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 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.
q1 q2
q1 q2
= const.
; (N)
2
r
jr1 r2 j2
(1.1)
where r = jr1 r2 j 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 by
F
I1 I2
= B1 I2 = 0
; (N/m)
(1.2)
l
2 d
the magnetic permeability 0 = 4 10 7 H/m in MKS-Ampere unit system is an assigned constant
introduced 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 approximately 9:0 109 N m2 C 2 : It is customary to write the constant in the form
const. =
2
1
;
4 "0
Figure 1-2: In MKS unit system, I = 1 Ampere current is de…ned if the force per unit length between
10 7
in…nite parallel currents 1 m apart is 2 10 7 N/m. The magnetic permeability 0 = 4
H/m is an assigned constant to de…ne 1 Ampere current.
where
"0 = 8:85
10
12
C2
N m2
=
Farad
m
;
is the vacuum permittivity. Since
c2 =
1
"0
; m2 s
2
;
0
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
1 q1 (r2 r1 )
4 "0 jr1 r2 j3
= q2 Eq1 ; (N)
F =
where
Eq1 =
1 q1 (r2 r1 )
;
4 "0 jr1 r2 j3
q2
N
V
=
C
m
(1.3)
(1.4)
is the electric …eld produced by the charge q1 at a distance r1 r2 : The …eld is associated with the
charge 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
1 q1 q2 (r2 r1 )
dW = F dr2 =
dr2 :
4 "0 jr1 r2 j3
Noting
rr2
1
jr1
r2 j
=+
r1
jr1
r2
;
r2 j3
integration from r2 = 1 to r2 can be readily carried out,
W =
1
q1 q2
; (J)
4 "0 jr1 r2 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 a
hydrogen atom at the ground state, an energy of 13:6 eV is required. In a hydrogen atom, the total
system energy is given by
1
W = mv 2
2
where rB = 5:3
10
11
e2
=
4 "0 rB
e2
< 0;
8 "0 rB
m is the Bohr radius and
1
e2
mv 2 =
= 13:6 eV,
2
8 " 0 rB
is the electron kinetic energy. The electric potential energy is
e2
=
4 "0 rB
4
27:2 eV.
The potential energy of two-charge system may be written as
W =
q1
1
4 "0 jr1 r2 j
q2 ; (J)
which de…nes the potential associated with a point charge q1 at a distance r;
=
1 q
; (J C
4 "0 r
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
1
E(r) =
4 "0
Z
(r0 )(r r0 ) 0
dV =
jr r0 j3
V
1
r
4 "0
Z
(r0 )
dV 0 ;
jr r0 j
(1.8)
since the di¤erential electric …eld due to a point charge dq = (r0 )dV 0 located at r0 is
dE =
Note that
r
1
jr
r0 j
1 (r0 )dV 0
(r
4 "0 jr r0 j3
=
r0
1
jr
r0 j
=
r0 ):
r
jr
r0
;
r0 j3
where r0 is the gradient with respect to r0 . Eq. (1.8) allows one to calculate the electric …eld for a
given 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
I
E 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),
I
Z
E dS =
=
However, the function 1= jr
r0 j satis…es a singular Poisson’s equation,
r2
Therefore,
r EdV
Z
Z
(r0 )
1
2
dV r
dV 0 :
0j
4 "0 V
jr
r
0
V
V
S
I
1
E dS =
"0
S
Z
1
r0 j
jr
dV
Z
4
(r
(r
1
r ) (r )dV =
"0
0
V0
V
r0 ):
=
0
(1.9)
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. From
Z
Z
1
r EdV =
(r)dV;
(1.11)
"0
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,
1
E(r) =
4 "0
Z
V
(r0 )(r r0 ) 0
dV =
jr r0 j3
1
r
4 "0
Z
V
(r0 )
dV 0 ;
jr r0 j
its curl identically vanishes because
r
E=
1
r
4 "0
(Note that for any scalar function F; r
electrostatics is
r
rF
r
Z
(r0 )
dV 0
jr r0 j
0:
0:) Therefore, the second Maxwell’s equation in
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 be
found 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 B
which 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 de…nition, characterized by
r
Al = 0; r At = 0:
The longitudinal component can be calculated from
1
r
4
Al (r) =
since
r2
1
jr
r0 j
Z
=
r0 A (r0 ) 0
dV ;
jr r0 j
4
r
r0 :
The transverse component is given by
At (r) = A (r)
1
r
=
4
Al (r)
Z
A (r0 )
r
dV 0 ;
jr r0 j
where the identity
r
r
A =rr A
r2 A;
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 longitudinal
component. Likewise, r A =r At specifes the curl of the tarnverse component. As a concrete
example, let us consider the electric …eld. Its divergence is
r E=
"0
;
and the solution to this di¤erential equation is
1
r
4 "0
El (r) =
Z
V
(r0 )
dV 0 :
jr r0 j
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),
1
(r) =
4 "0
In terms of the scalar function
Z
V
(r0 )
dV 0 :
jr r0 j
(1.16)
(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 adiabatically 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 =
q
Z
r2
r1
E dl = q
Z
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 is
independent 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
= "0 r 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
@
+ r J = 0:
@t
(1.20)
The current density can be estimated from the equation of motion for electrons
m
@v
=
@t
eE
m e v;
(1.21)
where c (s 1 ) is the electron collision frequency with the lattice ions. Noting J =
the density of conduction electrons, we obtain
@
+
@t
c
J=
ne2
E:
m
nev with n
(1.22)
Therefore, the equation for the excess charge density in a conductor becomes
@
@t
@
+
@t
+ ! 2p = 0;
c
! 2p =
ne2
;
"0 m
(1.23)
which describes a damped plasma oscillation 0 e t i!p t with an exponential damping factor =
13
c =2: The typical electron collision frequency in metals is of order c ' 10 = 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 = mp c=~ will modify the wave equation for all potentials, including the Coulomb potential, in
the form
1 @2
2
=
:
(1.24)
r2 kC
c2 @t2
"0
9
Figure 1-3: Charge initially given to the inner conductor will all end up at the outer surface of the
outer conductor. The absence of the electric …eld inside the sphere is a consequence of Coulomb’s
law.
In static case @=@t = 0; this reduces to
r2
2
kC
=
"0
;
(1.25)
and for a point charge (r) = q (r); yields a Debye or Yukawa type potential,
(r) =
and electric …eld
Er =
1 q
exp ( kC r) ;
4 "0 r
q
4 "0
1
kC
+
2
r
r
e
kC r
(1.26)
:
(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,
1 q
(r) =
exp ( kD r) ;
(1.28)
4 "0 r
where
kD =
s
ne2
;
"0 T
(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
kL2 B = 0;
where kL = ! p =c is the London skin depth with ! p =
The potential energy of a two-charge system is
W =
p
1
q1 q2
=2
4 "0 jr1 r2 j
where
(1.30)
ne2 ="0 me the electron plasma frequency.)
q1 q2
1 1
;
2 4 "0 jr1 r2 j
q1 q2
1 1
;
2 4 "0 jr1 r2 j
(1.31)
(1.32)
is the potential energy associated with each charge. For many charge system, this can be generalized
in the form
1 1 X qi qj
1X
;
(1.33)
W =
j qj =
2
2 4 "0
jri rj j
j
i6=j
where
j
=
1 X
qi
;
4 "0
jri rj j
(1.34)
i6=j
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 =
Since
= "0 r E;
1
1
dq =
2
2
dV:
1
dW = "0 r EdV:
2
Integrating over the entire volume, we …nd
Z
1
W =
"0
r EdV
2
V
Z
1
=
"0
[r ( E) (r )E] dV
2
V
I
Z
1
1
=
"0
E dS +
"0 E 2 dV:
2
2
S
V
where use is made of Gauss’theorem,
Z
V
r FdV =
I
F dS;
(1.35)
(1.36)
(1.37)
S
for an arbitrary well de…ned vector …eld F. The surface integral vanishes because at in…nity, both
potential and electric …eld vanish. Therefore, the potential energy associated with a distributed
11
charge system is
W =
Z
1
"0 E 2 dV;
2
(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
1
ue = "0 E 2 ; (J/m3 )
(1.39)
2
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 e2
:
4 " 0 re
(1.40)
This should not exceed the rest energy of the electron me c2 : Equating these two, the following
estimate for the electron radius emerges,
re '
1 e2
= 2:85
4 "0 mc2
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 2
r ;
3 e
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
Z
Z
1
1
2
2
U=
E
dV
=
dV:
(1.42)
0
0 (r )
2
2
Indeed, the variation of this integral
U
=
=
Z
( 0r
Z
r
0r
2
12
) dV
+
dV;
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 = "0 r E; or in the case of surface charge on a conductor surface,
= "0 En ; (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) =
can be written down as
(r) =
1
4
0
Z
V
(r)
;
(1.43)
(r0 )
dV 0 :
jr r0 j
(1.44)
0
This is understandable because the di¤erential potential due to a point charge dq = dV is
d
=
1 (r0 )dV 0
;
4 "0 jr r0 j
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,
(r; t)
r2 C (r; t) =
;
(1.45)
"0
if one employs the Coulomb gauge characterized by the absence of the longitudinal vector potential
r A = 0:
13
(1.46)
due to a “point charge” dq = dV 0 .
Figure 1-4: Di¤erential potential d
In the Coulomb gauge, the transverse vector potential satis…es the wave equation,
1 @2
c2 @t2
r2
At =
0 Jt ;
(1.47)
where Jt is the transverse current. The solution for the scalar potential in the Coulomb gauge is
non-retarded,
Z
1
(r0 ; t)
dV 0 ;
(1.48)
C (r; t) =
4 0 V jr r0 j
and so is the resultant longitudinal electric …eld,
El (r; t) =
r
C
1
=
4 "0
Z
V
r
jr
r0
0
0
3 (r ; t)dV :
0
rj
(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
Z
@At
Jt (t
) 0
jr r0 j
0 @
Et (r; t) =
=
dV
;
=
;
@t
4 @t
jr r0 j
c
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 ) =
14
1
;
4 jr r0 j
(1.50)
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
r2 G =
r0 );
(r
(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 r2 G = 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
r0 ):
= q (r
(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 y 0 ) (z z 0 );
(1.53)
in the spherical coordinates (r; ; ),
(r
r0 ) =
(r r0 ) [r(
(r r0 )
(
rr0 sin
(r r0 )
(cos
rr0
=
=
0
0
0
)] [r sin (
0
) (
)]
)
(1.54)
(1.55)
cos 0 ) (
0
);
(1.56)
and in the cylindrical coordinates ( ; ; z),
(r
r0 ) =
0
(
(
=
0
) [ (
0)
0
(
z0)
)] (z
z 0 ):
) (z
(1.57)
(1.58)
The singular Poisson’s equation
r2 G =
(r
r0 );
(1.59)
can be solved formally as follows. Let the Fourier transform of G(r
g(k) =
Z
Its inverse transform is
G(r
0
r)=
G(r
1
(2 )3
r0 )e
Z
ik (r r0 )
g(k)eik (r
dr:
r0 ) 3
Noting that the operator r is Fourier transformed as ik and (r
15
r0 ) be g(k); namely,
(1.60)
d k:
(1.61)
r0 ) as unity, we readily …nd from
Eq. (1.59),
g(k) =
1
:
k2
(1.62)
Substitution into Eq. (1.61) yields
0
G(r
r) =
=
=
=
where the polar angle
r r0 ; and
Z
1 ik (r r0 ) 3
e
d k
(2
k2
Z 2
Z 1 Z
1
ikjr r0 j cos
d
e
sin
d
dk
(2 )3 0
0
0
Z
1
(jr r0 j cos ) sin d
(2 )2 0
1
;
4 jr r0 j
1
)3
(1.63)
in the k-space is measured from the direction of the relative position vector
d3 k = k 2 dk sin d d ;
Z 1
eikx dk = 2 (x);
(1.64)
(1.65)
1
(ax) =
1
(x);
jaj
(1.66)
are noted. (The same technique will be used in …nding a Green’s function for the less trivial wave
equation,
1 @2
r2
G(r; r0 ; t; t0 ) =
(r r0 ) (t t0 );
(1.67)
c2 @t2
in Chapter 6.)
If the charge density distribution is spatially con…ned within a small volume such that r
the inverse distance function 1=jr r0 j may be Taylor expanded as follows:
1
jr
r0 j
=
=
1
1
1
1
r
r0 + rr
: r0 r0
r
r
2!
r
1 X
1 r r0
+ 3 + 5
(3ri rj r2 ij )ri0 rj0 +
r
r
2r
r0 ;
(1.68)
i;j
and so is the potential
(r) =
1
4
0
Z
V
(r0 )
1
dV 0 =
0
jr r j
4 0
1
r
Z
V
r
(r )dV + 3
r
0
0
Z
V
r0 (r0 )dV 0 +
3rr r2 1
:Q+
2r5
;
(1.69)
where 1 is the unit dyadic. The quantity q =
is the total charge contained in the (small)
volume V and the corresponding lowest order monopole potential is
R
(r0 )dV 0
monopole
=
16
1 q
:
4 0r
(1.70)
The vector quantity
p=
Z
r (r)dV;
(1.71)
is the dipole moment and the corresponding dipole potential is
dipole
=
The tensor
Q = Qij =
1 p r
:
4 0 r3
Z
(1.72)
ri rj (r)dV;
(1.73)
de…nes the quadrupole moment and the corresponding quadrupole potential is
quadrupole
=
1 3rr r2 1
1 1 X
:Q=
(3ri rj
5
4 0
2r
4 0 2r5
r2
ij )Qij :
(1.74)
i;j
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
p
cos :
2
0r
The equipotential surfaces are described by
cos
= const.
r2
The electric …eld can be found from
E=
r
=
1
4
0
p sin
2p cos
er +
e
3
r
r3
:
(1.75)
The equation to describe an electric …eld line is, by de…nition,
dr
rd
=
;
Er
E
(1.76)
which gives
dr
= 2 cot d ;
r
and thus
r = const. sin2 :
(1.77)
Evidently, the electric …eld lines are normal to the equipotential surfaces. In Fig.1-5, one equipotential surface and three electric …eld lines are shown.
Example 2 Linear quadrupole
17
1
0.5
-1
00
-0.5
0.5
1
-0.5
-1
Figure 1-5: Equipotential surface and electric …eld lines of an electric dipole pz : The z-axis is
vertical.
A linear quadrupole consists of charges q at z = a and 2q at z = 0 as shown in Fig.1-6. The
charge density is
(r) = q[ (z a) 2 (z) + (z + a)] (x) (y):
Therefore, only Qzz is nonvanishing,
Qzz =
and the quadrupole potential at r
(r) =
1
4
1
(3z 2
5
0 2r
Z
z 2 (r)dV = 2a2 q;
a is given by
r2 )Qzz =
1 2qa2 3 cos2
4 0 r3
2
where
P2 (cos ) =
3 cos2
2
1
1
=
1 2qa2
P2 (cos );
4 0 r3
;
is the Legendre function of order l = 2.
Alternatively, the potential can be found from the direct superposition of the potentials produced by each charge,
(r; ) =
q
4
0
p
1
r2
p
For r > a; the functions 1= r2 + a2
+
a2
2ar cos
2
1
+p
2
2
r
r + a + 2ar cos
:
2ar cos can be expanded in terms of the Legendre functions
18
Figure 1-6: Linear quadrupole.
Pl (cos ) as
p
1
r2
+
a2
2ar cos
=
1
1X a
r
r
l
( 1)l Pl (cos ); r > a:
l=0
Retaining terms up to l = 2 (quadrupole), we recover
(r; ) '
1 2qa2
P2 (cos );
4 0 r3
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 integration
The potential is symmetric about the z-axis because of the uniform charge distribution. Therefore, 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
-1
0
-0.5
0.5
1
-0.5
-1
Figure 1-7: Equipotential surface of a linear quadrupole. The z axis ( = 0) is in the vertical
direction. The thick line is at +1 unit potential, and the thin line is at 1:
dq = qd 0 =2 located at angle
d
where
to
=
0
is
1
4
dq
0
jr
1
=
r0 j
4
0
r2
2
Integrating over
0
0
= cos cos
+
a2
2ar cos
= =2) and r0 = (a;
is the angle between the vectors r = (r; ;
cos
q
p
0
+ sin sin
0
cos(
d 0;
0
) = sin sin
(1.78)
= =2;
0
0
): cos
:
reduces
(1.79)
from 0 to 2 ; we …nd
1
(r; ) =
Changing the variable from
0
4
to
Z
q
02
2
d
p
r2 + a2
0
through
2 =
0
+
2
0
:
(1.80)
4
K(k 2 );
r2 + a2 + 2ar sin
(1.81)
2ar sin sin
0
;
the integral can be reduced to
4
p
r2 + a2 + 2ar sin
Z
0
=2
1
p
1
k 2 sin2
20
d =p
Figure 1-8: Uniform ring charge.
where
k2 =
4ar sin
;
r2 + a2 + 2ar sin
21
(1.82)
is the argument of the complete elliptic integral of the …rst kind de…ned by
Z
2
K(k )
=2
1
0
The …nal form of the potential is
(r; ) =
q
2
2
f (x) =
p
0
1
p
Z
r2
+
a2
=2
p
1
0
k 2 sin2
d :
(1.83)
1
K(k 2 ):
+ 2ar sin
1
x sin2
(1.84)
d
4.5
4
3.5
3
2.5
2
0
0.2
0.4
0.6
0.8
1
Figure 1-9: K(x)
p the complete elliptic integral of the …rst kind. It diverges logarithmically at x . 1;
K(x) ' ln 4= 1 x :
The function K(k 2 ) is shown in 1-9. It diverges as x = k 2 approaches unity, that is, near the
ring itself, as expected. However, the divergence is only logarithmic,
lim
k2 !1
= ln
4
p
:
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 k 2 approaches unity,
k2 =
(a
)2
4a(a
)
2
+ a + 2a(a
)
2
'1
2a
;
and the ring potential becomes
ring
'
q
4
2
0a
8a
ln
:
(1.85)
Then the self-capacitance of the ring can be found from
C=
q
=
ring
4
2
ln
0a
8a
;
a
:
(1.86)
This formula is fairly accurate even for a not-so-thin ring because of the mere logarithmic dependence 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 Coordinates
As the second method, we directly solve the Poisson’s equation in the spherical coordinates,
r2
=
(r)
0
=
1
q
(r
2
a2
0
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@
@2
1 1 @
+
+ 2
@r2
r @r
r sin @
Assuming that the potential is separable in the form
r2
R
d2 R 2 dR
+
dr2
r dr
+
sin
@
@
= 0:
(1.88)
(r; ) = R(r)F ( ), we …nd
1 1 d
F sin 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;
d2 R 2 dR
+
dr2
r dr
1 d
sin d
which can be written as
d
d
2
)
(1.90)
dF
d
+ l(l + 1)F = 0;
(1.91)
dF
d
+ l(l + 1)F = 0;
(1.92)
sin
(1
l(l + 1)
R = 0;
r2
where = cos : Eq. (1.92) is known as the Legendre’s di¤erential equation. The solutions for
R(r) are
1
R(r) = rl and l+1 ;
(1.93)
r
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) =
1
1+x
Ql (x) = Pl (x) ln
2
1 x
Wl
1 (x);
W
3x2 1
5x3 3x
; P3 (x) =
;
2
2
1 (x)
3
= 0; W0 (x) = 1; W1 (x) = x;
2
(1.95)
(1.96)
For a general complex variable z; the de…nition for Ql (z) is modi…ed as follows:
1
z+1
Ql (z) = Pl (z) ln
2
z 1
Wl
1 (z);
W
1 (z)
3
= 0; W0 (z) = 1; W1 (z) = z;
2
(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
P
r
>
>
Al
>
l
<
a
l
Pl (cos );
r<a
(1.98)
>
>
>
: P Bl a
l
r
l+1
Pl (cos ); r > a
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
2@
1 1 @
+
+ 2
@r2
r @r
r sin @
sin
@
@
=
1
q
(r
2
a2
0
a) (cos );
to obtain
X
l
2 d
d2
+
2
dr
r dr
l(l + 1)
r2
Al Rl (r)Pl (cos ) =
where
Rl (r) =
8
>
>
>
<
>
>
>
:
r
a
l
a
r
l+1
;
1
q
(r
2
02 a
a) (cos );
(1.99)
r<a
(1.100)
; r>a
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
0
Pl (cos )Pl0 (cos ) sin d =
Z
1
1
Pl ( )Pl0 ( )d =
2
2l + 1
ll0 :
(1.101)
We thus obtain
Al
d2
2 d
+
2
dr
r dr
l(l + 1)
r2
Rl (r) =
1
q
(r
2
02 a
a)
2l + 1
Pl (0):
2
(1.102)
The LHS should contain a delta function (r a) to be compatible with that in the RHS. This can
be seen as follows. The radial function Rl (r) has a discontinuity in its derivative at r = a;
dRl
dr
=
r=a+0
l + 1 dRl
;
a
dr
r=a 0
l
=+ :
a
(1.103)
Therefore, the second order derivative d2 Rl =dr2 yields a delta function,
d 2 Rl
=
dr2
2l + 1
(r
a
25
a):
(1.104)
We thus …nally …nd the expansion coe¢ cient Al ;
Al =
q
4
0a
Pl (0) =
and the potential,
(r; ) =
q
4
q
4
( 1)l=2 l!
; (l = 0; 2; 4; 6;
l
2
0 a 2 [(l=2)!]
X ( 1)l=2 l!
Rl (r)Pl (cos ):
2l [(l=2)!]2
0a
);
(1.105)
(1.106)
l even
Figure 1-11: The derivative of the radial function R(r) is discontinuous at r = r0 and thus its
second 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
a
;
0a r
followed by the quadrupole potential,
l=2 (r;
)=
q
4
1 a
0a 2 r
3
P2 (cos );
and so on. The dipole potential (l = 1) vanishes because of the symmetric charge distribution,
( r) = (r);
Z
p = r (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
p
1
:
+ a2
(1.107)
z2
For jzj > a; this can be expanded as
(z) =
q
4
1
0 jzj
1
1 a
2 z
2
+
3 a
8 z
4
:
(1.108)
Therefore, at arbitrary point (r > a; ); the potential is
(r; ) =
q 1
4 0r
1
1 a
2 r
2
P2 (cos ) +
3 a
8 r
4
3 r
8 a
4
P4 (cos )
; r > a:
(1.109)
P4 (cos )
; r < a:
(1.110)
For interior region (r < a); the potential becomes
(r; ) =
q
4
1
0a
1
1 r
2 a
2
P2 (cos ) +
They agree with the potential given in Eq. (1.106).
27
Method 3: Cylindrical Coordinates
In the cylindrical coordinates ( ; ; z); the Poisson’s equation for the potential due to a ring
charge becomes
q (
a)
@2
1 @
@2
( ; z) =
+
+
(z):
(1.111)
2
2
@
@
@z
2 0
a
Since the z-coordinate extends from
1 to 1; we seek a solution in the form of Fourier transform,
1
( ; z) =
2
where
Since
Z
1
( ; k)eikz dk;
(1.112)
1
( ; k) is the one-dimensional Fourier transform of the potential having dimensions of V m.
@
( ; z);
@z
is Fourier transformed as
ik ( ; k);
and (z) as unity, the equation for
( ; k) becomes an ordinary di¤erential equation,
d2
1 d
+
d 2
d
k2
( ; k) =
Elementary solutions of the di¤erential equation at
d2
1 d
+
2
d
d
k2
q
2
(
a)
a
0
:
(1.113)
6= a;
( ; k) = 0;
(1.114)
are the zero-th order modi…ed Bessel functions,
( ; k) = I0 (k ); K0 (k );
shown in Fig.1-12.
The Fourier potential ( ; k) which is continuous at
may be constructed in the following form,
( ; k) =
8
>
< AI0 (k )K0 (ka);
>
:
AI0 (ka)K0 (k );
(1.115)
= a and remains bounded everywhere
<a
(1.116)
>a
The coe¢ cient A can be determined readily from the discontinuity in the derivative of
= a;
d
d
= AkI0 (ka)K00 (ka);
= AkI00 (ka)K0 (ka);
d =a+0
d =a 0
28
( ; k) at
(1.117)
5
4
3
2
1
0
0.5
1
x
1.5
2
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
= Ak I0 (ka)K00 (ka)
I00 (ka)K0 (ka)
(
a) =
=a
A
(
a
a);
(1.118)
where the Wronskian of the modi…ed Bessel functions
I0 (ka)K00 (ka)
1
;
ak
I00 (ka)K0 (ka) =
(1.119)
has been exploited. From
d2
d 2
A
(
a
=
=a
a) =
we …nd
A=
q
2
q
2
(
a)
a
0
;
(1.120)
;
(1.121)
0
and the …nal form of the potential is
( ; z) =
q
4
2
0
Z
1
8
9
>
>
I
(k
)K
(ka)
0
0
<
=
1>
:
I0 (ka)K0 (k )
>
;
<a
ikz
e
dk;
(1.122)
>a
Noting that the modi…ed Bessel functions are even with respect to the argument, this may be
further rewritten as
8
9
I0 (k )K0 (ka) >
<a
Z 1>
<
=
q
( ; z) = 2
cos(kz)dk;
(1.123)
>
2 0 0 >
:
;
I0 (ka)K0 (k )
>a
29
The convergence of the integral is rather poor since the function I0 K0 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
kjzj
( ; k)e
dk:
(1.124)
0
The Laplace’s equation thus reduces to
1 d
d2
+
+ k2
2
d
d
( ; 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
Z
@2
1 @
@2
+
+
@ 2
@
@z 2
1
A(k)J0 (k )e
kjzj
q
dk =
2
0
(
0
a)
a
(z):
(1.127)
Noting
d2
e
dz 2
we thus …nd
Z
1
kjzj
= k2 e
kjzj
kA(k)J0 (k )dk =
0
2k (z)
q
4
(
(1.128)
a)
a
0
:
(1.129)
However, the Bessel function forms an orthogonal set according to
Z
1
kJ0 (ka)J0 (k )dk =
(
0
a)
a
;
(1.130)
which uniquely determines the function A(k);
A(k) =
q
4
J0 (ka):
(1.131)
0
Therefore, the …nal form of the potential is
( ; z) =
q
4
0
Z
1
J0 (ka)J0 (k )e
kjzj
dk:
(1.132)
0
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 Coordinates
The toroidal coordinates ( ; ; ) are related to the cartesian coordinates (x; y; z) through the
following transformation,
8
R sinh cos
>
> x=
;
>
>
cosh
cos
>
>
>
>
>
>
<
R sinh sin
;
y=
(1.133)
>
cosh
cos
>
>
>
>
>
>
>
>
R sin
>
: z=
:
cosh
cos
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 ( ):
Figure 1-13: Cross-section of the toroidal coordinates ( ; ; ):
radius R:
(1.134)
! 0 corresponds to a thin ring of
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 d
sinh d
sinh
dF1
d
1
4
+
m2
sinh2
l2
F1 = 0;
(1.135)
d2
+ m2 F3 = 0:
d 2
d2
+ l2 F2 = 0;
d 2
Comparing Eq. (1.135) with the standard form of the di¤erential equation for the associated
Legendre function Plm (cos ); Qm
l (cos );
1 d
sin d
d
d
sin
m2
sin2
+ l(l + 1)
(Plm ; Qm
l ) = 0;
(1.136)
we see that solutions for F1 ( ) are
(cosh ):
F1 ( ) = Plm 1 (cosh ); Qm
l 1
2
(1.137)
2
The functions F2 and F3 are elementary,
F2 ( ) = eil ; F3 ( ) = eim ;
(1.138)
and the general solution for the potential may be written as
( ; ; )=
p
cosh
cos
X
Alm Pl
1
2
l;m
(cosh ) + Blm Ql
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,
( ; ; ) =
p
cosh
cos
1
X
Al Pl
l=0
1
X
1
2
(cosh ) + Bl Ql
1
2
(cosh ) (Cl cos l + Dl sin l )
(El cos m + Fl sin m ) :
m=0
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
:
(1.140)
0
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
1
X
l=0
32
Al P l
1
2
(cosh ) cos l ;
(1.141)
where the Legendre function of the second kind Ql 1 (cosh ) has been discarded because it diverges
2
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 =
or
1
X
p
cosh
1
X
cos
0
Al P l
Al P l
1
2
l=0
(cosh
=p
0 ) cos l
Multiplying both sides by cos l0 and integrating over
Ql
1
2
Pl
1
2
= 1;
l
Al = V
(cosh
1
2
l=0
0 ) cos l
V
cosh
cos
0
;
(1.142)
:
from 0 to ; we obtain
(cosh
0)
(cosh
0)
p
2
l;
(1.143)
(x):
(1.144)
where
0
= 2 (l
1);
and the following integral representation has been exploited,
Z
p
0
p
cos l
d = 2Ql
x cos
1
2
Then, the …nal form of the potential is
( ; )=
p
2V p
cosh
cos
1 Q
X
l
l=0
(cosh
0)
(cosh
0)
1
2
Pl
1
2
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) =
At a large distance from the torus r
R; both
r2 = x2 + y 2 + z 2 =
p
cosh
4
and
0r
:
(1.146)
approach zero,
R2 (sinh2 + sin2 )
!
(cosh
cos )2
cos '
Pl
q
1
2
p
2+
p
2
2
(cosh ) ' 1:
33
'
p R
2 ;
r
4R2
;
2+ 2
(1.147)
(1.148)
(1.149)
Therefore, the asymptotic potential is
(r
R) =
1
1
2
l=0
1
2
2V R X Ql
r
Pl
(cosh
0)
(cosh
0)
=
q
4
0r
;
(1.150)
from which the capacitance of the torus can be found,
C
1
X Ql
q
= 8 0R
V
Pl
1
2
1
2
l=0
(cosh
(cosh
0)
l
0)
= 8 0 a tanh
0
1 Q
X
l
l=0
where a is the major radius of the torus. The function Ql
(1.144) and Pl 1 (cosh 0 ) from
1
2
Pl
(cosh
1
2
1
2
(cosh
0)
(cosh
0)
0)
l;
(1.151)
can be evaluated from Eq.
2
Pl
1
2
(cosh
0)
=
1
Z
d
(cosh
0
0
+ sinh
0 cos
1
)l+ 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
0
1
2
Ql
1
2
(5)
Pl
1:00108
0:05063
0:00384
1
2
(5)
0:74575
2:03557
13:32184
Ql
1
2
Pl
1
2
(5)
(5)
1:34234
0:02487
0:00029
The ratio Ql 1 (5)=Pl 1 (5) rapidly converges as l increases and it su¢ ces to truncate the series at
2
2
l = 2: The capacitance is
C = 8"0 R(1:3423 + 2
0:0249 + 2
0:0003 +
) ' 8 0R
The approximate formula worked out earlier gives
C'
4
ln
2
0a
8a
b
= 8 0a
1:34;
which is in reasonable agreement with the numerical result.
34
1:393 = 8 0 a
1:36:
1.7
Spherical Multipole Expansion of the Scalar Potential
The potential due to a prescribed charge distribution,
Z
1
(r) =
4 "0
(r0 )
dV 0 ;
jr r0 j
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
1
;
4 jr r0 j
(1.153)
which satis…es the singular Poisson’s equation,
r2 G =
r0 ):
(r
(1.154)
The Green’s function G satis…es Laplace equation except at r = r0 ;
r2 G = 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
1 @
1 1 @
@2
+
+
@r2 r @r r2 sin @
sin
+
@2
1
2
r2 sin @ 2
l(l + 1)
r2
R(r) = 0;
(1.158)
m2
sin2
(1.159)
@
@
G(r; ; ) = 0;
(1.157)
we obtain three ordinary equations,
d2
1 d
+
2
dr
r dr
1 d
sin d
sin
d
d
+ l(l + 1)
d2
+ m2 F2 ( ) = 0;
d 2
F1 ( ) = 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 ( ) = Plm (cos ); Qm
l (cos );
where
(1.161)
dm
Pl (x);
dxm
dm
x2 )m=2 m Ql (x):
dx
Plm (x) = (1
x2 )m=2
Qm
l (x) = (1
(1.162)
(1.163)
That Plm (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+2 Pl
d m+2
2 (m + 1)
dm+1 Pl
+ [l(l + 1)
d m+1
m(m + 1)]
d m Pl
= 0:
d m
Let us assume
2 m=2
F1 ( ) = (1
)
f ( );
and substitute this into Eq. (1.159). f ( ) satis…es the same equation as
(1
2
)
d2 f
d 2
2 (m + 1)
df
+ [l(l + 1)
d
dm Pl
d m ;
m(m + 1)] f = 0:
m
Pl
Therefore, f ( ) = dd 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. Plm (cos )eim ;
where the constant is chosen from the following normalization,
Z
Ylm ( ; )Ylm ( ; )d = 1:
Noting
Z
0
[Plm (cos )]2 sin d =
we …nd
const. =
s
2 (l + m)!
;
2l + 1 (l m)!
2l + 1 (l m)!
;
4
(l + m)!
36
(1.164)
(1.165)
(1.166)
Ylm =
s
2l + 1 (l m)! m
P (cos )eim :
4
(l + m)! l
(1.167)
For historical reasons, it is customary to write Ylm ( ; ) in the form
Ylm ( ; ) = ( 1)m
Yl;
m(
s
2l + 1 (l m)! m
P (cos )eim ; for 0
4
(l + m)! l
; ) = ( 1)m Ylm ( ; ) ; for
l
m
m
l;
0;
or for arbitrary m; positive or negative,
Ylm ( ; ) = ( 1)(m+jmj)=2
s
2l + 1 (l jmj)! jmj
P (cos )eim :
4
(l + jmj)! l
(1.168)
Some low order forms of Ylm ( ; ) are:
l=0 m=0
l=1 m=0
m=
l=2 m=0
m=
m=
1
Y00 = p
4
Y10 =
1 Y1;
1
Y20 =
1 Y2;
2 Y2;
r
3
cos
4
r
=
r
3
sin e
8
5 3 cos2
4
2
1
=
r
2
1
=
4
r
i
1
15
sin cos e
8
15
sin2 e
2
i
2i
Having found the general solutions of the Green’s function, we now assume the following expansion for G;
X
G(r; r0 ) =
Alm gl (r; r0 )Ylm ( ; );
lm
where the radial function gl
(r; r0 )
is
gl (r; r0 ) =
8
rl
>
>
>
; r < r0 ;
>
< r0l+1
>
>
0l
>
>
: r ;
rl+1
37
r > r0 :
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
r2 G =
(r r0 )
(cos
rr0
r0 ) =
(r
cos 0 ) (
0
);
(1.169)
by Yl0 m0 ( ; ) and integrating the result over the entire solid angle,
Alm
(r r0 )
Ylm ( 0 ;
rr0
d2
gl (r; r0 ) =
dr2
0
);
where orthogonality of Ylm ( ; );
Z
Ylm ( ; )Yl0 m0 ( ; )d
=
ll0 mm0 ;
(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 = r0 ; it follows that
d2
gl (r; r0 ) =
dr2
(2l + 1)
r0 )
(r
r2
:
(1.171)
Therefore,
Alm =
1
Y ( 0;
2l + 1 lm
0
);
(1.172)
and the desired spherical harmonic expansion of the Green’s function is given by
G(r; r0 ) =
X 1
1
=
gl (r; r0 )Ylm ( ; )Ylm ( 0 ;
0
4 jr r j
2l + 1
0
):
(1.173)
lm
For a given charge distribution (r); this allows evaluation of the potential in the form of spherical
harmonics,
(r) =
=
=
Z
1
(r0 )
dV 0
4 "0
jr r0 j
Z
1 X 4
1
Ylm ( ; ) r0l Ylm ( 0 ; 0 ) (r0 ; 0 ; 0 )dV 0
4 "0
2l + 1 rl+1
lm
1 X 4
1
Ylm ( ; )qlm ; in the region r > r0 ;
l+1
4 "0
2l + 1 r
(1.174)
lm
where
qlm
Z
r0l Ylm ( 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
38
q alternatively placed at each corner of a
square of side a as shown in Fig.1-14. The charge density may be written as
(r) = q
(r)
p
(r
2a)
(r a)
(cos ) ( ) +
(cos )
2
2
a
2a
The potential in the far …eld region r
(r; ; ) =
where
q2;
2
=
Z
2
r Y2;
(r a)
(cos )
a2
4
2
a is of quadrupole nature and given by
1 1
(Y2;2 q2;2 + Y2;
"0 5r3
1
2 ( ; ) (r)dV =
2
r
2 q2; 2 ) ;
15 2
a qe
2
i =2
:
The potential reduces to
(r; ; ) '
3 a2
sin2 sin(2 ); r
8 "0 r 3
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
1.8
y plane.
Collection of Dipoles, Dielectric Properties
Dielectric properties of material media originate from dipole moments carried by atoms and molecules. 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 external 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
1 (r r0 ) p
:
4 0 jr r0 j3
(r) =
(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
4
0
Integration of this yields
(r) =
Since
r0 ) P(r0 ) 0
dV :
jr r0 j3
1 (r
=
1
4
r
jr
0
Z
(1.177)
r0 ) P(r0 ) 0
dV :
jr r0 j3
(r
r0
= r0
r0 j3
1
r0 j
jr
(1.178)
;
(1.179)
where r0 means di¤erentiation with respect to r0 ; the integral in Eq. (1.178) can be rewritten as
Z
(r
r0 ) P(r0 ) 0
dV =
jr r0 j3
Z
P(r0 )
jr r0 j
r0
dV 0
Z
r0 P (r0 ) 0
dV :
jr r0 j
The …rst term in the right can be converted into a surface integral through the Gauss theorem,
Z
V
r
0
P(r0 )
jr r0 j
0
dV =
I
S
P(r0 )
dS0 ;
jr r0 j
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 ) 0
dV :
jr r0 j
Comparing with the standard form of the potential,
1
(r) =
4 "0
Z
40
(r0 )
dV 0 ;
jr r0 j
(1.180)
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) =
Z
1
4
0
0
free (r )
dV 0
jr r0 j
1
4
0
Z
r0 P(r0 ) 0
dV :
jr r0 j
(1.181)
Noting
r2
1
jr
r0 j
r E=
r2
=
4
(r
we …nd
In this form too, it is evident that
(1.181) can be rearranged as
=
r0 );
free
r P
0
0
:
(1.182)
r P can be regarded as an e¤ective charge density. Eq.
free :
r ( 0 E + P) =
(1.183)
Introducing a new vector D (displacement vector) by
D=
0E
+ P;
(1.184)
r D=
free ;
(1.185)
we write Eq. (1.183) as
which is equivalent to the original Maxwell’s equation
r E=
all
;
(1.186)
0
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 Section 2, Maxwell’s equations would not be consistent with the charge conservation principle if the
displacement current,
@D
@("E)
=
;
@t
@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 = "0 E + P ="E:
(1.187)
In some solid and liquid crystals, the permittivity takes a tensor form,
D=
E or Di =
ij Ej ;
(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 displacement 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! 20 x = 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 the
electric …eld is
e2
E:
p = ex =
m! 20
If the atomic number density is n; the dipole moment density P is
P = np =
42
ne2
E;
m! 20
(1.189)
and the displacement vector is given by
D = "0 E + P
= "0 1 +
ne2
"0 m! 20
E = "E:
(1.190)
;
(1.191)
The permittivity can therefore be de…ned by
" = "0
where
!p =
! 2p
1+ 2
!0
s
!
ne2
;
"0 m
(1.192)
is an e¤ective 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
e2
=
n
;
! 20
m"0 ! 20
the quantity
e2
;
4 "0 m! 20
is of the order of r03 where r0 is the Bohr radius. This can be seen from the force balance for the
electron,
e2
mr0 ! 20 =
;
4 "0 r02
which indeed gives
e2
= r03 :
4 "0 m! 20
The quantity
= 4 "0 r03 =
e2
;
m! 20
(1.193)
is called the atomic polarizability. A more rigorous quantum mechanical calculation yields for this
quantity
9
=
4 "0 r03 ;
(1.194)
2
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 "0 r03 :
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,
"r 1
n
=3
;
"0
"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,
e2
p = ex =
E = Eext ;
(1.196)
m! 20
produces an electric …eld
p
;
4 "0 r 3
E0 =
(1.197)
at a distance r: If the number density of atoms is n; the average inter-atom distance R can be found
from
4 3
1
R = :
(1.198)
3
n
Then,
E0 =
4 "0 R 3
Eext ;
(1.199)
and the total electric …eld is
E =
1
=
1
Eext
4 "0 R 3
n
Eext :
3"0
(1.200)
(1.201)
Therefore, using this total …eld in calculation of the polarization …eld P; we …nd
n
P =
1
n ="0
n E=
n "0 E;
1
3"0
3"0
(1.202)
which de…nes the relative permittivity "r as
"r
1=
n ="0
n :
1
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,
@2
+ ! 20 x =
@t2
m
eE0 e
i!t
;
(1.204)
which yields
x(t) =
eE0 e i!t
:
! 2 ! 20
A resultant permittivity is
! 2p
1
"(!) = "0
!2
! 20
(1.205)
!
:
(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
kB T
= n0 exp
pE cos
kB T
;
(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 number
density oriented in the direction of the electric …eld can be calculated from
R
n(cos ) cos d
n= R
:
(1.209)
exp pEkBcos
d
T
In practice, the potential energy is much smaller than the thermal energy pE
the exponential function can be approximated by
exp
pE cos
kB T
and we obtain
n'
'1+
n0 p
E:
3kB T
45
pE cos
;
kB T
kB T: Therefore,
(1.210)
(1.211)
Resultant dipole moment density is
P = np =
n 0 p2
E:
3kB T
(1.212)
This de…nes a permittivity in a medium consisting of molecules having a permanent dipole moment
p;
n 0 p2
:
(1.213)
" = "0 1 +
3"0 kB T
Note that the correction term
"=
n 0 p2
;
3kB T
(1.214)
is inversely proportional to the temperature. This property can be exploited in separating contributions 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 number 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 approximately " ' 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=
where
free
free ;
(1.215)
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 the
pancake in the direction perpendicular to the surface dS) on the boundary yields
Z
r DdV =
I
D dS =
Z
free dV:
(1.216)
In the limit of in…nitesimally thin pancake, this reduces to
(Dn1
Dn2 )dS =
free dndS
46
=
free dS;
(1.217)
Figure 1-16: Gauss’law applied to a thin pancake at a boundary between two dielectrics.
where free = free dn 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)
H
Figure 1-17: E 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 as
well.
47
The equation r E = 0 demands that the tangential component of electric …eld be continuous
across a boundary of dielectrics. This can be seen by integrating r E = 0 over a small rectangular
area on the boundary,
Z
I
r
E dS =
E dl = 0;
(1.221)
E2 ) dl = 0; E1t = E2t :
(1.222)
which yields
(E1
In fact, the continuity of the tangential component of the electric …eld holds for general time varying
case,
@B
r E=
;
(1.223)
@t
for the area integral
Z
B dS;
S
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 q
:
4 "0 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 = E0 ez : 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 may
be expanded in the spherical harmonics,
(r; ) =
0
+
X
r
a
Al
l
(r; ) =
0
+
X
Bl
l
a
r
l
Pl (cos );
l+1
Pl (cos );
r < a (interior);
r > a (exterior);
(1.225)
(1.226)
where
0 (r;
)=
E0 z =
E0 r cos ;
(1.227)
is the potential associated with the external electric …eld E0 = E0 ez . 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,
1@
E =
:
(1.228)
r@
The continuity of the normal (radial) component of the displacement vector requires
"
E0 cos +
X
l
!
l
Al Pl (cos ) =
a
"0
E0 cos +
X
l
!
l+1
Al
Pl (cos ) :
a
(1.229)
Since cos = P1 (cos ); only the l = 1 terms are non-vanishing and we readily …nd
A1 =
" "0
aE0 :
" + 2"0
Then the solutions for the potentials are
8
" "0
>
>
E0 r cos ;
r<a
0 (r; ) +
>
>
" + 2"0
<
(r; ) =
>
>
" "0
a3
>
>
: 0 (r; ) +
E0 2 cos ; r > a
" + 2"0 r
(1.230)
(1.231)
The electric …eld in the dielectric sphere is uniform and given by
Eiz =
3"0
E0 ;
" + 2"0
r < a:
(1.232)
This is smaller than the external …eld E0 since " > "0 : The interior displacement vector is
Diz = "Eiz =
3"
" 0 E0 ;
" + 2"0
49
(1.233)
which is larger than D0 = "0 E0 : The perturbed exterior potential,
" "0 a3
E0 cos ;
" + 2"0 r2
(1.234)
is of dipole form with an e¤ective dipole moment
pz = 4 " 0
" "0 3
a E0 ;
" + 2"0
(1.235)
located at the center of the sphere.
In the limit of "
"0 ; the potential reduces to
(r; ) =
8
0;
>
>
<
>
>
:
r<a
a3
0 (r; ) + 2 E0 cos ; r > a
r
(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 = "0 Ez + Pz ;
Pz =
3(" "0 )
" 0 E0 ;
" + 2"0
(1.237)
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 )
"0 E0 cos
" + 2"0
(r
The total dipole moment carried by the sphere is
pz =
=
4 a3
Pz
3
4 (" "0 )a3
" 0 E0 ;
" + 2"0
which agrees with that deduced from the exterior potential.
50
a):
(1.239)
1.10
Electric Force
The energy density associated with an electric …eld
1
"0 E 2 ;
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 ("0 EE)
"0 E rE:
(1.241)
1
In electrostatics, r E = 0; and thus the curvature E rE and gradient ErE = rE 2 of the
2
electric …eld are identical,
r(E E) = 2E
r
rE 2 = 2E rE:
E + 2E rE
(1.242)
Therefore,
f =r
"0 EE
1
"0 E 2 1 ;
2
(1.243)
1
"0 E 2
2
(1.244)
where 1 is the unit sensor. The tensor
Tij = "0 Ei Ej
ij ;
is called the Maxwell’s 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 =
@
Tij :
@xj
(1.245)
Let us consider a simple case: an electric …eld in the z-direction, E = Ez ez : The tensor Tij is
diagonal with the following components,
0
B
B
T=B
B
@
1
"0 Ez2
2
0
0
0
0
1
"0 Ez2
0
2
1
0
+ "0 Ez2
2
51
1
C
C
C:
C
A
(1.246)
The force in the directions perpendicular to the …eld are
fx =
@
@x
1
"0 Ez2 ;
2
(1.247)
fy =
@
@y
1
"0 Ez2 ;
2
(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
"0 Ez2 ;
2
(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
Z
I
F =
f dV =
T dS
V
S
I
1
=
"0 E(E n)
"0 E 2 n dS;
(1.250)
2
S
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
"0 Er2 ; (N/m2 );
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
is
Fz =
"E(E n)
I
"Ez En
1 2
"E n dS
2
1 2
"E nz dS;
2
(1.251)
where on the spherical surface of one of the hemispheres,
Ez = E0
3
" cos2 + "0 sin2
" + 2"0
52
;
(1.252)
Er = E0
E =
3"
cos ;
" + 2"0
E0
(1.253)
3"0
sin ;
" + 2"0
E 2 = Er2 + E 2
9
= E02
"2 cos2 + "20 sin2
(" + 2"0 )2
(1.254)
:
(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) shown
are relevant to the preceding Example as well.
Ezgap =
"
3"
Eiz =
E0 ;
"0
" + 2"0
where
Eiz =
3"0
E0 ;
" + 2"0
(1.256)
(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 a2
Z
0
=2
" 0 Ez Er
1
"0 E 2 cos
2
sin d
Z =2
9"0 E02
"2 cos2 + ""0 sin2
(" + 2"0 )2 0
9"0 E02 1 2
a2
(" + 2""0 "20 ):
(" + 2"0 )2 4
= 2 a2
=
53
1 2
(" cos2 + "0 sin2 ) cos sin d
2
(1.258)
This force is repelling. The contribution from the ‡at surface in the gap is
9"0 E02 "2
;
(" + 2"0 )2 2
(1.259)
9(" "0 )2
"0 E02 :
4(" + 2"0 )2
(1.260)
a2
Fz2 =
and the net force is
a2
Fz =
The minus sign indicates an attractive force between the hemispheres. In the limit "
force becomes
9 2
Fz =
a "0 E02 :
4
"0 ; the
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
residing on each electrode are related through the capacitance C;
Q = CV:
Q
(1.261)
An incremental energy required to increase the charge by dQ is
dU = V dQ =
1
QdQ = CV dV:
C
(1.262)
Therefore the amount of energy stored in a capacitor is
U=
1 Q2
1
= CV 2 :
2 C
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 "0 a:
(1.264)
If it carries a charge Q; the amount of potential energy associated with it is
U=
Q2
:
8 "0 a
(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
a
1
Q2
1
"0 Er2 4 r2 dr =
;
2
8 "0 a
54
(1.266)
where
Er =
Q
;
4 "0 r 2
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,
1 @C( )
F = V2
; if the voltage is held constant,
2
@
F =
1 2@
Q
2 @
1
C( )
; if the charge is held constant.
(1.268)
(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 =
a2
1
"0 2 V 2 =
2 d
The radial force is
Fa = " 0
1 Q2
:
2 "0 a2
d
a 2
V =
Q2 ;
d
"0 a3
(1.271)
(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
(r
r0 ) =
(r
x0 ) (y
0)
(
0
the volume integral is unity,
Z
0
(
(r r0 )
(cos
rr0
r0 ) =
y 0 ) (z
) (z
cos 0 ) (
z 0 );
z 0 );
0
);
r0 )dV = 1:
(r
1.2 The electron cloud in a hydrogen atom is described by the charge density distribution
e
exp
a3
(r) =
where a = 5:3
atom is
10
11
2r
a
;
m is the Bohr radius. Show that the potential energy of a hydrogen
U=
1 e2
:
4 "0 a
1.3 A planar quadrupole in the x y plane consists of four charges, +q and q alternatively
placed 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;
1
X
rY1;m ( ; )q1;m =
m= 1
56
3
r p;
4
and a resultant dipole potential is consistent with the direct expansion of the potential,
dipole
=
1 r p
:
4 "0 r 3
1.6 An octupole consists of eight charges (four +q and 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 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
I
X a l+1
0 0
0 0
0
(r) =
Ylm ( ; )
s ( ; )Ylm ( ; )d ; r > a;
r
lm
where
s(
; ) is the surface potental and d
1.8 The potential due to a long line charge
= sin 0 d 0 d 0 :
(C/m) is
( )=
where
0
2 "0
ln + constant,
is the distance from the line charge. Verify this from the basic formula,
(r) =
=
Z
(r0 )
1
dV 0
4 "0
jr r0 j
Z 1
1
p
dz 0 :
2 + z 02
4 "0 1
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
; (F/m).
d
ln
a
57
What is the inductance per unit length of the transmission line? (Hint: The product
CL
;
l l
is constant and equal to "0 0 : For a coaxial cable …lled with insulating material having
permittivity "; the capacitance is
2 "
C
=
;
l
ln(a=b)
and the product
CL
l l
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 dl 2
(x
2l l! dxl
1)l ; (Rodrigues’formula).
Using this repeatedly, verify the orthogonality of Legendre functions,
Z
1
1
Pl (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
! 2p
ne2
"
=
const:
;
0
! 20
me ! 20
" = const:
where n is the molecule density, me = 9:1 10 31 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 1
4 "0 r 3
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,
q
( ; z) =
4 "0
Z
1
J0 (ka)J0 (k )e
kjzj
dk:
0
Show that the solution above satis…es the Poisson’s equation,
1 @
@2
@2
+
+
@ 2
@r @z 2
( ; z) =
q
(
2 "0 a
a) (z):
1.20 A conducting sphere of radius a is placed in a uniform electric …eld E0 = E0 ez whose potential
is 0 = E0 z: The dipole moment induced on the sphere is
pz = 4 "0 a3 E0 ;
59
which produces a potential
0
=
1 pz
cos :
4 "0 r 2
(a) The exterior electric …eld is
r 0
pz
= E 0 ez +
(2 cos er + sin e ) ; r > a:
4 "0 r 3
E(r; ) =
r
0
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=
Recover this result from the direct integral,
I
1
U=
2
4 "0 a3 E02 :
0 dS;
over the sphere surface. Note that at a conductor surface,
= "0 Er (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 you
will …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
60
" "0 q
" + "0 jr + aj
; z>0
and that in the dielectric is
dielectric
=
2"
q
1
; z < 0;
4 " " + "0 jr aj
where a = aez :
1.22 The permittivity of an unmagnetized plasma is
" (!) = "0
1
! 2pe
!2
!
;
p
where ! pe = ne2 =me "0 is the plasma frequency. Show that a spherical plasma exhibits a
dipole (l = 1) ocsillation at a frequency
! pe
!= p :
3
What are the frequencies of higher multipole modes?
61