lec2 electrostatics

8/2/2019 Lec2 Electrostatics

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Chapter 1

Electrostatics I. Potential due to

Prescribed Charge Distribution,

Dielectric Properties, Electric Energy

and Force

1.1 Introduction

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

through Coulombs 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 havingno 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 diraction phenomenon occurs through

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

1.2 Coulombs 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. (Cavendishs work preceded

Coulombs. However, Coulombs 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

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the well known Coulombs law,

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

F = const.q1q2

r2= const.

q1q2jr1 r2j2 ; (N) (1.1)

where r = jr1 r2j is the relative distance between the charges. In CGS-ESU (cm-gram-secondelectrostatic 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 = 105 N on each other, the charge is dened

as 1 ESU ' 13 109 C. The electronic charge in ESU is e = 4:8 1010 ESU. In MKS-Ampere(or SI) unit system, 1 Coulomb of electric charge is dened from

1 Coulomb = 1 Ampere 1 sec,

where 1 Ampere of electric current is dened as follows. If two innitely long parallel currents of

equal amount separated by 1 meter exert a force per unit length of 2 107 N/m on each other(attracting if the currents are parallel and repelling if antiparallel), the current is dened 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 = 4107 H/m in MKS-Ampere unit system is an assigned constantintroduced to dene 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 Coulombs law is approx-

imately 9:0 109 N m2 C2: It is customary to write the constant in the form

const. =1

4"0;

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Figure 1-2: In MKS unit system, I = 1 Ampere current is dened if the force per unit length betweeninnite parallel currents 1 m apart is 2 107 N/m. The magnetic permeability 0 = 4 107H/m is an assigned constant to dene 1 Ampere current.

where

"0 = 8:85 1012 C2

N m2

=Farad

m

;

is the vacuum permittivity. Since

c2 =1

"00; m2s2;

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 Coulombs 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 ;

N

C=

V

m

(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

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

planar coordinates. Corresponding Maxwells equation is

r E = "0

; in MKS-Ampere,

orr 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 dierential work needed to move the charge q2 against the Coulomb force

is

dW = F dr2 = 14"0

q1q2(r2 r1)jr1 r2j3 dr2:

Notingrr2

1

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 Wis negative, an energy jWj 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 e

2

4"0rB= e

2

8"0rB< 0;

where rB = 5:3 1011 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.

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

W =1

4"0

q1jr1 r2j q2; (J)

which denes the potential associated with a point charge q1 at a distance r;

=1

4"0

q

r; (J C1 = V); (1.6)

and the electric eld is related to the potential through

E = r; (V m1): (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 Maxwells 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 = 14"0

rZ

(r0)

jr r0jdV0; (1.8)

since the dierential electric eld due to a point charge dq= (r0)dV0 located at r0 is

dE =1

4"0

(r0)dV0

jr r0j3 (r r0):

Note that

r 1jr r0j = r0 1

jr r0j = r r0jr 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 dierential equation to be satised 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,

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not to be confused with Gauss law),

ISE dS =

ZVr EdV

=

1

4"0ZV

dV

r2 Z

V0

(r0)

jr r0

jdV0:

However, the function 1= jr r0j satises a singular Poissons equation,

r2 1jr r0j = 4(r r0): (1.9)

Therefore, ISE dS = 1

"0

ZV

dV

ZV0

(r r0)(r0)dV0 = 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

Coulombs inverse square law. FromZr EdV = 1

"0

Z(r)dV; (1.11)

it also follows that

r E = "0

: (1.12)

This is the dierential form of the Gauss law for the longitudinal electric eld and constitutes one

of Maxwells equations.

For the electric eld due to a static charge distribution,

E(r) = 14"0

ZV

(r0)(r r0)jr r0j3 dV

0 = 14"0

rZV

(r0)jr r0jdV

0;

its curl identically vanishes because

rE = 14"0

rrZ

(r0)

jr r0jdV0 0:

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

rE = 0: (1.13)

This is a special case of the more general Maxwells equation

rE = @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 specied. (This is known as Helmholtzs

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theorem.) In electrodynamics, two vector elds, the electric eld E and magnetic eld B, are to be

found for given charge and current distributions (r; t) and J(r; t): Therefore it is not accidental

that four Maxwells equations emerge specifying the four functions rE; rE; r B; and rBwhich determine the longitudinal and transverse components of E and B.

Digression: A vector A can be decomposed into the longitudinal and transverse components Al

and At which are, by denition, characterized by

rAl = 0; r At = 0:

The longitudinal component can be calculated from

Al (r) = 14rZr0 A (r0)

jr r0j dV0;

since

r2 1

jr r0j = 4r r0 :The transverse component is given by

At (r) = A (r)Al (r)=

1

4rr

ZA (r0)

jr r0jdV0;

where the identity

rrA =rr Ar2A;

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

be specied. Since r A =r Al; divergence of a vector spcies the divergence of the longitudinalcomponent. Likewise, rA =rAt 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 dierential equation is

El (r) = 14"0

rZV

(r0)

jr

r0

j

dV0:

The tranverse component is specied by the magnetic eld as

rE = @B@t

:

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

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gradient of a scalar potential ,

E = r; (1.15)

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

rr 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 Poissons equation. In general, solving the scalar dierential equation for iseasier than solving vector dierential 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 dierence 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 aecting the electric eld.

In a conductor, an electric eld drives a current ow according to the Ohms law,

J = E; (A m2) (1.19)

where (S m1) 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

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= "0r E = 0: A charge given to a conductor can only appear as a surface charge (C m2)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= eEmev; (1.21)

where c (s1) is the electron collision frequency with the lattice ions. Noting J = nev with n

the 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 0eti!pt with an exponential damping factor =

c=2: The typical electron collision frequency in metals is of order c ' 1013= sec and the timeconstant to establish electrostatic state is indeed very short. (If the low frequency Ohms law is

used, an unrealistically short transient time emerges,

@@t

+ 0

= 0; = 0et=;

where = 0= ' 1019 sec.)The principle that a charge given to a conductor must reside entirely on its outer surface is also

a consequence of Coulombs 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 Coulombs law is correct. Experiments

have been conducted to measure residual electric elds inside and it has been established that thepower in the Coulombs 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, in

the form r2 k2C

1

c2@2

@t2

=

"0: (1.24)

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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 Coulombs

law.

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

ekCr: (1.27)

The power in the Coulombs law Er _ q=ra experimentally established is 2 = O(1016);

and this corresponds to an upper limit of photon mass of mp < 1051 kg. The lower limit of

photon Compton wavelength is C > 4 105 km at which distance a signicant deviation from theCoulombs 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

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

r2 k2LB = 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

1

2

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

2Xj

jqj =1

2

1

4"0Xi6=j

qiqj

jri

rj j

; (1.33)

where

j =1

4"0

Xi6=j

qijri rjj ; (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 dierential charge dq = dV, which

can be regarded as a point charge if dV is suciently small, is

dW =1

2dq =

1

2dV:

Since = "0r E;dW =

1

2"0r 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 I

S

EdS + ZV

1

2"0E

2dV: (1.36)

where use is made of Gauss theorem,ZVr FdV =

ISFdS; (1.37)

for an arbitrary well dened vector eld F. The surface integral vanishes because at innity, both

potential and electric eld vanish. Therefore, the potential energy associated with a distributed

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charge system is

W =

Z1

2"0E

2dV; (1.38)

which is positive denite. 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 ' 14"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 ' 14"0

e2

mc2= 2:85 1015 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

3r2e ;

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

1015 m, the analogy obviously breaks down.

The Poissons equation in Eq. (1.17) is the Eulers equation to make the energy-like integral

stationary

U =

Z1

20E

2

dV =

Z1

20(r)2

dV: (1.42)

Indeed, the variation of this integral

U =

Z(0r r ) dV

= Z

0r2 +

dV;

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becomes stationary (U = 0) when the Poissons 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 Poissons 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 m2);

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

1.5 Formal Solution to the Poissons Equation

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

the solution for the Poissons equation

r2(r) = (r)0

; (1.43)

can be written down as

(r) =

1

40ZV

(r0)

jr r0jdV0

: (1.44)

This is understandable because the dierential potential due to a point charge dq = dV is

d =1

4"0

(r0)dV0

jr r0j ;

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

It is noted that the Poissons 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)

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Figure 1-4: Dierential potential d due to a point charge dq = dV0.

In the Coulomb gauge, the transverse vector potential satises 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) = 140

ZV

(r0; t)jr r0jdV

0; (1.48)

and so is the resultant longitudinal electric eld,

El(r; t) = rC = 14"0

ZV

r r0jr r0j3 (r

0; t)dV0: (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) = 14jr r0j ; (1.50)

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

as a solution to the following singular Poissons equation

r2G = (r r0); (1.51)

subject to the boundary condition that G = 0 at r = 1: (G subject to this boundary conditionis called the Greens function in free space. It is a particular solution to the singular Poissons

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

surface by adding general solutions satisfying r2G = 0:) It is evident that the Greens functionis 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 the

cartesian 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 Poissons 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)eik(rr0)dr: (1.60)

Its inverse transform is

G(r r0) = 1(2)3

Zg(k)eik(rr

0)d3k: (1.61)

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

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Eq. (1.59),

g(k) =1

k2: (1.62)

Substitution into Eq. (1.61) yields

G(r r0) =1

(2)3Z 1

k2 eik(rr0

)d3k

=1

(2)3

Z10

dk

Z0

eikjrr0j cos sin d

Z20

d

=1

(2)2

Z0

(jr r0j cos )sin d

=1

4jr 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 dd; (1.64)Z11

eikxdk = 2(x); (1.65)

(ax) =1

jaj(x); (1.66)

are noted. (The same technique will be used in nding a Greens 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 conned 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

rr

1

r

r0 + 1

2!rr

1

r

: r0r0

=1

r+r r0

r3+

1

2r5

Xi;j

(3rirj r2ij)r0ir0j + (1.68)

and so is the potential

(r) =1

40

ZV

(r0)

jr r0jdV0 =

1

40

1

r

ZV

(r0)dV0 +r

r3ZVr0(r0)dV0 +

3rr r212r5

: Q +

;

(1.69)

where 1 is the unit dyadic. The quantity q =R

(r0)dV0 is the total charge contained in the (small)

volume V and the corresponding lowest order monopole potential is

monopole =1

40

q

r: (1.70)

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The vector quantity

p =

Zr(r)dV; (1.71)

is the dipole moment and the corresponding dipole potential is

dipole =1

40

pr

r3 : (1.72)

The tensor

Q = Qij =

Zrirj(r)dV; (1.73)

denes the quadrupole moment and the corresponding quadrupole potential is

quadrupole =1

40

3rr r212r5

: Q =1

40

1

2r5

Xi;j

(3rirj r2ij)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

40

p

r2cos :

The equipotential surfaces are described by

cos

r2= const.

The electric eld can be found from

E = r = 140

2p cos

r3er +

p sin

r3e

: (1.75)

The equation to describe an electric eld line is, by denition,

dr

Er=

rd

E; (1.76)

which givesdr

r= 2cot 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


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10.50-0.5-1

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

40

1

2r5(3z2 r2)Qzz = 1

40

2qa2

r33cos2 1

2=

1

40

2qa2

r3P2(cos );

where

P2(cos ) =3cos2 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

40

1p

r2 + a2 2ar cos 2

r+

1pr2 + a2 + 2ar cos

:

For r > a; the functions 1=p

r2 + a2 2ar cos can be expanded in terms of the Legendre functions

18


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Figure 1-6: Linear quadrupole.

Pl(cos ) as

1pr2 + a2 2ar cos =

1

r

1Xl=0

ar

l(1)lPl(cos ); r > a:

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

(r; ) ' 140

2qa2

r3P2(cos ); r a:

At r a; the leading order potential is of quadrupole. The total charge is zero, thus no monopolepotential at r a. Also, the charge system consists of two dipoles of equal magnitude oriented inopposite directions. Therefore, the dipole potential vanishes at r a as well. Equipotential proleis shown in Fig. ??.

1.6 Potential due to a Ring Charge: Several Methods

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

is unique, and answers found by dierent 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 integration

The 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 dierential potential due to a point charge

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-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 = qd0=2 located at angle 0 is

d =1

40

dq

jr r0j =1

40

q

2p

r2 + a2 2ar cos d0; (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

40

q

2

Z20

d0pr2 + 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=20

1p1 k2 sin2

d =4p

r2 + a2 + 2ar sin K(k2); (1.81)

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Figure 1-8: Uniform ring charge.

where

k2 =4ar sin

r2 + a2 + 2ar sin ; (1.82)

21


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is the argument of the complete elliptic integral of the rst kind dened by

K(k2) Z=20

1p1 k2 sin2

d: (1.83)

The nal form of the potential is

(r; ) =q

220

1pr2 + a2 + 2ar sin

K(k2): (1.84)

f(x) =

Z=20

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

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the argument k2 approaches unity,

k2 =4a(a )

(a )2 + a2 + 2a(a ) ' 1

2a

2;

and the ring potential becomes

ring ' q420a

ln

8a

: (1.85)

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

C =q

ring=

420a

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 thecapacitance 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 Coordinates

As the second method, we directly solve the Poissons equation in the spherical coordinates,

r2 = (r)0

= 10

q

2a2(r a)(cos ): (1.87)

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

satises 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;

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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 dierential equations for R and F;

d2R

dr2+

2

r

dR

dr l(l + 1)

r2R = 0; (1.90)

1sin

dd

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 Legendres dierential 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 1

2; P3(x) =

5x3 3x2

; (1.95)

Ql(x) =1

2Pl(x) ln

1 + x

1 x Wl1(x); W1(x) = 0; W0(x) = 1; W1(x) =3

2x; (1.96)

For a general complex variable z; the denition for Ql(z) is modied as follows:

Ql(z) =1

2Pl(z) ln

z + 1

z 1 Wl1(z); W1(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


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

(r; ) =

8>>>>>:

Pl Al

ra

lPl(cos ); r < a

Pl Bla

rl+1

Pl(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 Poissons equation,

@2

@r2+

2

r

@

@r+

1

r21

sin

@

@

sin

@

@

= 1

0

q

2a2(r a)(cos );

to obtain

Xl

d

2

dr2 + 2r ddr l(l + 1)r2

AlRl(r)Pl(cos ) = 10 q2a2 (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 =

Z1

1Pl()Pl0()d =

2

2l + 1ll0 : (1.101)

We thus obtain

Al

d2

dr2+

2

r

d

dr l(l + 1)

r2

Rl(r) = 1

0

q

2a2(r a) 2l + 1

2Pl(0): (1.102)

The LHS should contain a delta function (ra) 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=a0

= + la

: (1.103)

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

d2Rldr2

= 2l + 1a

(r a): (1.104)

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We thus nally nd the expansion coecient Al;

Al =q

40aPl(0) =

q

40a

(1)l=2l!2l[(l=2)!]2

; (l = 0; 2; 4; 6; ); (1.105)

and the potential,

(r; ) =q

40a

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

l=0(r) =q

40a

a

r;

followed by the quadrupole potential,

l=2(r; ) = q40a

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 sucient 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


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charge, the axial potential can be readily found,

(z) =q

40

1pz2 + a2

: (1.107)

For

jz

j> a; this can be expanded as

(z) =q

40

1

jzj

1 12

az

2+

3

8

az

4

: (1.108)

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

(r; ) =q

40

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

40

1

a

11

2r

a2

P2(cos ) +3

8 r

a4

P4(cos ) ; r < a: (1.110)They agree with the potential given in Eq. (1.106).

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Method 3: Cylindrical Coordinates

In the cylindrical coordinates (;;z); the Poissons equation for the potential due to a ring

charge becomes @2

@2+

1

@

@+

@2

@z2

(; z) = q

20

( a)a

(z): (1.111)

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

(; z) =1

2

Z11

(; k)eikzdk; (1.112)

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

@

@z(; z);

is Fourier transformed as

ik(; k);

and (z) as unity, the equation for (; k) becomes an ordinary dierential equation,

d2

d2+

1

d

d k2

(; k) = q20

( a)a

: (1.113)

Elementary solutions of the dierential equation at 6= a;

d2

d2+

1

d

d k2

(; k) = 0; (1.114)

are the zero-th order modied 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 coecient 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

=a0

= AkI00(ka)K0(ka); (1.117)

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x21.510.50

5

4

3

2

1

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

from which it follows that

d2

d2

=a

= Ak

I0(ka)K00(ka) I00(ka)K0(ka)

( a) = A

a( a); (1.118)

where the Wronskian of the modied Bessel functions

I0(ka)K00(ka) I00(ka)K0(ka) =

1

ak; (1.119)

has been exploited. From

d2

d2=a

= Aa

( a) = q20

( a)a

; (1.120)

we nd

A =q

20; (1.121)

and the nal form of the potential is

(; z) =q

420

Z11

8>:

I0(k)K0(ka)

I0(ka)K0(k)

9>=>;

eikzdk;

< a

> a

(1.122)

Noting that the modied Bessel functions are even with respect to the argument, this may be

further rewritten as

(; z) =q

220

Z10

8>:

I0(k)K0(ka)

I0(ka)K0(k)

9>=>; cos(kz)dk;

< a

> a

(1.123)

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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) = Z1

0

(; k)ekjzjdk: (1.124)

The Laplaces equation thus reduces to

d2

d2+

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 Poissons equation becomes

@2

@2+

1

@

@+

@2

@z2

Z10

A(k)J0(k)ekjzjdk = q

20

( a)a

(z): (1.127)

Notingd2

dz2ekjzj = k2ekjzj 2k(z) (1.128)

we thus nd Z10

kA(k)J0(k)dk =q

40

( a)a

: (1.129)

However, the Bessel function forms an orthogonal set according to

Z10

kJ0(ka)J0(k)dk =( a)

a; (1.130)

which uniquely determines the function A(k);

A(k) =q

40J0(ka): (1.131)

Therefore, the nal form of the potential is

(; z) =

q

40Z10 J0(ka)J0(k)e

kjzj

dk: (1.132)

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

suitable for numerical evaluation.

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Method 4: Toroidal Coordinates

The 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

(; ; ) =p

cosh 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, ! 0describes 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

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

1

sinh

d

d

sinh

dF1d

+

1

4 l2 m

2

sinh2

F1 = 0; (1.135)

d2d2 + l

2

F2 = 0; d2

d2 + m2

F3 = 0:

Comparing Eq. (1.135) with the standard form of the dierential equation for the associated

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

1

sin

d

d

sin

d

d

+ l(l + 1) m

2

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

(; ; ) =p

cosh cos Xl;m

AlmPl 1

2

(cosh ) + BlmQl 12

(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,

(; ; ) =p

cosh cos 1Xl=0

AlPl 1

2

(cosh ) + BlQl 12

(cosh )

(Cl cos l + Dl sin l)

1Xm=0

(El cos m + Fl sin m) :

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

(; ) =p

cosh cos 1Xl=0

AlPl 12

(cosh )cos l; (1.141)

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where the Legendre function of the second kind Ql 12

(cosh ) has been discarded because it diverges

at ! 0 which corresponds to the zaxis. (The potential along the zaxis should be bounded.)The expansion coecients Al can be determined from the boundary condition, = V at = 0;

V =p

cosh 0 cos

1

Xl=0 AlPl

1

2 (cosh 0)cos l; (1.142)

or1Xl=0

AlPl 12

(cosh 0)cos l =Vp

cosh 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)

where0 = 1; l = 2 (l 1);

and the following integral representation has been exploited,

Z0

cos lpx cos d =

p2Ql 1

2

(x): (1.144)

Then, the nal form of the potential is

(; ) =

p2V

pcosh cos 1

Xl=0

Ql 12

(cosh 0)

Pl 12

(cosh 0)Pl 1

2

(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) = q40r

: (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 '

p2 + 2p

2'p

2R

r; (1.148)

Pl 12

(cosh ) ' 1: (1.149)

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Therefore, the asymptotic potential is

(r R) = 2V Rr

1Xl=0

Ql 12

(cosh 0)

Pl 12

(cosh 0)=

q

40r; (1.150)

from which the capacitance of the torus can be found,

C qV

= 80R

1Xl=0

Ql 12

(cosh 0)

Pl 12

(cosh 0)l = 80a 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

Z0

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 dened 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 12

(5)Ql 1

2

(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 12

(5) rapidly converges as l increases and it suces to truncate the series at

l = 2: The capacitance is

C = 8"0R(1:3423 + 2 0:0249 + 2 0:0003 + ) ' 80R 1:393 = 80a 1:36:

The approximate formula worked out earlier gives

C' 42

0aln

8a

b

= 80a 1:34;

which is in reasonable agreement with the numerical result.

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1.7 Spherical Multipole Expansion of the Scalar Potential

The potential due to a prescribed charge distribution,

(r) =1

4"0Z

(r0)

jrr0

jdV0;

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

sucient to expand the Greens function,

G(r; r0) =1

4

1

jr r0j ; (1.153)

which satises the singular Poissons equation,

r2G = (r r0): (1.154)

The Greens function G satises 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) m

2

sin2

F1() = 0; (1.159)

d2

d

2 + m2F2() = 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 Greens function must be periodic with respect to the azimuthal angle : Therefore,

F2() = eim with m being an integer.

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Equation (1.159) is known as the modied Legendre equation and its solutions are

F1() = Pml (cos ); Q

ml (cos ); (1.161)

where

Pml (x) = (1 x2)m=2 dm

dxmPl(x); (1.162)

Qml (x) = (1 x2)m=2dm

dxmQl(x): (1.163)

That Pml (cos ) given in Eq. (1.162) satises the modied Legendre equation can be seen as follows.

The ordinary Legendre function Pl(cos ) = Pl() satises

d

d

(1 2) dPl()

d

+ l(l + 1)Pl() = 0:

Dierentiating m times yields

(1 2) dm+2Pl

dm+2 2(m + 1) d

m+1Pldm+1

+ [l(l + 1) m(m + 1)] dmPl

dm= 0:

Let us assume

F1() = (1 2)m=2f();

and substitute this into Eq. (1.159). f() satises the same equation as dmPldm ;

(1 2) d2f

d2 2(m + 1) df

d+ [l(l + 1) m(m + 1)] f = 0:

Therefore, f() = dmPldm :

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() andintroduce the spherical harmonic function dened by

Ylm(; ) = const. Pml (cos )e

im; (1.164)

where the constant is chosen from the following normalization,

ZYlm(; )Ylm

(; )d = 1: (1.165)

Noting Z0

[Pml (cos )]2 sin d =

2

2l + 1

(l + m)!

(l m)! ; (1.166)

we nd

const. =

s2l + 1

4

(l m)!(l + m)!

;

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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 + 14

(l m)!(l + m)!

Pml (cos )eim; for 0 m l;Yl;m(; ) = (1)mYlm(; ) ; for l m 0;

or for arbitrary m; positive or negative,

Ylm(; ) = (1)(m+jmj)=2s

2l + 1

4

(l jmj)!(l + jmj)! P

jmjl (cos )e

im: (1.168)

Some low order forms of Ylm(; ) are:

l = 0 m = 0 Y00 = 1p4

l = 1 m = 0 Y10 =

r3

4cos

m = 1 Y1;1 = r

3

8sin ei

l = 2 m = 0 Y20 =r5

4

3cos2 12

m = 1 Y2;1 = r

15

8sin cos ei

m = 2 Y2;2 = 14

r15

2sin2 e2i

Having found the general solutions of the Greens 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:

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

be determined by multiplying the both sides of

r2G = (r r0) = (r r0)

rr0(cos cos 0)( 0); (1.169)

by Yl0m0(; ) and integrating the result over the entire solid angle,

Almd2

dr2gl(r; r

0) = (r r0)

rr0Ylm(

0; 0);

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

l0m0(; )d = ll0mm0 ; (1.170)

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

in its derivative at r = r

0

; it follows that

d2

dr2gl(r; r

0) = (2l + 1) (r r0)

r2: (1.171)

Therefore,

Alm =1

2l + 1Ylm(

0; 0); (1.172)

and the desired spherical harmonic expansion of the Greens function is given by

G(r; r0) =1

4 jr r0j =Xlm1

2l + 1gl(r; r

0)Ylm(; )Ylm(

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(; )

Zr0lYlm(

0; 0)(r0; 0; 0)dV0

=1

4"0

Xlm

4

2l + 1

1

rl+1Ylm(; )qlm; in the region r > r

0; (1.174)

where

qlm Z

r0lYlm(0; 0)(r0; 0; 0)dV0; (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

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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 =

Zr2Y2;2(; )(r)dV =

1

2

r15

2a2qei=2:

The potential reduces to

(r;;) ' 38"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 1030 Cm due to deviation of the center of electron cloud from thecenter of proton charges. When water is placed in an external electric eld, the dipoles tend to be

39


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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

40

(r r0) pjr r0j3 : (1.176)

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

density P =np (Cm/m3 = C/m2) where n is the number density of dipoles. An incrementalpotential due to an incremental point dipole dp = P(r0)dV0 is

d =1

40

(r r0) P(r0)jr r0j3 dV

0: (1.177)

Integration of this yields

(r) =1

40

Z(r r0) P(r0)jr r0j3 dV

0: (1.178)

Sincer r0jr r0j3 = r

0

1

jr r0j

; (1.179)

where

r0 means dierentiation with respect to r0; the integral in Eq. (1.178) can be rewritten as

Z(r r0) P(r0)jr r0j3 dV

0 =

Zr0 P(r0)

jr r0j

dV0 Zr0 P(r0)

jr r0j dV0:

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

ZVr0 P(r0)

jr r0j

dV0 =

IS

P(r0)

jr r0j dS0;

which vanishes on a closed surface with innite extent on which all sources should be absent.

Therefore, the potential due to distributed dipole moments is

dipole(r) = 140

Zr0 P(r0)jr r0j dV

0: (1.180)

Comparing with the standard form of the potential,

(r) =1

4"0

Z(r0)

jr r0jdV0;

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we see that

e = r P;

can be regarded as an eective charge density. To distinguish it from the free charge density,

e dened 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 themonopole potential due to a free charge density free(r); we nd the total potential

(r) =1

40

Zfree(r

0)

jr r0j dV0 1

40

Zr0 P(r0)jr r0j dV

0: (1.181)

Noting

r2 1jr 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 eective 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 Maxwells equation

r E =all0

; (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 briey discussed in Sec-

tion 2, Maxwells 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

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allows us to introduce an eective 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)

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and the displacement vector is given by

D = "0E + P

= "0

1 +

ne2

"0m!20

E = "E: (1.190)

The permittivity can therefore be dened by

" = "0

1 +

!2p!20

!; (1.191)

where

!p =

sne2

"0m; (1.192)

is an eective plasma frequency. (The plasma frequency pertains to free electrons in a plasma. !p

introduced here is so called only for dimensional convenience.)In the incremental contribution to the permittivity normalized by "0,

!2p!20

= ne2

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:

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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 + 2:7 104 "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 = p4"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"0R3

Eext ; (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 n3"0

E =n="0

1 n3"0

"0E; (1.202)

which denes the relative permittivity "r as

"r 1 = n="01 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 reects, somewhat surprisingly, microscopic atomic polarizability through a simple

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relationship.

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

m

@2

@t2+ !20

x = eE0ei!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 = pEcos ; (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

pEcos

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 1023 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 dR

exppEcos 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

pEcos kBT

' 1 + pEcos

kBT; (1.210)

and we obtain

n ' n0p3kBT

E: (1.211)

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Resultant dipole moment density is

P = np =n0p

2

3kBTE: (1.212)

This denes 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

1030 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 atomicpolarizability is thus dominant.

1.9 Boundary Conditions for E and D

In electrostatics, the electric eld E obeys the Maxwells equations,

r E = "0

;

and

rE = 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 ofr 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 yields

Zr DdV =

ID dS =

ZfreedV: (1.216)

In the limit of innitesimally thin pancake, this reduces to

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

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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 thevolume 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.

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

rE 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,

rE = @B@t

; (1.223)

for the area integral ZSB dS;

vanishes in the limit of innitesimally 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,Z2(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. Inpower 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

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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 = 1r

@@

: (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"0aE0: (1.230)

Then the solutions for the potentials are

(r; ) =

8>>>>>>>:

0(r; ) +" "0

" + 2"0E0r cos ; r < a

0(r; ) +" "0

" + 2"0E0

a3

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)

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

" "0" + 2"0

a3

r2E0 cos ; (1.234)

is of dipole form with an eective dipole moment

pz = 4"0" "0

" + 2"0a3E0; (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 innite 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 =4a3

3Pz

=4(" "0)a3

" + 2"0"0E0;

which agrees with that deduced from the exterior potential.

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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 r

E and gradient Er

E =1

2rE2 of the

electric eld are identical,

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

Therefore,

f = r

"0EE 12

"0E21

; (1.243)

where 1 is the unit sensor. The tensor

Tij = "0EiEj 12

"0E2ij; (1.244)

is called the Maxwells stress tensor associated with the electric eld. For a linear dielectric, "0

may be replaced with its permittivity ": The force vector is

f= r T; or fi = @@xj

Tij: (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@1

2"0E

2z 0 0

0 12

"0E2z 0

0 0 +1

2"0E

2z

1CCCCA : (1.246)

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The force in the directions perpendicular to the eld are

fx = @@x

1

2"0E

2z

; (1.247)

fy = @

@y1

2 "0E2z

; (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 "E2

n

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)

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Er = E03"

" + 2"0cos ; (1.253)

E = E0 3"0" + 2"0

sin ; (1.254)

E2 = E2r + E2

= E209

(" + 2"0)2

"2 cos2 + "20 sin2

: (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 normal

to 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 = 2a2

Z=20

"0EzEr 1

2"0E

2 cos

sin d

= 2a29"0E

20

(" + 2"0)2

Z=20

"2 cos2 + ""0 sin

2 12

("2 cos2 + "0 sin2 )

cos sin d

= a29"0E

20

(" + 2"0)21

4("2 + 2""0 "20): (1.258)

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

Fz2 = a2 9"0E20

(" + 2"0)2"2

2; (1.259)

and the net force is

Fz = a2 9(" "0)24(" + 2"0)2

"0E20 : (1.260)

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

Fz = 94

a2"0E20 :

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

1.11 Force in Capacitor

For two electrode systems, the potential dierence 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

2CV2: (1.263)

This is applicable for a single electrode as well if the second electrode is at innity (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 =

Z1a

1

2"0E

2r4r

2dr =Q2

8"0a; (1.266)

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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

2V2

@C()

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

F = 12

Q2@

@

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) = "0a2

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 = 12

"0a2

d2V2 = 1

2

Q2

"0a2: (1.271)

The radial force is

Fa = "0a

dV2 =

d

"0a3Q2; (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.

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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) = ea3 exp2ra ;

where a = 5:3 1011 m is the Bohr radius. Show that the potential energy of a hydrogenatom is

U = 14"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:Find

the electric dipole and quadrupole moments.

1.5 Show that for l = 1;1X

m=1

rY1;m(; )q1;m =3

4r p;

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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 +qand four

q) alternatively placed at the corners

of a cube of side a: Determine the far eld potential (r;;) at r a and also the potentialenergy 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 expansionis

(r) =Xlm

ar

l+1Ylm(; )

Is(

0; 0)Ylm(0; 0)d0; r > a;

where s (; ) is the surface potental and d0 = sin 0d0d0:

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

() = 2"0

ln + constant,

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

(r) =1

4"0

Z(r0)

jr r0jdV0

= 14"0

Z11

p2 + 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).

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

C

l

L

l;

is constant and equal to "00: 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

Poissons equation,

r2 = "0

;

is satised 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,

Z11

Pl(x)Pl0(x)dx =2

2l + 1ll0 :

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 + 2:7 104 "0:

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When " is written in the form

" = const:!2p!20

"0 = const:ne2

me!20;

where n is the molecule density, me = 9:1

1031 kg is the electron mass, and !0 is the

frequency 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 m1) 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

Z10

J0(ka)J0(k)ekjzjdk:

Show that the solution above satises the Poissons equation,

@2

@2+

1

@

@r+

@2

@z2(; z) = q

2"0

a(

a)(z):

1.20 A conduct

lec2 electrostatics

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