Question-no.docxChapter7Question no’s 2,3,4,5,6,8,10,13,14,.docx

Question-no.docx
Chapter7
Question no’s:
2,3,4,5,6,8,10,13,14,15,17,18,19,20,21,27,28,29,31,32,33,36
Chapter 8
Question no’s:
1,2,3,4,6,7,9,13,14,15,19,20,21,22,24,26,28,29,30
ch7.pdf
Chapter 7 Laplace’s Equation: The Potential Produced by
Surface Charge
7.13 Problems
7.1 Finding Charge From Potential
The potential in a spherical region r < R is '(x, y, z) = '0(z/R)
3. Find a volume charge density
Ω(r, µ) in the region r < R and a surface charge density æ(µ) on
the surface r = R which together
produce this potential. Express your answers in terms of
elementary trigonometric functions.
7.2 A Periodic Array of Charged Rings
Let the z-axis be the symmetry axis for an infinite number of
identical rings, each with charge
Q and radius R. There is one ring in each of the planes z = 0, z
= ±b, z = ±2b, etc. Exploit

the Fourier expansion in Example 1.6 to find the potential
everywhere in space. Check that your
solution makes sense in the limit that the cylindrical variable Ω
¿ R, b. Hint: If IÆ(y) and KÆ are
modified Bessel functions,
I
0
Æ(y)KÆ(y) ° IÆ(y)K0Æ(y) = 1/y.
7.3 Two Electrostatic Theorems
Use the orthogonality properties of the spherical harmonics to
prove the following for a function
'(r) which satisfies Laplace’s equation in and on an origin-
centered spherical surface S of radius
R:
(a)
R
S
dS '(r) = 4ºR2'(0)
(b)
Z
S
dSz'(r) =
4º
3
R
4 @'

@z
ØØØØ
r=0
7.4 Make a Field Inside a Sphere
Find the volume charge density Ω and surface charge density æ
which much be placed in and on a
sphere of radius R to produce a field inside the sphere of
E = °2V0
xy
R3
x̂ +
V0
R3
(y
2 ° x2)ŷ ° V0
R
ẑ.
There is no other charge anywhere. Express your answer in
terms of trigonometric functions of µ
and ¡.
7.5 Green’s Formula
Let n̂ be the normal to an equipotential surface at a point P . If
R1 and R2 are the principal
radii of curvature of the surface at P . A formula due to George
Green relates normal derivatives
(@/@n ¥ n̂ · r) of the potential '(r) (which satisfies Laplace’s

equation) at the equipotential surface
to the mean curvature of that equipotential surface ∑ = 1
2
(R°11 + R
°1
2 ):
@2'
@n2
+ 2∑
@'
@n
= 0.
Derive Green’s equation by direct manipulation of Laplace’s
equation.
7.6 The Channeltron
c∞2009 Andrew Zangwill 278
Surface Charge
The parallel plates of a channeltron are segmented into
conducting strips of width b so the po-
tential can be fixed on the strips at staggered values. We model
this using infinite-area plates, a
finite portion of which is shown below. Find the potential '(x, y)
between the plates and sketch

representative field lines and equipotentials. Note the
orientation of the x and y axes.
1 1
02 2
x
y
d
b
7.7 The Calculable Capacitor
The figure below shows a circle which has been divided into
two pairs of segments with equal
arc length by a horizontal bisector and a vertical line. The
positive x-axis bisects the segment
labelled “1” and the polar angle ¡ increases counterclockwise
from the x-axis as indicated . Now let
the segmented circle be the cross section of a segmented
conducting cylinder (with tiny insulating
regions to separate the segments).
1
2
4
3
r
x

O
1
2
(a) Let segment 1 have unit potential and ground the three
others. If the angle Æ subtends
segment 1 as viewed from the origin O, show that the charge
density induced on the inside
surface of segment 3 is
æ(¡) =
≤0
2ºR
∑
sin( 1
2
Æ + ¡)
1 ° cos( 1
2
Æ + ¡)
+
sin( 1
2
Æ ° ¡)
1 ° cos( 1
2
Æ ° ¡)

∏
.
(b) Enclose the segmented cylinder by a coaxial, grounded,
conducting cylindrical shell whose
radius is infinitesimally larger than R. This guarantees that that
no charge is induced on the
outside of segment 3. In that case, show that the cross-
capacitance per unit length between
segments 1 and 3 is
C13 = °
≤0
º
ln 2.
The non-trivial fact that C13 depends only on defined constants
(and not on R) is exploited
worldwide to “realize” the farad—the fundamental unit of
capacitance.
7.8 An Incomplete Cylinder
The figure below shows an infinitely long cylindrical shell from
which a finite angular range has
been removed. Let the shell be a conductor raised to a potential
corresponding to a charge per unit
length ∏. Find the fraction of charge which resides on the inner
surface of the shell in terms of ∏
and the angular parameter p. Hint: Calculate Qin ° Qout.

Surface Charge
2
p
7.9 Picht’s Equation
This problem addresses the focusing properties of cylindrically
symmetric potentials '(Ω, z) which
satisfy Laplace’s equation.
(a) Let V (z) = '(0, z). Use separation of variables to show that
E(Ω, z) º 1
2
V 00(z)Ω Ω̂ ° V 0(z)ẑ
for points near the symmetry axis where Ω ø
p
|V 0(z)/V 000(z)|. This is called the paraxial
regime in charged particle optics.
(b) Regard Ω(z, t) as the trajectory of a particle with charge q
and mass m and derive the trajectory
equation
Ω̈ = z̈ Ω
0
+ ż
2
Ω
00
=

q
2m
ΩV
00
(z).
(c) Use Newton’s second law and an approximate form of
conservation of energy (valid when vz
is large) to derive the trajectory equation
d2Ω
dz2
+
1
2
V 0
V
dΩ
dz
+
Ω
4
V 00
V
= 0.
(d) Show that a change of variables to R(z) = Ω(z)V 1/4(z)
transforms the equation in part (c) to

Picht’s equation,
d2R
dz2
= ° 3
16
R(z)
∑
V 0(z)
V (z)
∏2
.
(e) Integrate Picht’s equation and explain why it predicts
focusing for particles which enter the
potential parallel to the z-axis.
7.10 A Dielectric Wedge in Polar Coordinates
Two wedge-shaped dielectrics meet along the ray ¡ = 0. The
opposite edge of each wedge is held
at a fixed potential by a metal plate. The system is invariant to
translations perpendicular to the
diagram.
(a) Explain why the potential '(Ω, ¡) between the plates does not
depend on the polar coordinate
Ω.
(b) Find the potential everywhere between the plates.
2

!
1
"
2
"
0" #
2
V$#
1
!
1
V$#
Surface Charge
7.11 A Split Conical Conductor
An electron deflector takes the form of an infinite, segmented,
conducting cone whose apex is at
the origin and whose opening angle is 2Æ. The symmetry axis
inside the cone is the positive z-axis
and the two segments are held at the potentials ± V as shown
below.
x

VV
1. Use a separation of variables argument in spherical
coordinates to show that the potential
inside the cone is independent of the radial variable
1. Use the result of part (a) to show that Laplace’s equation can
be rewritten as
∫
2 @'
@∫2
+ ∫
@'
@∫
+
@2'
@¡2
= 0
where ∫ = tan 1
2
µ .
1. Separate variables and show that
'(µ, ¡) =
4V
º
1X
m=1,3,5,···

(°1)(m°1)/2
m
∑
tan µ/2
tan Æ/2
∏m
cos m¡
1. Exploit the expansion ln(1 ± z) = ±z ° 1
2
z2 ± 1
3
z3 ° 1
4
z4 + · · · to sum the series and show
that
' =
4V
º
tan
°1
Ω
2 tan 1
2
µ tan 1

2
Æ
tan2 1
2
µ ° tan2 1
2
Æ
cos ¡
æ
.
7.12 Practice with Bessel Functions
A grounded metal tube with radius R is coaxial with the z-axis.
The bottom of the tube at z = 0
is closed by a circular metal plate held at potential V . The top
of the tube is open and extends to
infinity. If J0(kmR) = 0, show that the potential inside the tube
is
'(Ω, z) =
2V
R
1X
m=1
exp(°kmz)
km
J0(kmΩ)
J1(kmR)

.
7.13 The Capacitance of an OÆ-Center Capacitor
A spherical conducting shell centered at the origin has radius
R1 and is maintained at potential V1.
A second spherical conducting shell maintained at potential V2
has radius R2 > R1 but is centered
at the point sẑ where s << R1.
Surface Charge
(a) To lowest order in s, show that the charge density induced
on the surface of the inner shell is
æ(µ) = ≤0
R1R2(V2 ° V1)
R2 ° R1
∑
1
R21
° 3s
R32 ° R31
cos µ
∏

.
(b) To lowest order in s, show that the force exerted on the
inner shell is
F =
Z
dS
æ2
2≤0
n̂ = ẑ2ºR
2
1
ºZ
0
dµ sin µ
æ2(µ)
2≤0
cos µ = ° Q
2
4º≤0
sẑ
R32 ° R31
.
(c) Integrate the force in (b) to find the capacitance of this

structure to second order in s.
7.14 The Plane-Cone Capacitor
A capacitor is formed by the infinite grounded, plane z = 0 and
an infinite, solid, conducting cone
with interior angle º/4 held at potential V . A tiny insulating
spot at the cone vertex (the origin
of coordinates) isolates the two conductors.
4
0
V
(a) Explain why '(r, µ, ¡) = '(µ) in the space between the
capacitor “plates”.
(b) Integrate Laplace’s equation explicitly to find the potential
between the plates.
7.15 The Near-Origin Potential of Four Point Charges
Four identical positive point charges sit at (a, a), (°a, a), (°a,°a),
and (a,°a) in the z = 0 plane.
Very near the origin, the electrostatic potential can be written in
the form
'(x, y, z) = A + Bx + Cy + Dz + Exy + F xz + Gyz + Hx
2
+ Iy
2
+ Jz

2
.
(a) Deduce the non-zero terms in this expansion and the
algebraic sign of their coe±cients. Do
not calculate the exact value of the non-zero coe±cients.
(b) Sketch electric field lines and equipotentials in the z = 0
plane everywhere inside the square
and a little bit outside the square. Do not miss any important
features of the patterns.
7.16 U-Shaped Electrodes
Two semi-infinite blocks of matter share a common interface as
shown below. The matter with
dielectric constant ∑2 is completely surrounded by a æ-shaped
electrode which is grounded. The
matter with dielectric constant ∑1 is completely surrounded by
a Ω-shaped electrode which is held
at potential V .
Surface Charge
1 2 d
V 0
y
x

a
(a) Determine '(x, y) everywhere between the two electrodes.
(b) Find the polarization charge on the interface when ∑1 is
slightly greater than ∑2 and also when
∑1 is slightly less than ∑2.
(c) Sketch electric field lines when ∑1 ¿ ∑2 and also when ∑1 ø
∑2.
7.17 The Potential Inside an Ohmic Duct
The z-axis runs down the center of an infinitely long heating
duct with a square cross section. For
a real metal duct (not a perfect conductor), the electrostatic
potential '(x, y) varies linearly along
the sidewalls of the duct. Suppose that the duct corners at (±a,
0) are held at potential +V and
the duct corners at (0, ±a) are held at potential °V. Find the
potential inside the duct beginning
with the trial solution
'(x, y) = A + Bx + Cy + Dx
2
+ Ey
2
+ F xy.
7.18 A Potential Patch By Separation of Variables
The square region defined by °a ∑ x ∑ a and °a ∑ y ∑ a in the z
= 0 plane is a conductor held

at potential ' = V . The rest of the z = 0 plane is a conductor
held at potential ' = 0. The plane
z = d is also a conductor held at zero potential.
V 2a
2a
d
(a) Find the potential for 0 ∑ z ∑ d in the form of a Fourier
integral.
(b) Find the total charge induced on the upper surface of the
lower (z = 0) plate. The answer is
very simple. Do not leave it in the form of an unevaluated
integral or infinite series.
(c) Sketch field lines of E(r) between the plates.
7.19 Poisson’s Integral Formula
The Poisson integral formula
'(r) =
(R2 ° r2)
4ºR
Z
|y
S
|=R
dyS

'̄(yS )
|r ° yS|3
|r| < R
gives the potential at any point r inside a sphere if we specify
the potential '̄(yS ) at every point
on the surface of the sphere. Derive this formula by summing
the general solution of Laplace’s
equation inside the sphere using the derivatives (with respect to
r and R) of the identity
Surface Charge
1
|r ° yS|
=
1X
`=0
r`
R`+1
P`(r̂ · ŷS ).
7.20 An Electrostatic Analog of the Helmholtz Coil
A spherical shell of radius R is divided into three conducting
segments by two very thin air gaps

located at latitudes µ0 and º °µ0. The center segment is
grounded. The upper and lower segments
are maintained at potentials V and °V , respectively. Find the
angle µ0 such that the electric field
inside the shell will be as nearly constant as possible near the
center of the sphere.
0
0
V
V
0
R
7.21 A Conducting Sphere at a Dielectric Boundary
A conducting sphere with radius R and charge Q sits at the
origin of coordinates. The space outside
the sphere above the z = 0 plane has dielectric constant ∑1. The
space outside the sphere below
the z = 0 plane has dielectric constant ∑2.
R
Q
1
2
(a) Find the potential everywhere outside the conductor.

(b) Find the distributions of free charge and polarization charge
wherever they may be.
7.22 Bumps and Pits on a Flat Conductor
A flat metal plate occupies the z = 0 plane. When raised to a
non-zero potential, the plate develops
a uniform surface charge density æ0 and a uniform field E0 =
(æ0/≤0)ẑ in the space z > 0.
(a) Place a hemispherical metal bump of radius R on the plate as
shown in part (a) of the figure
below. Ground the plate and bump combination and demand that
E(z ! 1) ! E0. Show
that E for this problem diÆers from E0 by the field of a suitably
placed point dipole. Calculate
the charge density induced on the conducting surface.
(b) Replace the hemispherical metal bump by a hemispherical
metal crater as shown in part (b) of
the figure below. Ground the plate and crater combination and
demand that E(z ! 1) !
E0. Why is it less straightforward to find the potential for this
problem as for the bump
problem? How would you set up to solve for '(r, µ) outside the
crater? Numerical results
show that E for the crater problem diÆers from E0 by the field
of a dipole placed at the same
point as in part (a). However, the dipole moment is reversed in
direction and has a magnitude
only 1/10 as large as the bump problem. Rationalize both of
these results qualitatively.

Surface Charge
R
R( )a ( )b
z
7.23 A Conducting Slot
The figure shows an infinitely long and deep slot formed by two
grounded conductor plates at x = 0
and x = a and a conductor plate at z = 0 held at a potential '0.
Find the potential inside the slot.
0x! x a!
0" "!
0" !
z
x
7.24 A Corrugated Conductor
A flat metal plate occupies the z = 0 plane. When raised to a
non-zero potential '0, the plate
develops a uniform surface charge density æ0 and a uniform
field E0 = (æ0/2≤0)ẑ in the space z > 0.
(a) Corrugate the plate slightly so z(x) = b sin kx with kb ø 1

describes the free surface. Demand
that E(z ! 1) ! E0 and show that the charge density induced on
the metal surface is
æ(x) º æ(0)[1 + kz(x)].
(b) Discuss the behavior of æ(x) at the peaks and valleys of the
surface in connection with the
results of Section 7.10.
7.25 Unisphere Potential
Let '0 be the value of the potential applied to the metallic
Unisphere in Section 6.8.1. Outline
a procedure (other than direct integration of the Coulomb
integral) which gives the potential at
every point in space. The procedure may be partly numerical.
7.26 Potential of a Cylindrical Capacitor
An infinitely long conducting tube (radius Ω1) is held at
potential '1. A second, concentric tube
(radius Ω2 > Ω1) is held at potential '2. Integrate Laplace’s
equation and find the capacitance per
unit length.
7.27 Axially Symmetric Potentials
Surface Charge
Let V (z) be the potential on the axis of an axially symmetry

electrostatic potential in vacuum.
Show that the potential at any point in space is
V (Ω, z) =
1
º
ºZ
0
d≥ V (z + iΩ cos ≥).
Hint: Show that the proposed solution satisfies Laplace’s
equation and exploit uniqueness.
7.28 A Segmented Cylinder
The figure below is a cross section of an infinite, conducting
cylindrical shell. Two infinitesimally
thin strips of insulating material divide the cylinder into two
segments. One segment is held at
unit potential. The other segment is held at zero potential. Find
the electrostatic potential inside
the cylinder. Hint: Z 2º
0
d¡ cos m¡ cos n¡ = º±mn (m 6= 0)
D
DR
y
x

1M
0M
7.29 A 2D Potential Problem in Cartesian Coordinates
Two flat conductor plates (infinite in the x and y directions)
occupy the planes z = ±d. The x > 0
portion of both plates is held at ' = +'0. The x < 0 portion of
both plates is held at ' = °'0.
Derive an expression for the potential between the plates using
a Fourier integral to represent the
x variation of '(x, z).
x
z
d
d!
0M! 0M
7.30 Target Field in a Dielectric Sphere
An origin-centered sphere with radius R and dielectric constant
∑1 is embedded in an infinite
medium with dielectric constant ∑2. The electric field inside
the sphere is
E1 = (V0/R
2
)(zx̂ + xẑ).
(a) Find the electric field outside the sphere, E2(x, y, z),
assuming that E2 ! 0 as r ! 1.

Surface Charge
(b) Calculate the density of charge (free or polarization) at the
interface between the two media.
7.31 The Two-Cylinder Electron Lens
Two semi-infinite, hollow cylinders of radius R are coaxial with
the z-axis. Apart from an insulating
ring of thickness d ! 0, the two cylinders abut one another at z =
0 and held at potentials VL and
VR. Find the potential everywhere inside both cylinders. You
will need the integrals
∏
Z 1
0
ds s J0(∏s) = J1(∏) and 2
Z 1
0
ds s J0(xns)J0(xms) = J
2
1 (xn)±nm.
The real numbers xm satisfy J0(xm) = 0.
R

d
L
V
R
V
7.32 Contact Potential
The x > 0 half of a conducting plane at z = 0 is held at zero
potential. The x < 0 half of the plane
is held at potential V . A tiny gap at x = 0 prevents electrical
contact between the two halves.
0!"V! "
#
$
x
z
(a) Use a change of scale argument to conclude that the z > 0
potential '(Ω, ¡) in plane polar
coordinates cannot depend on the radial variable Ω.
(a) Find the electrostatic potential in the z > 0 half-space.
(b) Make a semi-quantitative sketch of the electric field lines
and use words to describe the most
important features.

7.33 Circular Plate Capacitor
Consider a parallel plate capacitor with circular plates of radius
a separated by a distance 2L.
z
!2L
aV"
V#
Surface Charge
A paper published in 1983 proposed a solution for the potential
for this situation of the form
'(Ω, z) =
1Z
0
dk A(k)f (k, z)J0(kΩ),
where J0 is the zero-order Bessel function and
A(k) =
2V
1 ° e°2kL

sin(ka)
ºk
.
(a) Find the function f (k, z) so the proposed solution satisfies
the boundary conditions on the
surfaces of the plates. You may make use of the integral
1Z
0
dk
sin(ka)
k
J0(kΩ) =
8<
:
º/2 0 ∑ Ω ∑ a
sin°1(a/Ω) Ω ∏ a.
(b) Show that the proposed solution nevertheless fails to solve
the problem because the electric
field it predicts is not a continuous function of z when Ω > a.
7.34 A Slightly Dented Spherical Conductor
The surface of a slightly dented spherical conductor is given by
the equation r = a[1 + ≤PN (cos µ)]
where ≤ ø 1. Let the conductor be grounded and placed in a
constant electric field E0 parallel to
the polar axis, Show that the induced surface charge density is

æ(µ) = æ0 + ≤
Ω
3N ≤0E
2N + 1
[(N + 1)PN+1(cos µ) + (N ° 2)PN°1(cos µ)]
æ
where æ0 is the induced charge density for ≤ = 0. Along the
way, confirm and use the fact that the
normal to the surface is n̂ = r̂ ° ≤ @Pn
@µ
µ̂ + O(≤
2
). Hint: (2N + 1)PN (x)P1(x) = N PN°1(x) + (N +
1)PN+1(x).
7.35 A Conducting Duct
Solve the conducting duct problem treated in Section 7.5.1
using the method indicated in the
penultimate paragraph of that section.
7.36 The Force on an Inserted Conductor
A set of known constants Æn parameterizes the potential in a
volume r < a as
'ext(r, µ) =
1X

n=1
Æn
≥ r
R
¥n
Pn(cos µ).
Let ẑ point along µ = 0 and insert a solid conducting sphere of
radius R < a at the origin. Show
that the force exerted on the sphere when it is connected to
ground is in the z direction and
Fz = 4º≤0
1X
n=1
(n + 1)ÆnÆn+1.
Hint: The Legendre polynomials satisfy (n + 1)Pn+1(x) +
nPn°1(x) = (2n + 1)xPn(x).
ch8.pdf
Chapter 8 Poisson’s Equation: The Potential Produced by
Volume Charge
8.9 Problems

8.1 Debye’s Model for the Work Function
In 1910, Debye suggested that the work function W of a metal
could be computed as the work
performed against the electrostatic image force when an
electron is removed from the interior of
a finite piece of metal to a point infinitely far outside the metal.
Model the metal as a perfectly
conducting sphere with a macroscopic radius R and suppose that
the image force only becomes
operative at a microscopic distance d outside the surface of the
metal. Show that
W =
e2
8º≤0
∑
2
R + d
° R
(R + d)2
+
R
(R + d)2 ° R2
∏
.
Let x = d/R and take the limit R ! 1 to find the Debye model
prediction for the work function of
a semi-infinite sample. Today, it is well understood that the

image force plays an insignificant role
in the physics of the work function [See, e.g., A. Zangwill,
Physics at Surfaces (Cambridge, 1988)].
8.2 Images in Spheres I
A point charge q is placed at a distance 2R from the center of an
isolated, conducting sphere of
radius R. The force on q is observed to be zero at this position.
Now move the charge to a distance
3R from the center of the sphere. Show that the force on q at its
new position is repulsive with
magnitude
F =
1
4º≤0
173
5184
q2
R2
.
Hint: A spherical equipotential surface remains an equipotential
surface if an image point charge
is placed at it center.
8.3 Images in Spheres II
Positive charges Q and Q0 are placed on opposite sides of a
grounded sphere of radius R at distances
of 2R and 4R, respectively, from the sphere center. Show that

Q0 is repelled from the sphere if
Q0 < (25/144)Q.
8.4 The Charge Induced by Induced Charge
Maintain the plane z = 0 at potential V and introduce a
grounded conductor somewhere into the
space z > 0. Use the “magic rule” for the Dirichlet Green
function to find the charge density
æ(x, y) induced on the z = 0 plane by the charge æ0(r) induced
on the surface S0 of the grounded
conductor.
8.5 Images and Interaction Energy
Place a point charge q at a distance z from the center of a
grounded, origin-centered conducting
sphere of radius R < z.
(a) Integrate the image force to find the potential energy of
interaction between q and the grounded
conductor.
(b) Compare the answer in part (a) to the potential energy of
interaction between the real point
charge and the image point charge for this problem.
8.6 Using a Cube to Simulate a Point Charge
Volume Charge

(a) Use completeness relations to represent ±(x ° x0)±(y ° y0)
and then the method of direct
integration for the inhomogeneous diÆerential equation which
remains to find the interior
Dirichlet Green function for a cubical box with sidewalls at x =
±a, y = ±a, and z = ±a.
(b) Use the result of part (a) to find the charge density that must
be glued onto the surfaces of
an insulating box with sides walls at x = ±a, y = ±a, and z = ±a
so that the electric field
everywhere outside the box is identical to the field of a
(fictitious) point charge Q located at
the center of the box. It is su±cient to calculate æ(x, y) for the z
= a face. Give a numerical
value (accurate to 0.1%) for æ(0, 0).
(c) This problem was solved in the textbook by a diÆerent
method. Check that both methods
give the same (numerical) answer for æ(0, 0).
8.7 Green Function for a Sphere by Direct Integration
(a) Use the completeness relation
X
`m
Y`m(r̂ )Y
§
`m(r̂ 0) =
1
sin µ
±(µ ° µ0)±(¡ ° ¡0).

and the method of direct integration to show that
G(r, r
0
) =
1
4º≤0
1X
`=0
Ω
r`<
r`+1>
° r
`
<r
`
>
R2`+1
æ
P`(r̂ · r̂ 0)
is the interior Dirichlet Green function for a sphere of radius R.
(b) Show that G(r, r0) above is identical to the image solution
for this problem.
8.8 Practice with Complex Potentials

Show that
f (z) = ° ∏
2º≤0
ln tan
ºz
a
can be used as the complex potential for an array of equally
spaced, parallel, charged lines in the
y = 0 plane. Let n be an integer and let x = na and x = (n + 1
2
)a be the positions of the positive
and negatively charged lines, respectively. Show that the
asymptotic behavior (|y| ! 1) of the
physical potential is consistent with previous results.
8.9 The Image Force and Its Limits
(a) The text asserts that the attractive force F between an
origin-centered, grounded, conducting
sphere of radius R and a point charge located at a point s > R on
the positive z-axis varies
as 1/s3 when s ¿ R. Show this explicitly.
(b) Replace the sphere by a grounded conductor of any shape.
Use Green’s reciprocity principle to
show that the force in part (a) still varies as 1/s3 when s is large
compared to the characteristic
size of the conductor.
8.10 Images for a Hemispherical Boss

A grounded conductor consists of the x-y plane and a
hemispherical boss of radius R centered on
the z-axis. A point charge q sits at a point z0 > R on the
positive z-axis. Use the method of images
to show that the total charge induced on the flat portion of the
conductor is qind = °q cos 2µ sec µ.
Volume Charge
z
0
z
R
8.11 Free Space Green Function in 1D
Show that the free-space Green function in one dimension is
G0(x, x
0
) = ° 1
2≤0
|x ° x0|.
(a) using an argument based entirely on the physical meaning of
the Green function.

(b) by verifying that the stated function satisfies the correct
diÆerential equation and the correct
matching conditions.
8.12 Electrostatics of a Cosmic String
A cosmic string is a one-dimensional object with an
extraordinarily large linear mass density (
µ ª 1022 kg/m) which (in some theories) forms during the initial
cool-down of the Universe after
the Big Bang. In two-dimensional (2D) general relativity, such
an object distorts flat space-time
into an extremely shallow cone with the cosmic string at its
apex. Alternatively, one can regard
flat 2D space as shown below: undistorted but with a tiny
wedge-shaped region removed from the
physical domain. The usual angular range 0 ∑ ¡ < 2º is thus
reduced to 0 ∑ ¡ < 2º/p where
p°1 = 1 ° 4Gµ
±
c2 , G is Newton’s gravitational constant, and c is the speed of
light. The two
edges of the wedge are indistinguishable so any physical
quantity f (¡) satisfies f (0) = f (2º/p) .
unphysical region2 p
(a) Begin with no string. Show that the free-space Green
function in 2D is
G0(Ω, Ω
0
) = ° 1

2º≤0
ln |Ω ° Ω0|.
(b) Now add the string so p 6= 1. To find the modified free-
space Green function Gp0(Ω, Ω
0) , we
need a representation of the delta function which exhibits the
proper angular behavior. Show
that a suitable form is
±(¡ ° ¡0) = p
2º
1X
m=°1
e
imp(¡°¡0)
.
(c) Exploit the ansatz
G
p
0(Ω, ¡, Ω
0
, ¡
0
) =
p
2º

1X
m=°1
e
imp(¡°¡0)
Gm(Ω, Ω
0
)
to show that
G
p
0(Ω, ¡, Ω
0
, ¡
0
) =
1
2º
1X
m=1
cos[mp (¡ ° ¡0)] 1
m
µ
Ω<
Ω>
∂mp
° p

2º
ln Ω>
Volume Charge
(d) Perform the indicated sum and find a closed form expression
for Gp0. Check that G
1
0(Ω, Ω
0)
correctly reproduces your answer in part (a).
(e) Show that a cosmic string at the origin and a line charge q at
Ω are attracted with a force
F = (p ° 1) q
2Ω̂
4º≤0Ω0
8.13 Green Function Inequalities
The Dirichlet Green function for any volume V can always be
written in the form
GD(r, r
0
) =

1
4º≤0
1
|r ° r0|
+ §(r, r
0
) r, r
0 2 V.
The function §(r, r0) satisfies r2§(r, r0) = 0.
(a) Use the physical meaning of the Dirichlet Green function to
prove that
GD(r, r
0
) <
1
4º≤0
1
|r ° r0|
.
(b) Use Earnshaw’s theorem to prove that
GD(r, r
0
) > 0.
8.14 Image Energy and Real Energy

Suppose that a collection of image point charges q1, q2, . . . ,
qN exists to find the force on a point
charge q at position rq due the presence of a conductor held at
potential 'C . Call the electrostatic
potential energy between q and the conductor. Let UB be the
electrostatic energy of q in the
presence of the image charges. Find the general relation
between UA and UB and confirm that
UA =
1
2
UB when 'C = 0.
8.15 Free Space Green Functions by Eigenfunction Expansion
Find the free space Green function G
(d)
0 (x, x
0) in d = 1, 2, 3 space dimensions by the method of
eigenfunction expansion.
8.16 Weyl’s Formula
Use direct integration to derive the Weyl Formula for the free-
space Green function in three dimen-
sions,
G0(r, r
0
) =
1
2≤0

Z
d2k?
(2º)2
e
ik?·(r?°r
0
?)
1
k?
e
°k?|z°z
0|
.
8.17 A Dipole Near a Conducting Plane
A point dipole p sits a distance d from a flat, grounded, metal
plate. How much work is required
to rotate the dipole from perpendicular orientation (pointed at
the plane) to a parallel orientation?
8.18 A Line Charge Inside a Grounded Wedge
A line with uniform charge per unit length ∏ passes through the
point (a, 0) in the x-y plane. In
addition, two grounded, conducting planes (infinite in the z-
direction) extend from Ω = 0 to Ω = 1
at angles ¡ = ±º/4 with respect to the positive x-axis.

Volume Charge
(a) Use a multiple image technique to find the electrostatic
potential '(Ω, ¡) in the space between
the planes which satisfies the boundary conditions on the
conducting planes.
(b) Show that the charge per unit area induced on each plane is
æ(Ω) = ° ∏ap
2º
∑
1
Ω2 °
p
2Ωa + a2
° 1
Ω2 +
p
2Ωa + a2
∏
.
(c) Comment on the value of æ(0).
8.19 Inversion in a Cylinder

(a) Let ©(Ω, ¡) be a solution of Laplace’s equation in a
cylindrical region Ω < R. Show that the
function ™(Ω, ¡) = ©(R2/Ω, ¡) is a solution of Laplace’s
equation in the region Ω > R.
(b) Show that a suitable linear combination of the functions ©
and ™ in part (b) can be used
to solve Poisson’s equation for a line charge located anywhere
inside or outside a grounded
conducting cylindrical shell.
(c) Show that a linear combination of the functions © and ™
can be used to solve Poisson’s
equation for a line charge located inside or outside a solid
cylinder of radius R and dielectric
constant ∑1 when the cylinder is embedded in a space with
dielectric constant ∑2.
(d) Comment on the implications of this problem for the method
of images.
8.20 Green Function for a Dented Beer Can
An empty beer can is bounded by the surfaces z = 0, z = h, and
Ω = R. By slamming it against
his forehead, a frustrated football fan dents the can into the
shape shown below. Our interest is
the interior Dirichlet Green function of the dented can.
2
p
!

(a) Show that suitable choices for the allowed values of ∞ and Ø
in the sums makes the ansatz,
GD(r, r
0
) =
X
∞
X
Ø
sin(∞z) sin(∞z
0
) sin(Ø¡) sin(Ø¡
0
)g∞Ø (Ω, Ω
0
),
satisfy the boundary conditions at z = 0, z = h, ¡ = 0, and ¡ =
2º/p, and thus reduces
the defining equation for the Green function to a one-
dimensional diÆerential equation for
gÆØ (Ω, Ω
0).
(b) Complete the solution for GD(r, r
0).
8.21 The Charge Induced on a Conducting Tube

(a) Find the charge density æ(¡, z) induced on the outer surface
of a conducting tube of radius R
when a point charge q is placed at a perpendicular distance s >
R from the symmetry axis
of the tube.
(b) Confirm that the point charge induces a total charge °q on
the tube surface.
Volume Charge
(c) The angle-averaged induced charge/length is ∏(z) = R
R 2º
0
d¡ æ(¡, z) . Show that
∏(z) ª ° q ln(s/R)
z ln2(z/R)
as z ! +1. That is, the charge density falls oÆ extremely slowly
with distance along the
length of the tube.
8.22 Free Space Green Function in Polar Coordinates
The free-space Green function in two dimensions (potential of a
line charge) is G
(2)
0 (r, r

0) =
° ln |r ° r0|/2º≤0. Use the method of direct integration to reduce
the two-dimensional equation
≤0r2G(r, r0) = °±(r ° r0) to a one-dimensional equation and
establish the alternative representa-
tion
G
(2)
0 (r, r
0
) = ° 1
2º≤0
ln Ω> +
1
2º≤0
1X
m=1
1
m
Ωm<
Ωm>
cos m(¡ ° ¡0)
8.23 Poisson’s Formula for a Sphere
The Poisson integral formula
'(r) =

(R2 ° r2)
4ºR
Z
|r0
S
|=R
dS
'̄(r0S)
|r ° r0S|3
|r| < R
gives the potential at any point r inside a sphere if we specify
the potential '̄(rS ) at every point
on the surface of the sphere. Derive this formula by summing
the general solution of Laplace’s
equation inside the sphere using the derivatives (with respect to
r and R) of the identity
1
|r ° r0S|
=
1X
`=0
r`
R`+1
P`(r̂ · r̂ 0S).

8.24 Symmetry of the Dirichlet Green Function
Prove that GD(r, r
0) = GD(r
0, r).
8.25 Capacitance of a Sphere Above a Grounded Plane
A conducting sphere with radius R and potential V is centered
at a height h above the grounded
plane z = 0. We are interested in how the charge-to-potential
ratio C = Q/V for the sphere
varies as a function of h/R. Show that the boundary conditions
can be satisfied using an infinite
number of image charges (some inside the sphere and some in
the space z < 0. Use a finite number
of images for a numerical solution and plot C/C0 versus h/R
where C0 is the capacitance of the
isolated sphere. Analyze your data suitably to find the analytic
dependence when h/R º 1 and
when h/R ¿ 1.
8.26 Force Between a Line Charge and a Conducting Cylinder
Let b the perpendicular distance between an infinite line with
uniform charge per unit length ∏ and
the center of an infinite conducting cylinder with radius R =
b/2.

Volume Charge
R
!
b
"
(a) Show that the charge density induced on the surface of the
cylinder is
æ(¡) = ° ∏
2ºR
µ
3
5 ° 4 cos ¡
∂
.
(b) Find the force per unit length on the cylinder by an
appropriate integration over æ(¡).
(c) Confirm you answer to (b) by computing the force per unit
length on the cylinder by another
method.
Hint: Let the single image line inside the sphere fix the
potential of the cylinder.
8.27 Charge and Plane and Dielectric Bump
A dielectric hemisphere with permittivity ≤ and radius R sits on

the flat surface of a conducting
half-space. A point charge q is placed above the hemisphere (on
the symmetry axis) at a great
distance d ¿ R above the plane. Find the electric field
everywhere following the steps below.
R
d
q
!
conductor
(a) Consider an origin-centered dielectric sphere with volume V
polarized by a uniform external
field E0. Show that the charge density induced on the surface of
the sphere is
æ =
p0 · r̂
V
where p0 = 3≤0
≤ ° ≤0
≤ + 2≤0
V E0.
(b) Use the information in part (a), and the fact that the electric
field is nearly uniform near the
midpoint between two equal and opposite point charges, to
construct an image system that
permits you to compute the electric field E everywhere for the

geometry indicated in the
figure.
8.28 Rod and Plane
The diagram below shows a rod of length L and net charge Q
(distributed uniformly over its length)
oriented parallel to a grounded infinite conducting plane at the
distance d from the plane.
Volume Charge
x
y
z
d
Q
L
(a) Evaluate a double integral to find the exact force exerted on
the rod by the plane.
(b) Simplify your result in part (b) in the limit d ¿ l. Give a
physical argument for your result.
(c) Find the charge density æ(x, y) induced on the conducting
plane.

(d) Find the total charge induced on the plane without
integrating æ(x, y).
8.29 Point Dipole in a Grounded Shell
A point electric dipole with moment p sits at the center of a
grounded, conducting spherical shell
of radius R. Use the method of images to show that the electric
field inside the shell is the sum of
the electric field produced by p and a constant electric field, E
= p/4º≤0R
3.
Hint: Use the formula for the charge density induced on a
grounded plane by a point charge q
located a distance z0 above the plane: æ(Ω) = °qz0/[2º(Ω2 + z20
)3/2].
8.30 A Dielectric Slab Intervenes
An infinite slab with dielectric constant ∑ = ≤/≤0 lies between z
= a and z = b = a + c. A point
charge q sits at the origin of coordinates. Let Ø = (∑ ° 1)/(∑ +
1) and use solutions of Laplace’s
equation in cylindrical coordinates to show that
'(z > c) =
q(1 ° Ø2)
4º≤0
1Z
0
dk

J0(kΩ) exp(°kz)
1 ° Ø2 exp(°2kc)
=
q(1 ° Ø2)
4º≤0
1X
n=0
Ø2np
(z + 2nc)2 + Ω2
.
Note: The rightmost formula is sum over image potentials, but it
is much more tedious to use
images from the start.
8.31 A Point Charge Near a Dielectric Sphere
A point charge q sits at a distance c from the center of a
dielectric sphere with radius R < c and
dielectric constant ∑.
(a) Find '(r) everywhere using Legendre expansions with respect
to the sphere center for both
the potential of the point charge and the potential produced by
the dielectric.
(b) Compute the force exerted by the sphere on q.
(c) Show that an image solution is possible only in the ∑ ! 1
(conductor) limit.

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