a: (1.238)

Since P = 0 outside the sphere, the divergence of the polarization vector yields the bound charge

density induced on the sphere surface,

�e¤ = �r �P

=2("� "0)"+ 2"0

"0E0 cos ��(r � a): (1.239)

The total dipole moment carried by the sphere is

pz =4�a3

3Pz

=4�("� "0)a3"+ 2"0

"0E0;

which agrees with that deduced from the exterior potential.

50

1.10 Electric Force

The energy density associated with an electric �eld

1

2"0E

2; (J/m3) (1.240)

manifests itself as either pressure or tensile stress depending on the direction of the force relative

to the electric �eld. A charge density � placed in an electric �eld E experiences a force per unit

volume,

f = �E

= "0(r �E)E= r � ("0EE)� "0E � rE: (1.241)

In electrostatics, r � E = 0; and thus the curvature E � rE and gradient ErE =1

2rE2 of the

electric �eld are identical,

r(E �E) = 2E�r�E+ 2E � rErE2 = 2E � rE: (1.242)

Therefore,

f = r ��"0EE�

1

2"0E

21

�; (1.243)

where 1 is the unit sensor. The tensor

Tij = "0EiEj �1

2"0E

2�ij ; (1.244)

is called the Maxwell�s stress tensor associated with the electric �eld. For a linear dielectric, "0may be replaced with its permittivity ": The force vector is

f = r �T; or fi =@

@xjTij : (1.245)

Let us consider a simple case: an electric �eld in the z-direction, E = Ezez: The tensor Tij is

diagonal with the following components,

T =

0BBBB@�12"0E

2z 0 0

0 �12"0E

2z 0

0 0 +1

2"0E

2z

1CCCCA : (1.246)

51

The force in the directions perpendicular to the �eld are

fx = �@

@x

�1

2"0E

2z

�; (1.247)

fy = �@

@y

�1

2"0E

2z

�; (1.248)

which appear as a pressure acting from a higher energy density region to lower energy density

region. The force in the direction of the electric �eld is

fz = +@

@z

�1

2"0E

2z

�; (1.249)

which appears as tension acting from a lower energy density region to higher energy density region.

The force to act on a volume V can be found from the integral

F =

ZVfdV =

IST � dS

=

IS

�"0E(E � n)�

1

2"0E

2n

�dS; (1.250)

where n is the unit normal vector on the closed surface S; dS = ndS: For a charged conducting

sphere, the electric �eld is radially outward everywhere. At the surface, the force per unit area is

1

2"0E

2r ; (N/m

2);

acting radially outward. Since E = 0 inside a conducting sphere, the force acts from lower energy

density region to higher energy density region as expected of tensile stress. Of course, no net force

acts on the sphere.

Example 7 Dielectric Hemispheres in an External Electric Field

As a less trivial example, we consider a dielectric sphere consisting of identical hemispheres with

a narrow gap at an equatorial plane. It is placed in an electric �eld with the gap plane perpendicular

to the �eld. We wish to �nd a force to act between the hemisphere. The force, if any, should be

axial in the direction perpendicular to the gap. The z-component of the integral,

F =

I �"E(E � n)� 1

2"E2n

�dS

is

Fz =

I �"EzEn �

1

2"E2nz

�dS; (1.251)

where on the spherical surface of one of the hemispheres,

Ez = E03

"+ 2"0

�" cos2 � + "0 sin

2 ��; (1.252)

52

Er = E03"

"+ 2"0cos �; (1.253)

E� = �E03"0

"+ 2"0sin �; (1.254)

E2 = E2r + E2�

= E209

("+ 2"0)2�"2 cos2 � + "20 sin

2 ��: (1.255)

In the gap, the �eld is uniform,

Figure 1-18: A dielectric sphere with a negligible gap at an equator placed in an electric �eld normalto the gap surface. The hemispheres attract each other. The D-�led lines (not E �eld lines) shownare relevant to the preceding Example as well.

Egapz ="

"0Eiz =

3"

"+ 2"0E0; (1.256)

where

Eiz =3"0

"+ 2"0E0; (1.257)

is the internal electric �eld in the dielectric sphere given in Eq. (1.232). The integral over the

hemispherical surface is

Fz1 = 2�a2Z �=2

0

�"0EzEr �

1

2"0E

2 cos �

�sin �d�

= 2�a29"0E

20

("+ 2"0)2

Z �=2

0

�"2 cos2 � + ""0 sin

2 � � 12("2 cos2 � + "0 sin

2 �)

�cos � sin �d�

= �a29"0E

20

("+ 2"0)21

4("2 + 2""0 � "20): (1.258)

53

This force is repelling. The contribution from the �at surface in the gap is

Fz2 = ��a29"0E

20

("+ 2"0)2"2

2; (1.259)

and the net force is

Fz = ��a29("� "0)24("+ 2"0)2

"0E20 : (1.260)

The minus sign indicates an attractive force between the hemispheres. In the limit " � "0; the

force becomes

Fz = �9

4�a2"0E

20 :

This corresponds to the case of solid or closed conducting hemispheres.

1.11 Force in Capacitor

For two electrode systems, the potential di¤erence between the electrodes V and the charges �Qresiding on each electrode are related through the capacitance C;

Q = CV: (1.261)

An incremental energy required to increase the charge by dQ is

dU = V dQ =1

CQdQ = CV dV: (1.262)

Therefore the amount of energy stored in a capacitor is

U =1

2

Q2

C=1

2CV 2: (1.263)

This is applicable for a single electrode as well if the second electrode is at in�nity (self-capacitance).

For example, a conducting sphere of radius a has a self-capacitance of

C = 4�"0a: (1.264)

If it carries a charge Q; the amount of potential energy associated with it is

U =Q2

8�"0a: (1.265)

Of course, the energy is stored in the space surrounding the sphere, and the energy can alternatively

be calculated from the integral,

U =

Z 1

a

1

2"0E

2r4�r

2dr =Q2

8�"0a; (1.266)

54

where

Er =Q

4�"0r2; r > a (1.267)

is the electric �eld.

If a capacitance is described as a function of geometrical factor �; such as electrode separation

distance and electrode size, the electric force tends to act in such a way to increase the capacitance,

F� =1

2V 2@C(�)

@�; if the voltage is held constant, (1.268)

F� = �1

2Q2

@

@�

�1

C(�)

�; if the charge is held constant. (1.269)

For example, the capacitance of parallel plates capacitor consisting of circular disks is

C(d; a) = "0�a2

d; (1.270)

where a is the disk radius and d is the separation distance. The disks evidently attract each other

so as to reduce the distance d or increase the capacitance with a force

Fd = �1

2"0�a2

d2V 2 = �1

2

Q2

"0�a2: (1.271)

The radial force is

Fa = "0�a

dV 2 =

d

"0�a3Q2; (1.272)

which acts so as to increase the radius of the disks. Note that when the charge is �xed, the force

acts to reduce the energy stored in the capacitor, while when the voltage is �xed, the force acts

to increase the energy. A power supply is a large reservoir of energy and a capacitor connected to

a power supply (the case of �xed voltage) is not a closed system. The earlier statement based on

the variational principle that an electrostatic equilibrium is the minimum energy state of course

pertains to closed systems.

55

Problems

1.1 Verify that for each of the expressions of the three dimensional delta functions,

�(r� r0) = �(x� x0)�(y � y0)�(z � z0);

�(r� r0) = �(�� �0)�0

�(�� �0)�(z � z0);

�(r� r0) = �(r � r0)rr0

�(cos � � cos �0)�(�� �0);

the volume integral is unity, Z�(r� r0)dV = 1:

1.2 The electron cloud in a hydrogen atom is described by the charge density distribution

�(r) = � e

�a3exp

��2ra

�;

where a = 5:3 � 10�11 m is the Bohr radius. Show that the potential energy of a hydrogen

atom is

U = � 1

4�"0

e2

a:

1.3 A planar quadrupole in the x � y plane consists of four charges, +q and �q alternativelyplaced at the corners of a square of side a: What is the potential energy of the quadrupole?

1.4 Four charges e; e; �e and �e are placed at the corners of a tetrahedron having side a:Findthe electric dipole and quadrupole moments.

1.5 Show that for l = 1;1X

m=�1rY1;m(�; �)q1;m =

3

4�r � p;

56

and a resultant dipole potential is consistent with the direct expansion of the potential,

�dipole =1

4�"0

r � pr3:

1.6 An octupole consists of eight charges (four +q and four �q) alternatively placed at the cornersof a cube of side a: Determine the far �eld potential �(r; �; �) at r � a and also the potential

energy of the octupole.

1.7 Equal octants on a spherical surface of radius a are maintained alternatively at potentials V

and �V: Determine the lowest order far �eld potential �(r; �; �) at r � a: Useful expansion

is

�(r) =Xlm

�ar

�l+1Ylm(�; �)

I�s(�

0; �0)Y �lm(�0; �0)d0; r > a;

where �s (�; �) is the surface potental and d0 = sin �0d�0d�0:

1.8 The potential due to a long line charge � (C/m) is

�(�) = � �

2�"0ln �+ constant,

where � is the distance from the line charge. Verify this from the basic formula,

�(r) =1

4�"0

Z�(r0)

jr� r0jdV0

=1

4�"0

Z 1

�1

�p�2 + z02

dz0:

Then, show that the capacitance per unit length of a parallel-wire transmission line with wire

separation distance d and common wire radii a (� d) is approximately given by

C

l' �"0

ln

�d

a

� ; (F/m).57

What is the inductance per unit length of the transmission line? (Hint: The product

C

l

L

l;

is constant and equal to "0�0: For a coaxial cable �lled with insulating material having

permittivity "; the capacitance isC

l=

2�"

ln(a=b);

and the product ClLl is equal to "�0:)

1.9 An insulating circular disk of radius a carries a total charge q uniformly distributed over

its area �a2: By �rst �nding the potential on the axis of the disk, �(z); generalize it to a

potential at arbitrary position �(r; �) in terms of the spherical harmonic functions Pl(cos �):

(The case of charged conducting disk will be analyzed in Chapter 2.) Check whether the

Poisson�s equation,

r2� = � �"0;

is satis�ed by your solution. (It should be.)

1.10 Solve the preceding problem using the cylindrical coordinates (�; z).

1.11 The Legendre polynomial Pl(x) can be generated from

Pl(x) =1

2ll!

dl

dxl(x2 � 1)l; (Rodrigues�formula).

Using this repeatedly, verify the orthogonality of Legendre functions,Z 1

�1Pl(x)Pl0(x)dx =

2

2l + 1�ll0 :

1.12 Evaluate numerically the capacitance of a conducting torus having a major radius of 10 cm

and minor radius of 3 cm.

1.13 Two concentric circular rings of radii a and b in the same plane carry charges q and �q;respectively. Show that the far �eld potential is of quadrupole nature. What is the potential

energy of the system?

1.14 Two coaxial conductor rings of radii a and b and axial separation distance c carry charges q

and �q: What is the dominant far �eld potential? Find the mutual capacitance.

1.15 The permittivity of a molecular hydrogen (H2) gas under the standard condition, 0 C, one

atmospheric pressure, is

" = "0 +�" =�1 + 2:7� 10�4

�"0:

58

When �" is written in the form

�" = const:!2p!20"0 = const:

ne2

me!20;

where n is the molecule density, me = 9:1 � 10�31 kg is the electron mass, and !0 is thefrequency of bound harmonic motion of electron ~!0 = 27:2 eV, what should the constant

be? For atomic hydrogen, the constant is 4.5 as predicted by quantum mechanics.

1.16 Two dipole moments p1 and p2 are a distance r apart. Show that the potential energy of the

two dipole system is given by

U =1

4�"0

1

r3

�p1 � p2 � 3

(p1 � r)(p2 � r)r2

�:

1.17 Two parallel charge sheets of opposite polarity �� (C/m2) separated by a small distance �form a double layer. (a) Show that the potential jump across the double layer is

�� =��

"0=�

";

where � = �� (C m�1) is called the moment of double layer. (b) A circular double layer of

radius a and moment � is on the x� y plane. Find the potential �(r; �):

1.18 Concentric spherical capacitor of radii a (inner) and b (outer) is �lled with a nonuniform

dielectric material whose permittivity depends on the radius r; "(r): Determine "(r) if the

radial electric �eld between the electrodes is to be constant. Assume that the permittivity at

r = b is "0:

1.19 The problem of ring charge can alternatively be solved by the method of Laplace transform

in the cylindrical coordinates as shown in Chapter 1,

�(�; z) =q

4�"0

Z 1

0J0(ka)J0(k�)e

�kjzjdk:

Show that the solution above satis�es the Poisson�s equation,�@2

@�2+1

�

@

@r+@2

@z2

��(�; z) = � q

2�"0a�(�� a)�(z):

1.20 A conducting sphere of radius a is placed in a uniform electric �eld E0 = E0ez whose potential

is �0 = �E0z: The dipole moment induced on the sphere is

pz = 4�"0a3E0;

59

which produces a potential

�0 =1

4�"0

pzr2cos �:

(a) The exterior electric �eld is

E(r; �) = �r�0 �r�0

= E0ez +pz

4�"0r3(2 cos �er + sin �e�) ; r > a:

Find a change in the total electric energy and interpret your result. Note that E = 0 in

the sphere.

(b) A dipole p in an electric �eld E has a potential energy

U = �p �E = �4�"0a3E20 :

Recover this result from the direct integral,

U =1

2

I��0dS;

over the sphere surface. Note that at a conductor surface, � = "0Er(r = a):

1.21 A charge q1 is placed at a distance z = a from a dielectric plate having a permittivity ":

Another charge q2 is at the mirror position z = �a: Find the force on each charge. As youwill �nd, the forces are not equal in magnitude. Explain.

Hint: If a charge q is placed at a distance z = a from the surface of a dielectric plate, the

potential in the air region is

�air =1

4�"0

�q

jr� aj �"� "0"+ "0

q

jr+ aj

�; z > 0

60

and that in the dielectric is

�dielectric =1

4�"

2"

"+ "0

q

jr� aj ; z < 0;

where a = aez:

1.22 The permittivity of an unmagnetized plasma is

" (!) = "0

1�

!2pe!2

!;

where !pe =pne2=me"0 is the plasma frequency. Show that a spherical plasma exhibits a

dipole (l = 1) ocsillation at a frequency

! =!pep3:

What are the frequencies of higher multipole modes?

61