Classical Electrodynamics

Answers To a Selection of Problems from
Classical Electrodynamics
John David Jackson
by Kasper van Wijk
Center for Wave Phenomena
Department of Geophysics
Colorado School of Mines
Golden, CO 80401
Samizdat
Press

Published by the Samizdat Press
Center for Wave Phenomena
Department of Geophysics
Colorado School of Mines
Golden, Colorado 80401
and
New England Research
76 Olcott Drive
White River Junction, Vermont 05001
c Samizdat Press, 1996
Release 2.0, January 1999
Samizdat Press publications are available via FTP or WWW from
samizdat.mines.edu
Permission is given to freely copy these documents.

Contents
1 Introduction to Electrostatics
7
1.1 Electric Fields for a Hollow Conductor . . . . . . . . . . . . . . . . . . . . . . . . .
7
1.4 Charged Spheres . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
9
1.5 Charge Density for a Hydrogen Atom . . . . . . . . . . . . . . . . . . . . . . . . .
10
1.7 Charged Cylindrical Conductors . . . . . . . . . . . . . . . . . . . . . . . . . . . .
11
1.13 Green’s Reciprocity Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
12
2 Boundary-Value Problems in Electrostatics: 1
15
2.2 The Method of Image Charges . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
15
2.7 An Exercise in Green’s Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . .
17
2.9 Two Halves of a Conducting Spherical Shell . . . . . . . . . . . . . . . . . . . . . .
19
2.10 A Conducting Plate with a Boss . . . . . . . . . . . . . . . . . . . . . . . . . . . .
22
2.11 Line Charges and the Method of Images . . . . . . . . . . . . . . . . . . . . . . . .
24
2.13 Two Cylinder Halves at Constant Potentials . . . . . . . . . . . . . . . . . . . . . .
26
2.23 A Hollow Cubical Conductor . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
28
8 Waveguides, Resonant Cavities and Optical Fibers
31
8.1 Time Averaged Forces Per Unit Area on a Conductor . . . . . . . . . . . . . . . .
31
8.2 TEM Waves in a Medium of Two Concentric Cylinders . . . . . . . . . . . . . . .
33
8.3 TEM Waves Between Metal Strips . . . . . . . . . . . . . . . . . . . . . . . . . . .
36
3

4
CONTENTS
8.4 TE and TM Waves along a Brass Cylinder . . . . . . . . . . . . . . . . . . . . . . .
38
8.4.1
a. Cutoﬀ Frequencies . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
38
10 Scattering and Diﬀraction
41
10.3 Scattering Due to a Solid Uniform Conducting Sphere . . . . . . . . . . . . . . . .
41
10.14Diﬀraction from a Rectangular Opening . . . . . . . . . . . . . . . . . . . . . . . .
42
11 Special Theory of Relativity
47
11.3 The Parallel-velocity Addition Law . . . . . . . . . . . . . . . . . . . . . . . . . . .
47
11.5 The Lorentz Transformation Law for Acceleration . . . . . . . . . . . . . . . . . .
48
11.6 The Rocket Ship . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
49
12 Practice Problems
53
12.1 Angle between Two Coplanar Dipoles . . . . . . . . . . . . . . . . . . . . . . . . .
53
12.2 The Potential in Multipole Moments . . . . . . . . . . . . . . . . . . . . . . . . . .
54
12.3 Potential by Taylor Expansion . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
55
A Mathematical Tools
57
A.1 Partial integration . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
57
A.2 Vector analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
57
A.3 Expansions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
58
A.4 Euler Formula . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
59
A.5 Trigonometry . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
59

CONTENTS
5
Introduction
This is a collection of my answers to problems from a graduate course in electrodynamics. These
problems are mainly from the book by Jackson [4], but appended are some practice problems. My
answers are by no means guaranteed to be perfect, but I hope they will provide the reader with a
guideline to understand the problems.
Throughout these notes I will refer to equations and pages of Jackson and Duﬃn [2]. The latter is
a textbook in electricity and magnetism that I used as an undergraduate student. References to
equations starting with a “D” are from the book by Duﬃn. Accordingly, equations starting with
the letter “J” refer to Jackson.
In general, primed variables denote vectors or components of vectors related to the distance between
source and origin. Unprimed coordinates refer to the location of the point of interest.
The text will be a work in progress. As time progresses, I will add more chapters.

6
CONTENTS

Chapter 1
Introduction to Electrostatics
1.1
Electric Fields for a Hollow Conductor
a. The Location of Free Charges in the Conductor
Gauss’ law states that
ρ = ·E,
(1.1)
0
where ρ is the volume charge density and 0 is the permittivity of free space. We know that
conductors allow charges free to move within. So, when placed in an external static electric ﬁeld
charges move to the surface of the conductor, canceling the external ﬁeld inside the conductor.
Therefore, a conductor carrying only static charge can have no electric ﬁeld within its material,
which means the volume charge density is zero and excess charges lie on the surface of a conductor.
b. The Electric Field inside a Hollow Conductor
When the free charge lies outside the cavity circumferenced by conducting material (see ﬁgure
1.1b), Gauss’ law simpliﬁes to Laplace’s equation in the cavity. The conducting material forms a
volume of equipotential, because the electric ﬁeld in the conductor is zero and
E = − Φ
(1.2)
Since the potential is a continuous function across a charged boundary, the potential on the inner
surface of the conductor has to be constant. This is now a problem satisfying Laplace’s equation
with Dirichlet boundary conditions. In section 1.9 of Jackson, it is shown that the solution for this
problem is unique. The constant value of the potential on the outer surface of the cavity satisﬁes
Laplace’s equation and is therefore the solution. In other words, the hollow conductor acts like a
electric ﬁeld shield for the cavity.
7

8
CHAPTER 1. INTRODUCTION TO ELECTROSTATICS
a
b
q
q
q
Figure 1.1: a: point charge in the cavity of a hollow conductor. b: point charge outside the cavity
of a hollow conductor.
With a point charge q inside the cavity (see ﬁgure 1.1a), we use the following representation of
Gauss’ law:
q
E · dS =
(1.3)
0
Therefore, the electric ﬁeld inside the hollow conductor is non-zero. Note: the electric ﬁeld outside
the conductor due to a point source inside is inﬂuenced by the shape of the conductor, as you can
see in part c.
c. The Direction of the Electric Field outside a Conductor
An electrostatic ﬁeld is conservative. Therefore, the circulation of E around any closed path is
zero
E · dl = 0
(1.4)
This is called the circuital law for E (E4.14 or J1.21). I have drawn a closed path in four legs
2
1
4
3
Figure 1.2: Electric ﬁeld near the surface of a charged spherical conductor. A closed path crossing
the surface of the conductor is divided in four sections.
through the surface of a rectangular conductor (ﬁgure 1.2). Sections 1 and 4 can be chosen
negligible small. Also, we have seen earlier that the ﬁeld in the conductor (section 3) is zero. For

1.4. CHARGED SPHERES
9
the total integral around the closed path to be zero, the tangential component (section 2) has to
be zero. Therefore, the electric ﬁeld is described by
σ
E =
ˆ
r
(1.5)
0
where σ is the surface charge density, since – as shown earlier – free charge in a conductor is located
on the surface.
1.4
Charged Spheres
Here we have a conducting, a homogeneously charged and an in-homogeneously charged sphere.
Their total charge is Q. Finding the electric ﬁeld for each case in- and outside the sphere is an
exercise in using Gauss’ law
q
E · dS =
(1.6)
0
For all cases:
• Problem 1.1c showed that the electric ﬁeld is directed radially outward from the center of
the spheres.
• For r > a, E behaves as if caused by a point charge of magnitude of the total charge Q of
the sphere, at the origin.
Q
E =
ˆ
r
4π 0r2
As we have seen earlier, for a solid spherical conductor the electric ﬁeld inside is zero (see ﬁgure
1.3). For a sphere with a homogeneous charge distribution the electric ﬁeld at points inside the
sphere increases with r. As the surface S increases, the amount of charge surrounded increases
(see equation 1.6):
Qr
E =
ˆ
r
(1.7)
4π 0a3
For points inside a sphere with an inhomogeneous charge distribution, we use Gauss’ law (once
again)
2π
π
r
E4πr2 = 1/ 0
ρ(r )r 2sinθ dr dφ dθ
(1.8)
0
0
0
Implementing the volume charge distribution
ρ(r ) = ρ0r n,
the integration over r for n > −3 is straightforward:
ρ
E =
0rn+1
ˆ
r,
(1.9)
0(n + 3)

10
CHAPTER 1. INTRODUCTION TO ELECTROSTATICS
homogeneously charged
inhomogeneously charged: n=2
inhomogeneously charged: n=−2
conductor
electric field strength
0
a
distance from the origin
Figure 1.3: Electric ﬁeld for diﬀerently charged spheres of radius a. The electric ﬁeld outside the
spheres is the same for all, since the total charge is Q in all cases.
where
a
4πρ
Q =
4πρ
0an+3
0rn+2dr =
0
n + 3
⇔
Q(n + 3)
ρ0 =
(1.10)
4πan+3
It can easily be veriﬁed that for n = 0, we have the case of the homogeneously charged sphere
(equation 1.7). The electric ﬁeld as a function of distance are plotted in ﬁgure 1.3 for the conductor,
the homogeneously charged sphere and in-homogeneously charged spheres with n = −2,2.
1.5
Charge Density for a Hydrogen Atom
The potential of a neutral hydrogen atom is
q
e−αr
αr
Φ(r) =
1 +
(1.11)
4π 0
r
2
where α equals 2 divided by the Bohr radius. If we calculate the Laplacian, we obtain the volume
charge density ρ, through Poisson’s equation
ρ = 2Φ
0

1.7. CHARGED CYLINDRICAL CONDUCTORS
11
Using the Laplacian for spherical coordinates (see back-cover of Jackson), the result for r > 0 is
α3q
ρ(r) = −
e−αr
(1.12)
8πr2
For the case of r → 0
q
lim Φ(r) = lim
(1.13)
r→0
r→0 4π 0r
From section 1.7 in Jackson we have (J1.31):
2(1/r) = −4πδ(r)
(1.14)
Combining (1.13), (1.14) and Poisson’s equation, we get for r → 0
ρ(r) = qδ(r)
(1.15)
We can multiply the right side of equation (1.15) by e−αr without consequences. This allows for
a more elegant way of writing the discrete and the continuous parts together
α3
ρ(r) =
δ(r) −
qe−αr
(1.16)
8πr2
The discrete part represents the stationary proton with charge q. Around the proton orbits an
electron with charge −q. The continuous part of the charge density function is more a statistical
distribution of the location of the electron.
1.7
Charged Cylindrical Conductors
Two very long cylindrical conductors, separated by a distance d, form a capacitor. Cylinder 1 has
surface charge density λ and radius a1, and number 2 has surface charge density −λ and radius a2
(see ﬁgure 1.4). The electric ﬁeld for each of the cylinders is radially directed outward
a
-
λ
2
λ
a1
0
P
d
Figure 1.4: Top view of two cylindrical conductors. The point P is located in the plane connecting
the axes of the cylinders.
λ
E =
ˆ
r,
(1.17)
2π 0|r|

12
CHAPTER 1. INTRODUCTION TO ELECTROSTATICS
where ˆ
r is the radially directed outward unit vector. Taking a point P on the plane connecting
the axes of the cylinders, the electric ﬁeld is constructed by superposition:
λ
1
1
E =
+
(1.18)
2π 0
r
d − r
The potential diﬀerence between the two cylinders is
λ
a1
1
1
|Va
+
dr
2 − Va1| =
2π 0 d−a r d
2
− r
λ
=
[lnr + ln(d
2π
− r)]a1d
0
−a2
λ
=
[lna
2π
1 + ln(d − a1) + ln(d − a2) − lna2]
(1.19)
0
If we average the radii of the cylinders to a1 = a2 = a and assume d
a, then the potential
diﬀerence is
λ
d
|Va
ln
(1.20)
2 − Va1| ≈ π 0
a
The capacitance per unit length of the system of cylinders is given by
λ
π
C =
0
(1.21)
|Va
ln(d/a)
1 − Va2| ≈
From here, we can obtain the diameter δ of wire necessary to have a certain capacitance C at a
distance d:
δ = 2a ≈ 2d · e−π 0C ,
(1.22)
where the permittivity in free space 0 is 8.854 · 10−12F/m. If C = 1.2 · 10−11F/m and
• d = 0.5 cm, the diameter of the wire is 0.1 cm.
• d = 1.5 cm, the diameter of the wire is 0.3 cm.
• d = 5 cm, the diameter of the wire is 1 cm.
1.13
Green’s Reciprocity Theorem
Two inﬁnite grounded parallel conducting plates are separated by a distance d. What is the induced
charge on the plates if there is a point charge q in between the plates?
Split the problem up in two cases with the same geometry. The ﬁrst is the situation as sketched by
Jackson; two inﬁnitely large grounded conducting plates, one at x = 0 and one at x = d (see the
right side of ﬁgure 1.5). In between the plates there is a point charge q at x = x0. We will apply
Green’s reciprocity theorem using a “mirror” set-up. In this geometry there is no point charge but
the plates have a ﬁxed potential ψ1 and ψ2, respectively (see the left side of ﬁgure 1.5).

1.13. GREEN’S RECIPROCITY THEOREM
13
S3
S3
Plate 1
Plate 2
Plate 1
Plate 2
S1
S2
S1
S2
q
ψ
φ
φ
ψ
2
1
2
1
S4
S4
|
|
|
|
|
0
x
0
d
0
d
Figure 1.5: Geometry of two conducting plates and a point-charge. S is the surface bounding the
volume between the plates. The right picture is the situation of the imposed problem with the point
charge between two grounded plates. The left side is a problem with the same plate geometry, but
we know the potential φ on the plates.
Green’s theorem (J1.35) states
∂ψ
∂φ
φ 2ψ − ψ 2φ dV =
φ
dS
(1.23)
V
S
∂n − ψ ∂n
The volume V is the space between the plates bounded by the surface S. S1 and S2 bound the
plates and S3 and S4 run from plate 1 to plate 2 at + and - ∞, respectively. The normal derivative
∂
at the surface S is directed outward from inside the volume V.
∂n
When the plates are grounded, the potential in the plates is zero. The potential is continuous
across the boundary, so on S1 and S2 φ = 0. Note that if the potential was not continuous the
electric ﬁeld (E = − φ) would go to inﬁnity. At inﬁnite distance from the point source, the
potential is also zero:
∂ψ
φ
dS = 0
(1.24)
S
∂n
The remaining part of the surface integral can be modiﬁed according to Jackson, page 36:
∂φ = φ
∂n
· n = −E · n
(1.25)
The electric ﬁeld across a boundary with surface charge density σ is (Jackson, equation 1.22):
n · (Econductor − Evoid) = σ/ 0
(1.26)
However, inside the conductor E = 0, therefore
n · Evoid = −σ/ 0,
(1.27)

14
CHAPTER 1. INTRODUCTION TO ELECTROSTATICS
for each of the plates. The total surface integral in equation 1.23 is then
∂φ
σ
σ
ψ
dS =
ψ
1
2
1
dS +
ψ2
dS
(1.28)
S
∂n
S1
0
S2
0
In case of the plates of ﬁxed potential ψ, the legs S3 and S4 have opposite potential and thus
cancel. Using
σdS = Q,
(1.29)
S
the surface integral of Greens theorem is
∂φ
ψ
dS = 1/ 0(ψ1QS1 + ψ2QS2)
(1.30)
S
∂n
In the volume integral in equation (1.23), the mirror case of the charged boundaries includes no
free charges:
2ψ = −(total charge)/ 0 = 0
(1.31)
Applying Gauss’ law to the case with the point charge gives us
2φ = −(total charge)/ 0 = −q/ 0δ(x − x0),
(1.32)
where x0 is the x-coordinate of the point source location. The volume integral will be
−q/ 0δ(x − x0)ψ(x)dV = −q/ 0ψ(x0)
(1.33)
V
For two plates with ﬁxed potentials, the potential in between is a linear function
ψ
ψ(x
2 − ψ1
0) = ψ1 + (
)(1
d
− x0)d = x0ψ1 + ψ2(1 − x0)
(1.34)
Green’s theorem is now reduced to equation (1.30) and equation (1.34) in (1.33):
−q(x0ψ1 + (1 − x0)ψ2) = QS1ψ1 + QS2ψ2
(1.35)
Since this equality must hold for all potentials, the charges on the plates must be
QS1 = −qx0 and QS2 = −q(1 − x0)
(1.36)

Chapter 2
Boundary-Value Problems in
Electrostatics: 1
2.2
The Method of Image Charges
a. The Potential Inside the Sphere
This problem is similar to the example shown on pages 58, 59 and 60 of Jackson. The electric ﬁeld
due to a point charge q inside a grounded conducting spherical shell can also be created by the
point charge and an image charge q only. For reasons of symmetry it is evident that q is located
on the line connecting the origin and q. The goal is to ﬁnd the location and the magnitude of the
image charge. The electric ﬁeld can then be described by superposition of point charges:
q
q
Φ(x) =
+
(2.1)
4π 0|x − y| 4π 0|x − y |
In ﬁgure 2.1 you can see that x is the vector connecting origin and observation point. y connects
the origin and the unit charge q. Finally, y is the connection between the origin and the image
charge q . Next, we write the vectors in terms of a scalar times their unit vector and factor the
scalars y and x out of the denominators:
q/4π
q /4π
Φ(x) =
0
+
0
(2.2)
x|n − yn
y
n
x
|
|n − xy |
The potential for x = a is zero, for all possible combinations of n ·n . The magnitude of the image
charge is
a
q = − q
(2.3)
y
at distance
a2
y =
(2.4)
y
15

16
CHAPTER 2. BOUNDARY-VALUE PROBLEMS IN ELECTROSTATICS: 1
a
P
x
θ
0
y
q
y’
q’
Figure 2.1: A point charge q in a grounded spherical conductor. q is the image charge.
This is the same result as for the image charge inside the sphere and the point charge outside (like
in the Jackson example). After implementing the amount of charge (2.3) and the location of the
image (2.4) in (2.1), potential in polar coordinates is

a

q
1
y
Φ(r, θ) = 4π 
−

0  y2+r2−2yrcosθ
2
a2
+ r2
rcosθ
y
− 2 a2y
, (2.5)
where θ is the angle between the line connecting the origin and the charges and the line connecting
the origin and the point P (see ﬁgure 2.1). r is the length of the vector connecting the origin and
observation point P .
b. The Induced Surface Charge Density
The surface charge density on the sphere is
∂Φ
σ = 0
(2.6)
∂r
r=a
Diﬀerentiating equation (2.5) is left to the reader, but the result is
2
q
a
1 − ay
σ = −
(2.7)
4πa2
y
2
3/2
1 + a
cosθ
y
− 2ay

2.7. AN EXERCISE IN GREEN’S THEOREM
17
c. The Force on the Point Charge q
The force on the point charge q by the ﬁeld of the induced charges on the conductor is equal to
the force on q due to the ﬁeld of the image charge:
F = qE
(2.8)
The electric ﬁeld at y due to the image charge at y is directed towards the origin and of magnitude
q
|E | =
(2.9)
4π 0(y − y)2
We already computed the values for y and q in equation (2.4) and (2.3), respectively. The force
is also directed towards the origin of magnitude
1 q2
a
a 2 −2
|F| =
1
(2.10)
4π
−
0 a2
y
y
d. What If the Conductor Is Charged?
Keeping the sphere at a ﬁxed non-zero potential requires net charge on the conducting shell. This
can be imaged as an extra image charge at the center of the spherical shell. If we now compute
the force on the conductor by means of the images, the result will diﬀer from section c.
2.7
An Exercise in Green’s Theorem
a. The Green Function
The Green function for a half-space (z > 0) with Dirichlet boundary conditions can be found by the
method of images. The potential ﬁeld of a point source of unit magnitude at z from an inﬁnitely
large grounded plate in the x-y plane can be replaced by an image geometry with the unit charge
q and an additional (image) charge at z = −z of magnitude q = −q. The situation is sketched in
ﬁgure 2.2. The potential due to the two charges is the Green function GD:
1
1
(2.11)
(x − x )2 + (y − y )2 + (z − z )2 − (x − x )2 + (y − y )2 + (z + z )2
b. The Potential
The Green function as deﬁned in equation (2.11) can serve as the “mirror set-up” required in
Green’s theorem:
∂ψ
∂φ
φ 2ψ − ψ 2φ dV =
φ
dS
(2.12)
V
S
∂n − ψ ∂n

18
CHAPTER 2. BOUNDARY-VALUE PROBLEMS IN ELECTROSTATICS: 1
z
P
r
θ
q
V
y
z
a
φ
x
’
z
q’
Figure 2.2: A very large grounded surface in which a circular shape is cut out and replaced by a
conducting material of potential V .
with GD = ψ and Φ = φ. There are no charges the volume z > 0, so Laplace equation holds
throughout the half-space V:
2Φ = 0
(2.13)
Chapter one in Jackson (J1.39) showed
2GD = −4πδ(ρ − ρ )
(2.14)
leaving Φ(ρ ) after performing the volume integration. The Green function on the surface S (GD)
is constructed with the assumption that part of the surface (the base, if you will) is grounded, and
the other parts stretch to inﬁnity. Therefore
GDdS = 0
(2.15)
S
The potential Φ = V in the circular area with radius a, but everywhere else Φ = 0. Also:
∂G
G
D
D · ˆndS = −
dS,
(2.16)
S
S
∂z
since the normal ˆ
n is in the negative z-direction −ˆk. Thus we are left with the following remaining
terms in Green’s theorem (in cylindrical coordinates):
a
2π ∂G
Φ(r ) =
D
0V
ρdρdφ
(2.17)
0
0
∂z
From here on we will exchange the primed and unprimed coordinates. This is OK, since the
reciprocity theorem applies. Some algebra left to the reader leads to
V z
a
2π
ρ dρ dφ
Φ(r) =
(2.18)
2π 0
0
(ρ2 + ρ 2 + z2 − 2ρρ cos(φ − φ ))3/2

2.9. TWO HALVES OF A CONDUCTING SPHERICAL SHELL
19
c. The Potential on the z-axis
For ρ = 0, general solution (2.18) simpliﬁes to
V z
a
2π
ρ dρ dφ
Φ(ρ, φ)
=
2π 0
0
(ρ 2 + z2)3/2
a
1
= −V z
ρ 2 + z2 0
z
= V
1 − √
(2.19)
a2 + z2
d. An Approximation
Slightly rewriting equation (2.18):
V z
a
2π
ρ dρ dφ
Φ(r) =
(2.20)
2π (ρ2 + z2)3/2
3/2
0
0
1 + ρ 2−2ρρ cos(φ−φ )
ρ2+z2
The denominator in the integral can be approximated by a binomial expansion. The ﬁrst three
terms of the approximation give
V a2z
3a2
5a4
15a2ρ2
Φ(r) ≈
1 −
+
+
(2.21)
2 (ρ2 + z2)3/2
4 (ρ2 + z2)
8 (ρ2 + z2)2
8 (ρ2 + z2)2
Along the axis (ρ = 0) the expression simpliﬁes to
V a2
3a2
5a4
Φ(φ, z) ≈
1
+
(2.22)
2z2
− 4z2 8z4
This is the same result when we expand expression (2.19):
z
Φ(ρ, φ)
= V
1 − √a2 + z2
a2 −1/2
= V
1 − 1 + z2
V a2
3a2
5a4
≈
1
+
(2.23)
2z2
− 4z2 8z4
2.9
Two Halves of a Conducting Spherical Shell
A conducting spherical shell consists of two halves. The cut plane is perpendicular to the homoge-
neous ﬁeld (see ﬁgure 2.3). The goal is to investigate the force between the two halves introduced

20
CHAPTER 2. BOUNDARY-VALUE PROBLEMS IN ELECTROSTATICS: 1
r^
E
E
0
0
z
a
Figure 2.3: A spherical conducting shell in a homogeneous electric ﬁeld directed in the z-direction.
by the induced charges.
The electric ﬁeld due to the induced charges on the shell is (see [2], p. 51):
σ
Eind =
ˆ
r
(2.24)
2 0
You can see this as the resulting ﬁeld in a capacitor with one of the plates at inﬁnity. The electric
ﬁeld inside the conducting shell is zero. Therefore the external ﬁeld has to be of the same magnitude
(see ﬁgure 2.4).
The force of the external ﬁeld on an elementary surface dS of the conductor is:
σ2dS
dF = Eextdq =
,
(2.25)
2 0
directed radially outward from the sphere’s center. From the symmetry we can see that all forces
cancel, except the component in the direction of the external ﬁeld.1
a. An Uncharged Shell
The derivation of the induced charge density on a conducting spherical shell in a homogeneous
electrical ﬁeld E0 is given ([4], p. 64). The homogeneous ﬁeld is portrayed by point charges of
opposite magnitude at + and − inﬁnity. Next, the location and magnitude of the image charges
are computed. The result is
σ(θ) = 3 0E0cosθ
(2.26)
1 The external ﬁeld is a superposition of the homogeneous electric ﬁeld plus the electric ﬁeld due to the induced
charges excluding dq! This external ﬁeld is perpendicular to the surface of the conductor with magnitude σ/(2 0).
This is not addressed in Jackson.

2.9. TWO HALVES OF A CONDUCTING SPHERICAL SHELL
21
E
E
ext
ext
Eind
Eind
cavity
Figure 2.4: Zooming in on that small part of the conductor with induced charges, where the
external ﬁeld is at normal incidence.
When we plug this result into equation (2.25), we get for the horizontal component of the force
(dFz) on an elementary surface:
9
dFz = dFcosθ = 2 0E20cos3θdS.
(2.27)
Now we can integrate to get the total force on the sphere halves. From symmetry we can also see
that the force on the left half is opposite of that on the right half (see ﬁgure 2.3). So we integrate
over the right half and multiply by two to get the total net force:
2π
π/2 9
Fz = 2
0
0
2 0E20cos3θa2sinθdθdφ ˆ
z
π/2
= 9πa2 0E20
cos3θsinθdθ ˆ
z
0
π/2
= 9πa2 0E20 −1/4cos4θ
ˆ
z
0
9
=
πa2
4
0E2
0 ˆ
z
(2.28)
b. A Shell with Total Charge Q
When the shell has a total charge Q it changes the charge density of equation (2.26) to
Q
σ(θ) = 3 0E0cosθ +
(2.29)
4πa2

22
CHAPTER 2. BOUNDARY-VALUE PROBLEMS IN ELECTROSTATICS: 1
When we plug this expression into equation (2.25) and compute again the net (horizontal) com-
ponent of the force, we ﬁnd that
9
Q2
E
F
0Q
z =
πa2
+
ˆ
z
(2.30)
4
0E2
0 + 32π 0a2
2
The total force is bigger then for the uncharged case. This makes sense when we look at equation
(2.25); when the shell is charged there is more charge per unit volume to, hence the force is bigger.
2.10
A Conducting Plate with a Boss
a. σ On the Boss
By inspection it can be seen that the system of images as proposed in ﬁgure (2.5) ﬁts the geometry
and the boundary conditions of our problem. We can write the potential as a function of these
four point charges. This is done in Jackson (p. 63). It has to be noted that R has to be chosen at
inﬁnity to apply to the homogeneous character of the ﬁeld. The potential can then be described
by expanding “the radicals after factoring out the R2.”
Q
2
2a3
Φ(r, θ)
=
rcosθ +
cosθ + . . .
4π
−
0
R2
R2r2
a3
= −E0 r −
cosθ
(2.31)
r2
The surface charge density on the boss (r = a) is
∂Φ
σ = − 0
= 3
∂r
0E0cosθ
(2.32)
r=a
b. The Total Charge on the Boss
The total charge Q on the boss is merely an integration over half a sphere with radius a:
2π
π/2
Q = 3 0E0
cosθa2sinθdθdφ
0
0
π/2
= 3 0E02πa2 1 sin2θ
2
0
= 3 0E0πa2
(2.33)

2.10. A CONDUCTING PLATE WITH A BOSS
23
E0
P
P
r
r
-q
q’
-q’
q
0
θ
0
θ
z
z
R
a
a
E
V
0
0
D
Figure 2.5: On the left is the geometry of the problem: two conducting plates separated by a distance
D. One of the plates has a hemispheric boss of radius a. The electric ﬁeld between the two plates
is E0. On the right is the set of charges that image the ﬁeld due to the conducting plates. In part
a and b, R → ∞ to image a homogeneous ﬁeld. In c, R = d.
c. The Charge On the Boss Due To a Point Charge
Now we do not have a homogeneous ﬁeld to image, but the result of a point charge on a grounded
conducting plate with the boss. Again we use the method of images to replace the system with
the plate by one entirely consisting of point charges. Checking the boundary conditions leads to
the same set of four charges as drawn in ﬁgure 2.5. The only diﬀerence is that R is not chosen
at inﬁnity to mimic the homogeneous ﬁeld, but R = d. The potential is the superposition of the
point charge q at distance d and its three image charges:
q
1
1
Φ(r, θ) = 4π 0 (r2 + d2 + 2rdcosθ)1/2 − (r2 + d2 −2rdcosθ)1/2
a
a

−
+
1/2
1/2
d r2 + a4 + 2a2r cosθ
d r2 + a4
cosθ

(2.34)
d2
d
d2 − 2a2r
d
The charge density on the boss is
∂Φ
σ = − 0
(2.35)
∂r r=a
The total amount of charge is the surface charge density integrated over the surface of the boss:
π/2
Q = 2πa2
σsinθdθ
(2.36)
0

24
CHAPTER 2. BOUNDARY-VALUE PROBLEMS IN ELECTROSTATICS: 1
The diﬀerentiation of equation (2.34) to obtain the the surface charge density and the following
integration in equation (2.36) are left to the reader. The resulting total charge is2
d2
Q = −q 1 −
− a2
(2.37)
d√d2 + a2
2.11
Line Charges and the Method of Images
a. Magnitude and Position of the Image Charge(s)
Analog to the situation of point charges in previous image problems, one image charge of opposite
magnitude at distance b2 (see ﬁgure 2.6) satisﬁes the conditions the boundary conditions
R
lim Φ(r, φ)
= 0 and
r→∞ Φ(b,φ) = V0
P
r
r
b
1
r2
φ
- τ
τ
R
V0
Figure 2.6: A cross sectional view of a long cylinder at potential V0 and a line charge τ at distance
R, parallel to the axis of the cylinder. The image line charge −τ is placed at b2/R from the axis
of the cylinder to realize a constant potential V0 at radius r = b.
2 I have chosen to keep Q as the symbol for the total charge. Jackson calls it q . I ﬁnd this confusing since the
primed q has been used for the image of q.

2.11. LINE CHARGES AND THE METHOD OF IMAGES
25
b. The Potential
The potential in polar coordinates is simply a superposition of the line charge τ and the image line
charge τ with the conditions as proposed in section a. The result is
τ
(R2r2 + b4
Φ(r, φ) =
ln
− 2rRb2cosφ)
(2.38)
4π 0
R2(r2 + R2 − 2Rrcosφ)
For the far ﬁeld case (r >> R) we can factor out (Rr)2. The b4 and R2 in equation (2.38) can be
neglected:
τ
(Rr)2(1
cosφ)
Φ(r, φ) ≈
ln
− 2b2
Rr
(2.39)
4π 0
(Rr)2(1 − 2Rcosφ)
r
The ﬁrst order Taylor expansion is
τ
2b2
2R
Φ(r, φ)
≈
ln (1
cosφ)(1 +
cosφ)
4π
−
0
Rr
r
τ
2R
2b2
≈
ln (1 +
cosφ
cosφ)
4π
−
0
r
Rr
τ
(R2
≈
− b2)cosφ
(using ln(1 + x)
2π
≈ x)
(2.40)
0
Rr
c. The Induced Surface Charge Density
∂Φ
σ(φ) = − 0
(2.41)
∂r r=b
Diﬀerentiation of equation (2.38) and substituting r = b:
τ
2bR2
R2(2b
σ(φ)
= −
− 2Rb2cosφ
− 2Rcosφ
4π
R2b2 + R2 + b4 − 2b3Rcosφ − R2(b2 + R2 − 2Rbcosφ
τ
(R/b)2
= −
− 1
(2.42)
2π
(R/b)2 + 1 − 2(R/b)cosφ
When R/b = 2, the induced charge as a function of φ is
τ
3
σ(φ)|R=2b = −
(2.43)
2πb
5 − 4cosφ
When the position of the line charge τ is four radii from the center of the cylinder, the surface
charge density is
τ
15
σ(φ)|R=4b = −
(2.44)
2πb
17 − 8cosφ
The graphs for either case are drawn in ﬁgure 2.7.

26
CHAPTER 2. BOUNDARY-VALUE PROBLEMS IN ELECTROSTATICS: 1
1
2
3
4
5
6
φ(rad)
-0.5
-1
-1.5
-2
-2.5
R/b=2
R/b=4
-3
σ τ
( /2 πb)
Figure 2.7: The behavior of the surface charge density σ with angle for two diﬀerent ratios between
the radius of the cylinder and the distance to the line charge.
d. The Force on the Line Charge
The force per meter on the line charge is Coulomb’s law:
τ 2
1
F = τ E(R, 0) = −
ˆı per meter
(2.45)
2π 0 (R2 − b2)
where ˆı is the directed from the the axis of the cylinder to the line charge, perpendicular to the
line charge and the cylinder axis.
2.13
Two Cylinder Halves at Constant Potentials
a. The Potential inside the Cylinder
In this case (see ﬁgure 2.8) there are no free charges in the area of interest. Therefore the potential
Φ inside the cylinder obeys Laplace’s equation:
2Φ = 0
(2.46)
We can write the potential in cylindrical coordinates and separate the variables:
Φ(ρ, φ) = R(r)F (φ)
(2.47)

2.13. TWO CYLINDER HALVES AT CONSTANT POTENTIALS
27
The general solution is (see J2.71):
∞
Φ(ρ, φ)
= a0 + b0lnρ +
anρnsin(nφ + αn) + bnρ−ncos(nφ + αn)
(2.48)
n=1
From this geometry it is obvious that at the center ρ = 0 the solution may not blow up, so:
bn = b0 = 0
(2.49)
This results in a potential
∞
Φ(ρ, φ)
= a0 +
anρnsin(nφ + αn)
(2.50)
n=1
The next step is to implement the boundary conditions
V1
P
r
φ
b
V2
Figure 2.8: Cross-section of two cylinder halves with radius b at constant potentials V1 and V2.
∞
Φ(b, φ) = V2 = a0 +
anbnsin(nφ + αn)
for (−π/2 < φ < π/2)
n=1
∞
Φ(b, φ) = V1 = a0 +
anbnsin(nφ + αn)
for (π/2 < φ < 3π/2)
(2.51)
n=1
b. The Surface Charge Density
∂Φ
σ = − 0
(2.52)
∂r r=b

28
CHAPTER 2. BOUNDARY-VALUE PROBLEMS IN ELECTROSTATICS: 1
2.23
A Hollow Cubical Conductor
a. The Potential inside the Cube
Vz
z
a
y
a
Vz
a
x
Figure 2.9: A hollow cube, with all sides but z=0 and z=a grounded.
2Φ = 0
(2.53)
Separating the variables:
1 d2X
1 d2Y
1 d2Z
+
+
= 0.
(2.54)
X dx2
Y dy2
Z dz2
x and y can vary independently so each term must be equal to a constant −α2:
1 d2X + α2 = 0
X dx2
⇒ X = Acosαx + Bsinαx
(2.55)
1 d2Y + β2 = 0
Y dy2
⇒ Y = Ccosβy + Dsinβy
(2.56)
1 d2Z + γ2 = 0
Z dz2
⇒ Z = Esinh(γz) + Fcosh(γz),
(2.57)
where γ2 = α2 + β2. The boundary conditions determine the constants:
Φ(0, y, z) = 0 ⇒ A = 0

2.23. A HOLLOW CUBICAL CONDUCTOR
29
Φ(a, y, z) = 0 ⇒ αn = nπ/a (n = 1,2,3,...)
Φ(x, 0, z) = 0 ⇒ C = 0
Φ(x, a, z) = 0 ⇒ βm = mπ/a (m = 1,2,3,...)
⇒ γnm = π n2 + m2
The solution is thus reduced to
∞
nπx
mπy
γ
γ
Φ(x, y, z) =
sin
sin
A
nmz
+ B
nmz
(2.58)
a
a
nmsinh
a
nmcosh
a
n,m=1
Now, let’s use the last boundary conditions to ﬁnd the coeﬃcients Anm and Bnm. The top and
bottom of the cube are held at a constant potential Vz, so
∞
nπx
mπy
Φ(x, y, 0) = Vz =
Bnmsin
sin
(2.59)
a
a
n,m=1
This means that Bnm are merely the coeﬃcients of a double Fourier series (see for instance [1] on
Fourier series):
4V
a
a
nπx
mπy
B
z
nm =
sin
sin
dxdy
(2.60)
a2
0
0
a
a
It can be easily shown that the individual integrals in equation (2.60) are zero for even integer
values and 2a for n is odd. Thus B
nπ
nm is
16V
B
z
nm =
for odd (n, m)
(2.61)
π2nm
The top of the cube is also at constant potential Vz, so
Φ(x, y, 0) = Vz = Φ(x, y, a) ⇔
Bnm = Anmsinh(γnm) + Bnmcosh(γnm) ⇔
1
A
− cosh(γnm)
nm
= Bnm
(2.62)
sinh(γnm)
Substituting the expressions for Anm and Bnm into equation (2.58), gives us
16V
∞
1
nπx
mπy
1
γ
Φ(x, y, z) =
z
sin
sin
− cosh(γnm)sinh nmz
π2
nm
a
a
sinh (γnm)
a
n,m odd
γ
+ cosh
nmz
,
(2.63)
a
where γnm = π√n2 + m2.

30
CHAPTER 2. BOUNDARY-VALUE PROBLEMS IN ELECTROSTATICS: 1
b. The Potential at the Center of the Cube
The potential at the center of the cube is
a a a
16V
∞
1
nπ
mπ
1
γ
Φ( , , ) =
z
sin
sin
− cosh(γnm)sinh nm
2 2 2
π2
nm
2
2
sinh (γnm)
2
n,m odd
γ
+ cosh
nm
(2.64)
2
With just n, m = 1, the potential at the center is
16Vz 1 − cosh(√2π)
π
π
sinh
+ cosh
π2
sinh(√2π)
√2
√2 ≈ 0.347546Vz
(2.65)
When we add the two terms (n = 3, m = 1) and (n = 1, m = 3), the potential is 0.332498Vz.
c. The Surface Charge Density
The surface charge density on the top surface of the cube is given by
∂Φ
σ = − 0
(2.66)
∂z z=a
In the appendix it is shown that the diﬀerentiation of the hyperbolic sine is the hyperbolic cosine.
Furthermore
dcosh(az) = asinh(az)
(2.67)
dz
Using this equality in diﬀerentiating the expression for the potential in equation (2.63), we get
∂Φ
16V
∞
γ
nπx
mπy
1
γ
=
z
nm sin
sin
− cosh(γnm)cosh nmz
∂z
π2
nma
a
a
sinh (γnm)
a
n,m odd
γ
+ sinh
nmz
,
(2.68)
a
where γnm = π√n2 + m2. Now we evaluate this expression for z = a:
∂Φ
σ = − 0
=
∂z z=a
16
∞
γ
nπx
mπy
1
− 0Vz
nm sin
sin
− cosh(γnm)cosh(γ
π2
nma
a
a
sinh (γ
nm) + sinh (γnm)
=
nm)
n,m odd
16
∞
γ
nπx
mπy
− 0Vz
nm sin
sin
[(1
π2
nma
a
a
− cosh(γnm))coth(γnm) + sinh(γnm)] (2.69)
n,m odd
Further simpliﬁcation??

Chapter 8
Waveguides, Resonant Cavities
and Optical Fibers
8.1
Time Averaged Forces Per Unit Area on a Conductor
a. A Good Conductor
The relationship between force F on one side and current density J and magnetic ﬁeld H on the
other is
F = µc
J × H dV.
(8.1)
V
The force per unit area (or pressure) would then be
dF =
J
dA
−µc
× H dξ,
(8.2)
ξ
where ξ is the normal pointing into the conductor. This is opposite of the “normal” ˆ
n, resulting
in the change of sign. The time average of this quantity f is
dF
1
f =
=
µ
[J
dA
−2 c
× H∗] dξ.
(8.3)
ξ
See section 6.9 in Jackson for the time-averaging. When we insert the following properties (see
J8.9):
H = H eξ/δ(1−ı)
and
J = σE,
(8.4)
and the approximation for a good conductor that
µ
E ≈
cω (1
2σ
− ı) ˆn × H eξ/δ(ı−1),
(8.5)
31

32
CHAPTER 8. WAVEGUIDES, RESONANT CAVITIES AND OPTICAL FIBERS
into equation 8.3, we are left with a simple integration of the exponential
∞ e−2ξ/δ dξ = −δ
(8.6)
0
2
and a series of cross products (for which we have the equality as deﬁned in the appendix and in
the inside cover of Jackson)
ˆ
n × H × H∗ = −H∗ × ˆn × H = − H∗ · H ˆn + H · ˆn H .
(8.7)
The ﬁrst dot-product is the square of the magnetic ﬁeld magnitude and the last dot-product is
obviously zero, since the magnetic ﬁeld is perpendicular to the normal. Putting all this together,
gives us the time average force per unit area
µ
2
f = −ˆn c H .
(8.8)
4
b. A Perfect Conductor
dF = Idl × B.
(8.9)
I
dK =
(8.10)
dl
Therefore
dF
1
f =
=
K
dA
2
× B∗.
(8.11)
Using
B∗ = µH∗
and
K = ˆ
n × H,
(8.12)
we get in a perfect conductor for the time averaged force per unit area
µ
f = − ˆn
2 |H |2.
(8.13)
This is twice as large as the force found in a good conductor in part a. This has to do with the
discontinuity between the ﬁelds inside and outside the conductor. For an explaination, see [3].
c. A Superposition of Diﬀerent Frequencies
As in section 6.8 in Jackson, time averaging leads to
|H|2 = [H] · [H] = 1/2H · H∗,
(8.14)
where H =
H eıωkt. Therefore, the dot-product of the leads to an exponential dependence,
k
but due to averaging the individual frequencies over time, they all cancel:
1
1
1
|H|2 =
H
H
2
· H∗ eı(ωk−ωl)t = 2 · H∗ = 2|H |2.
(8.15)
k
l
The rest of the derivation stays the same, so for the result to stay the same we have to replace
|H |2 by 2 |H|2 in equation 8.8.

8.2. TEM WAVES IN A MEDIUM OF TWO CONCENTRIC CYLINDERS
33
8.2
TEM Waves in a Medium of Two Concentric Cylinders
a. Time Averaged Power Flow Along the Guide
For TEM waves, Ez and Hz are by deﬁnition zero. This simpliﬁes equations J8.23 and J8.25 to
t · ETEM = 0 and
t × ETEM = 0
(8.16)
This implies that our problem is reduced to an electrostatic potential Φ that obeys Laplace’s
equation in the cavity between the two cylindrical surfaces:
ET EM = − tΦ and 2tΦ = 0.
(8.17)
When we solve this equation in cylindrical coordinates and use the symmetry of the problem, the
solution is only a function of ρ:
Φ(ρ) = Alnρ + B
(8.18)
where A and B are arbitrary constants which will be deﬁned by the boundary conditions. The
electric ﬁeld ET EM is
Aˆ
ρ
ET EM (ρ) = − tΦ = − .
(8.19)
ρ
According to equation J8.28, the magnetic ﬁeld is
Aˆ
ρ
µ
× ˆz
HT EM (ρ) = ±
ˆ
z
.
(8.20)
µ × ET EM =
ρ
The boundary condition for this problem is that at ρ = a, H = H0. This means the constant is
deﬁned is
µ
A = aH0
.
(8.21)
Next, we compute the Poynting vector
1
a2
µ
S =
(E
|H0|2 (ˆρ× (ˆz × ˆρ)).
(8.22)
2
× H∗) =
2ρ2
Using the vector identity from the appendix and realizing that ˆ
ρ · ˆz = 0, we are left with
a2
µ
S =
|H0|2ˆz.
(8.23)
2ρ2
Equation J8.49 gives us the power ﬂow
2π
b
π
a2
µ
µ
b
P =
S · ˆzda =
dφdρdθ
|H0|2ρsinθ =
πa2|H0|2ln
.
(8.24)
A
0
a
0
2ρ2
a

34
CHAPTER 8. WAVEGUIDES, RESONANT CAVITIES AND OPTICAL FIBERS
b. Attenuation of the Transmitted Power
Attenuation is due to power lost through the walls of the conductor. According to equations J8.56,
J8.57 and J8.58, the power ﬂow is
1 dP
P (z) = P0e P dz ,
(8.25)
The goal is therefore to ﬁnd P , which is related to the Poynting vector as deﬁned in equation 8.22.
S follows from equation 8.22. All that is left is to determine the values for H and E. The magnetic
ﬁeld is continuous across the boundary and thus is the same as in part a:
aH ˆ
φ
H =
0
(8.26)
ρ
The electric ﬁeld can then be approximated as before in equation 8.5, evaluated at the surface
ξ = 0:
µ
E ≈
cω (1
2σ
− ı) ˆn × H .
(8.27)
The Poynting vector is
1
1
µ
a 2
S =
E
cω
2
× H∗ = −2 2σ ρ |H0|2ˆρ.
(8.28)
This Poynting vector is directed outward of the guide walls. This is the direction of the power loss.
This leaves us with the contributions of the two cylinders to the power loss P :
µ
1
1
P = P
cω
a + Pb = −π
a2
dz +
dz
(8.29)
2σ
|H0|2 ρ=a ρ
ρ=b ρ
Therefore
dP
µ
1
=
cω a2
+
(8.30)
dz
−π 2σ |H0|2 1b a
Using P from equation 8.24, we can write |H0|2 as function of P and substitute into equation 8.30:
dP
µ
P
1/a + 1/b
=
cω
=
dz
−2π 2σ µ π
ln b
−2γP.
(8.31)
a
So P is indeed P0e−2γz where
1
1/a + 1/b
γ =
(8.32)
2σδ
µ
ln ba
c. The Characteristic Impedance
The voltage diﬀerence between the cylinders is
a
b
∆V = −
E · dl =
E · dρˆρ.
(8.33)
b
a

8.2. TEM WAVES IN A MEDIUM OF TWO CONCENTRIC CYLINDERS
35
Using the electric ﬁeld that we computed in part a (equation 8.19), we get
µ
b
∆V = aH0
ln
.
(8.34)
a
By Ampere’s Law, the current is related to the magnetic ﬁeld as computed in equation 8.20.
aH
I =
H · dl =
0 ˆ
φ
ρ
· ρdφˆφ = 2πaH0.
(8.35)
d. Series Resistance and Inductance
To ﬁnd the resistance per unit length, we use that
dP
1
=
I2R
dz
2
l,
(8.36)
where Rl is the resistance per unit length. In part c, we found that
I = 2πaH0
(8.37)
and dP we obtained in equation8.30. Plugging these results into equation 8.36, gives
dz
1
1
1
Rl =
+
,
(8.38)
2πδσ
a
b
where we used that σδ =
2σ .
µcω
Finally, we’ll compute the inductance L. The inductance is related to the power loss as follows:
dP
1
Φ 2
=
R
dz
2
L
l,
(8.39)
where Φ is the ﬂux thorugh the walls of the conductor and medium, as
Φ =
Bcond · dA + Bmedium · dA
(8.40)
Using the resistance Rl as computed above, we get
R
2
L2 =
l
Bcond
(8.41)
2 dP
· dA + Bmedium · dA
dz

36
CHAPTER 8. WAVEGUIDES, RESONANT CAVITIES AND OPTICAL FIBERS
8.3
TEM Waves Between Metal Strips
a. Two Identical Thin Strips
The Power
In the previous problem we saw that for TEM waves, we can deﬁne a potential that satisﬁes
Laplace’s equation. Since b >> a, we consider only variations in electric ﬁeld in the x-direction.
The Laplace equation simpliﬁes to
∂Φ2 = 0.
(8.42)
∂2x
The solutions are a linearly varying ﬁeld and a constant. When we compute the electric ﬁeld by
taking the gradient, we ﬁnd that
ET EM = E0 ˆ
x.
(8.43)
From equation J8.28, we know the relation between the magnetic and electric ﬁeld for TEM waves
to be
HT EM = ±
ˆ
z
E
µ × ET EM = ±
µ 0 ˆ
y.
(8.44)
If H0 is the (peak) amplitude for the magnetic ﬁeld, then H0 =
E
µ
0. The Poynting vector is
deﬁned as
1
1
µ
S =
(E
2
× H∗) = 2
|H0|2ˆz.
(8.45)
a
b
ab
µ
P =
S · ˆzda =
Sdxdy =
|H0|2.
(8.46)
A
0
0
2
The Exponential Decay Factor
1 dP
γ = −
,
(8.47)
2P dz
where P was calculated in the ﬁrst part of this exercise and dP is
dz
dP
1
=
dz
−2σδ |ˆn × H|2dl.
(8.48)
C
The magnetic ﬁeld is oriented in the z-direction and the normal ˆ
n is in the x-direction. Now we
integrate around the plates individually to get the total power. Only between the plates there is
a magnetic ﬁeld. Therefore the closed line integral reduces to
dP
1
b
=
.
(8.49)
dz
−2σδ |ˆn × H|2dl =
|ˆx × H0ˆzdl = |H0|2b
C
0
2σδ
This is the ohmic power loss per plate. The exponential decay factor is thus
1 dP
1
γ = −
=
.
(8.50)
2P dz
µ aσδ

8.3. TEM WAVES BETWEEN METAL STRIPS
37
The Characteristic Impedance
THe characteristic impedance was deﬁned in problem 8.2 as the voltage frop between the plates
divided by the current ﬂowing along one of the plates. The voltage diﬀerence between the cylinders
is
a
a
a
µ
µ
∆V = −
E · dl =
E0dx =
H0 = a
H0.
(8.51)
b
0
0
The other leg is the current along one of the plates. To ﬁnd this, we use
b
I =
H0 ˆ
y · dl =
H0dy = H0b.
(8.52)
C
0
The characteristic impedance is therefore
dV
µ a
Z0 =
=
.
(8.53)
I
b
The Series Resistance
2 dP
2
b
2
R
|H0|2
l =
=
=
.
(8.54)
I20 dz
(H0b)2
σδ
bσδ
The Series Inductance
The inductance is
Φ
L =
,
(8.55)
I
where Φ =
B · ˆnda and we know I = H0b. In the conducting plate (and we are looking at only
one plate, like we did to compute the current, too), the ﬂux is
δ
Φc = µcH0
∞ˆy· dxdzˆy.
(8.56)
0
0
Therefore, the ﬂux per unit length in the conductor is
Φl,c = µcH0δ.
(8.57)
For the medium, we integrate from zero to a, and get
Φl,0 = µ0H0a.
(8.58)
The total inductance per unit length is the addition of the two, divided by the current:
µ
L
0a + µcδ
l =
.
(8.59)
b

38
CHAPTER 8. WAVEGUIDES, RESONANT CAVITIES AND OPTICAL FIBERS
8.4
TE and TM Waves along a Brass Cylinder
8.4.1
a. Cutoﬀ Frequencies
A brass cylinder with radius R is oriented with its axis along the z-direction. TE or TM waves
propagate along the cylinder. Both TE and TM waves have a z-component that satisﬁes
t + γ2 ψ = 0,
(8.60)
where γ2 = µ ω2 −k2 and ψ is either Ez for TM waves or Bz for TE waves. Solving this equation
in cylindrical coordinates (and recognizing that there is no z-dependence), we get solutions
ψ = eımφJm(γρ)eı(kz−wt).
(8.61)
We already discarded the other solution to the Bessel equation, because it has a singularity at the
origin. Lets apply the rest of the boundary conditions for the appropriate wave. For TM waves we
know that
ψ|s = ψ|ρ=R = 0.
(8.62)
Hence the Bessel function is zero when ρ = R:
γmnR = xmn,
(8.63)
where xmn is the n-th root of the m-th order Bessel function. The cutoﬀ frequency is the lowest
frequency for which the wave does not become inhomogeneous or evanescent. In other words, the
wave number should be real. The cutoﬀ frequency is therefore
γ
x
ω
mn
mn
mn = √ =
(8.64)
µ
R√µ
All that’s left to do is look up the four smallest roots. From page 114 in Jackson or any book
with tables of Bessel functions it is obvious that ω11 is the dominant frequency. The next three
are ω21, ω02, ω12, respectively.
For TE waves, the boundary condition is
∂ψ
= 0,
(8.65)
∂n ρ=R
where the normal is ˆ
n = ˆ
ρ. We’ll call
∂ψ
∂J
=
m(γρ)
= J
∂ρ
∂ρ
m(γR),
(8.66)
ρ=R
ρ=R
which is a function conveniently tabulated on page 370 of Jackson. The dominant mode is
1.841
ω11 =
.
(8.67)
R√µ

8.4. TE AND TM WAVES ALONG A BRASS CYLINDER
39
The next four higher modes have ratios:
ω21 = 1.656,
ω11
ω01 = 2.081,
ω11
ω31 = 2.282,
ω11
ω12 = 2.896.
ω11

40
CHAPTER 8. WAVEGUIDES, RESONANT CAVITIES AND OPTICAL FIBERS

Chapter 10
Scattering and Diﬀraction
10.3
Scattering Due to a Solid Uniform Conducting Sphere
a. The Magnetic Field around the Sphere
Since there are no currents present, we ﬁnd that
× H = J = 0 and
· H = 0.
(10.1)
Therefore the magnetic ﬁeld can be described by a magnetic scalar potential Φm that satisﬁes
Laplace’s equation:
H
= − Φm
2Φm = 0.
(10.2)
The general solution to Laplace’s equation, simpliﬁes for our boundary conditions to (J5.117):
∞ α
Φ
l
m = −Hincr cos θ +
P
rl+1 l(cos θ).
(10.3)
l=0
where αl is deﬁned by (J5.121). With a = 0 (our shell is a solid sphere), µ = 0, because we are
dealing with a perfectly conducting sphere, αl = 0 for l = 1 and α1 = −b3Hinc. This means that
2
the magnetic scalar potential is
R3
Φm = −Hinc cosθ r +
(10.4)
2r2
and
1 ∂Φ
R3
H = H
m ˆ
ˆ
r + Hθ = −
θ =
θ.
(10.5)
r ∂θ
−Hinc sinθ 1 + 2r3
41

42
CHAPTER 10. SCATTERING AND DIFFRACTION
Hr is zero inside the sphere H = 0 and the boundary conditions say that Hr is continuous across
the boundary. Just outside the sphere, the magnetic ﬁeld is
3
H|r=R = − H
2 inc sin θ ˆ
θ
(10.6)
b. The Absorption Cross Section
According to (J8.12), the power loss per unit area is
dPloss
1
2
9
=
µ
=
µ
da
4 cωδ H
16 cωδH2inc sin2 θ.
(10.7)
The loss of power is then just a simple integration over area:
dP
9
π
3
P
loss
loss =
da =
µcωδH2inc2πR2
sin2 θ sin θdθ =
πR2µcωδH2inc.
(10.8)
S
da
16
0
2
According to the experts (I could not ﬁnd this anywhere), the absorption cross section is deﬁned
as
P
σ
loss
abs =
,
(10.9)
dPinc/da
where the denominator is the total incidence power per unit area. For plane waves, Griﬃths writes:
dPinc
1
=
[ˆ
n
.
(10.10)
da
2
· Einc × H∗inc] = |Hinc|2
2c
Plugging this and the expression for the loss of power into equation 10.9 shows that the absorption
cross section is proportional to the square root of frequency:
3 πR2µcωδH2
3πR2µ
3πR2µ
3πR2
2µ
σ
2
inc
cωδ
cω√2
c √
abs =
=
=
=
ω.
(10.11)
|Hinc|2
c
c√µcωσ
c
σ
2c
10.14
Diﬀraction from a Rectangular Opening
This problem involves an inﬁnite conducting sheet in the x-y plane with diﬀraction from a rectan-
gular aperture as depicted in ﬁgure 10.1.
a. The Smythe-Kirchhoﬀ Relation
According to the vector Smythe-Kirchoﬀ relation, the electric ﬁeld at an arbitrary point O at
location x due to diﬀraction from an aperture S1 is
ıeıkr
E(x) =
k
n
2πr
×
× E(x )e−ık·x da ,
(10.12)
S1

10.14. DIFFRACTION FROM A RECTANGULAR OPENING
43
z
O
k
ko
-a/2
E
θ
i
β
y
φ
a/2
-b/2
b/2
x
Figure 10.1: A rectangular aperture in a conducting sheet in the x-y plane.
where the electric ﬁeld inside the aperture is assumed to be that of the incoming one:
E(x ) = E0e−ık·x (sin β ˆ
x + cos β ˆ
y) .
(10.13)
Since n = ˆ
z,
ˆ
z × E(x ) = E0e−ık·x (sinβˆy− cosβˆx).
(10.14)
The integration is now straightforward:
a/2
b/2
n × E(x )e−ık·x da = E0 (sinβˆy− cosβˆx)
e−ı(kxx +kyy )dx dy
S1
−a/2 −b/2
4
k
k
= E
xa
y b
0 (sin β ˆ
y − cosβˆx)
sin
sin
.
(10.15)
kxky
2
2
With k = k(sin θ cos φˆ
x + sin θ sin φˆ
y + cos θˆ
z), the total expression for the diﬀracted electric ﬁeld
is
2ıE
k
k
E(x)
=
0eıkr sin
xa
sin
y b
k
πrk
× (sinβˆy − cosβˆx)
xky
2
2
2ıE
k
k
=
0eıkr sin
xa
sin
y b
(ˆ
z (k
πrk
x sin β + ky cos β) − ˆxkz sin β − ˆykz cos β)
xky
2
2
2ıE
ka sin θ cos φ
kb sin θ sin φ
=
0eıkr
sin
sin
πrk sin2 θ cos φ sin φ
2
2
×
×(ˆz(sinθ cosφsinβ + sinθ sinφcosβ) − ˆxcosθ sinβ − ˆycosθ cosβ)
2ıE
ka sin θ cos φ
kb sin θ sin φ
=
0eıkr
sin
sin
πrk sin2 θ cos φ sin φ
2
2
×
×(ˆzsinθ sin(φ − β) − ˆxcosθ sinβ − ˆycosθ cosβ)
(10.16)
The magnetic ﬁeld for plane waves has a relation to the electric ﬁeld that can easily be understood
by applying the far ﬁeld (where waves are “plane”) approximation of
= ık:
∂B
× E = − ∂t ⇔ ıkˆk × E = ıωB ⇔

44
CHAPTER 10. SCATTERING AND DIFFRACTION
k
2ıE
ka sin θ cos φ
kb sin θ sin φ
B(x)
=
× E =
0eıkr
sin
sin
c
cπr sin2 θ cos φ sin φ
2
2
×
× ˆz sin2 θ cosφ(sin(φ − β)) − sinθ sinφ(cosθ sinβ) −
ˆ
x sin2 θ sin φ(sin(φ − β)) − cosθ cosθ cosβ −
ˆ
y sin2 θ cos φ(sin(φ − β)) − cosθ(cosθ sinβ) .
(10.17)
The power per unit angle is deﬁned as
dP
r2
=
dω
2Z |E|2 =
0
2E20
sin2
ka sin θ cos φ
sin2
kb sin θ sin φ
Z0k2π2 sin4 θ cos2 φ sin2 φ
2
2
×
× sin2 θ sin2(φ − β) + cos2 θ .
(10.18)
b. The Scalar Kirchhoﬀ Approximation
We can get a scalar approximation from the far ﬁeld version of the Kirchhoﬀ integral (J10.79),
which is (J10.108):
eıkr
ψ(x) =
e−ık·x (n
4πr
· ψ(x ) + ık · nψ(x ))da ,
(10.19)
S1
where the assumption is that the scalar ψ(x ) in the aperture is the magnitude of the incoming
ﬁeld:
ψ(x ) = E0eık0·x = E0eıkz .
(10.20)
Therefore
n · ψ(x )|
= ıkE
z =0
0.
(10.21)
And
ık · nψ(x )|
= ık cos θE
z =0
0.
(10.22)
So what we are left with is an integral we solved in part a:
ıkE
ψ(x)
=
0eıkr (1 + cos θ)
e−ık·x da
4πr
S1
ıE
ka sin θ cos φ
kb sin θ sin φ
=
0eıkr (1 + cos θ) sin
sin
.
(10.23)
πrk sin2 θ cos φ sin φ
2
2
The power per unit angle in the scalar approximation is
dP
E2
=
0 (1 + cos θ)2
sin2
ka sin θ cos φ
sin2
kb sin θ sin φ
.
(10.24)
dω
2Z0π2k2 sin4 θ cos2 φ sin2 φ
2
2

10.14. DIFFRACTION FROM A RECTANGULAR OPENING
45
c. A Comparison
For the case: b = a, β = π/4, φ → 0 and ka/2 = kb/2 = 2π the power per unit angle for the vector
and the scalar approximation is respectively
dP
2E2
=
0
sin2
ka sin θ cos φ
sin2
kb sin θ sin φ
dω
k2π2 sin4 θ cos2 φ sin2 φ
2
2
×
× sin2 θ sin2(φ − β) + cos2 θ
2E2
sin2 θ
=
0
sin2(2π sin θ) sin2(2π sin θφ)
+ cos2 θ .
(10.25)
k2π2 sin4 θφ2
2
Using that
sin2(2π sin θφ)
lim
φ→0
φ2
≈ (2π sinθ)2 = 4π2 sin2 θ,
(10.26)
we get for the vector approximation of the power per unit angle that
dP
8E2
1
cos2 θ
4E2
lim
=
0
sin2(2π sin θ)
+
=
0 sin2(2π sin θ) 1 + cos2 θ .
(10.27)
φ→0 dω
k2 sin2 θ
2
2
k2 sin2 θ
And for the scalar approximation:
dP
E2
2E2
lim
=
0 (1 + cos θ)2 sin2 (2π sin θ) sin2 (2π sin θφ) =
0 sin2 (2π sin θ) (1 + cos θ)2 .(10.28)
φ→0 dω
2π2k2 sin4 θφ2
k2 sin2 θ
The results are plotted (with E0 = 1) in ﬁgure 10.2. Note that for small scattering angle θ the
k2
results are identical.

46
CHAPTER 10. SCATTERING AND DIFFRACTION
scalar
vector
power per unit angle
14
12
10
8
6
4
2
rad
1
2
3
4
5
6
Figure 10.2: Scalar and vector approximation of the radiated power per unit angle. For θ close to
zero (i.e. close to the direction of k0, the approximations lead to the same result.

Chapter 11
Special Theory of Relativity
11.3
The Parallel-velocity Addition Law
Two reference frames K ,K move in the same direction from K with diﬀerent velocities v1 and
v2, respectively. The frames can be oriented such that the direction of propagation with respect
to K is ˆ
x. The Lorentz transform can then be written in matrix form x = AK→K x
v
v
1
2
K
K’
K’’
Figure 11.1: Three reference frames. K and K move in the same direction from K with velocity
v1 and v2, respectively.
 x0
γ
−γβ 0 0
x0
  
 
 x1
−γβ
γ
0 0
x1
 x  
 
2   0
0
1 0  x2
x  = 
 ,
(11.1)
3
0
0
0 1
x3
where
v
1
β =
and
γ =
(11.2)
c
1 − β2
47

48
CHAPTER 11. SPECIAL THEORY OF RELATIVITY
with c equal to the speed of light. The transform from K to K
can then be written as the
multiplication of the transform matrix AK→K and AK →K . Here we will show that that is the
same as a direct Lorentz transform AK→K .
 γ1 −γ1β1 0 0
γ2
−γ2β2 0 0



 −γ1β1 γ1 0 0 −γ2β2 γ2 0 0
 0 0 1 0  0 0 1 0 
0
0
0 1  0
0
0 1  =
 γ1γ2+γ1γ2β1β2 −γ1γ2β2−γ1γ2β1 0 0


 −γ2γ1β1−γ2γ1β2 γ1γ2β1β2+γ1γ2 0 0

0
0
1 0 
0
0
0 1 
(11.3)
We can deﬁne γ and β in the total transformation AK→K and ﬁnd the total transformation
velocity:
γ = γ1γ2 + γ1γ2β1β2
and
βγ = γ1γ2β2 − γ1γ2β1.
(11.4)
Therefore
γ
γ
β
v1 + v2
v
β
=
1γ2β2 − γ1γ2β1 = 1γ2β2 − γ1γ2β1 = 2 + β1 = c
c
=
γ
γ1γ2 + γ1γ2β1β2
1 + β1β2
1 + v1v2
c ⇔
c2
v
v
=
1 + v2 .
(11.5)
1 + v1v2
c2
11.5
The Lorentz Transformation Law for Acceleration
Reference frame K moves from K with velocity v. We arrange our reference frame that v = vˆ
x.
To ﬁnd the acceleration transform we ﬁrst ﬁnd how velocity u and time t transform, so that we
can deﬁne a = du . From equation (J11.18) it is clear that
dt
dx
γ
v
dt =
0 = (dx
dx
(11.6)
c
c
0 + βdx1 ) = γ
dt + c2
The derivative of the transformed displacement with respect to time (11.6) gives us (J11.31) which
states
u + v
u
u =
and u
⊥
.
(11.7)
1 + v·u
⊥ = γ 1 + v·u
c2
c2
The acceleration parallel to the relative velocity between the reference frames is
du
1 + v·u
du
du
c2
− u + v vc2
a =
=
.
(11.8)
dt
2
1 + v·u
γ dt + v dx
c2
c2

11.6. THE ROCKET SHIP
49
Use dx1 = u1 dx
c
0 (Jackson page 531) and a
= du /dt to conclude that
3/2
a
1 + v2
a
1
c2
− v2c2
a =
=
.
(11.9)
vu
3
vu
3
γ 1 +
1 +
c2
c2
The acceleration perpendicular to the direction of propagation is
d
u⊥
du
γ(1+ v·u )
a
⊥
c2
⊥ =
=
.
(11.10)
dt
γ dt + v dx
c2
After computing the diﬀerential of the numerator with the chain rule, we obtain
du
u
v·du
du
⊥ c2
⊥ =
⊥
.
(11.11)
γ 1 + v·u
−
2
c2
γ 1 + v·u
c2
Use v · du = vdu and we have
1 + v·u
du
u
c2
⊥ du
a
⊥ − vc2
⊥ =
.
(11.12)
3
γ2dt 1 + v·u
c2
du
Realizing that du⊥ = a
= a and 1/γ2 = 1
dt
⊥, dt
− v2/c2 simpliﬁes equation 11.12 to
1 − v2c2
v
v
a
· u
· a
⊥ =
a
a
u
3
⊥
.
(11.13)
1 + v·u
⊥ + c2
⊥ − c2
c2
Writing out the outer products v × (a × u ) shows that
1 − v2c2
v
a⊥ =
a
3
1 + v·u
⊥ + c2 × (a × u ) .
(11.14)
c2
11.6
The Rocket Ship
a. How long are they gone?
There a four legs to the trip. We’ll do the calculations on the ﬁrst and then use symmetry to ﬁnd
the total answer. The ﬁrst leg is ﬁve years to the people in the rocket ship: t1 = 5 years. They
experience an acceleration of a1 = g in the direction which I will call ˆx. What we want to know is

50
CHAPTER 11. SPECIAL THEORY OF RELATIVITY
how long are those ﬁve years in the rocket ship to the observers (the other part of the twins) on
earth. In other words, what is t? They are related by
β
t = γ t +
x
,
(11.15)
c
where
v
1
β =
and
γ =
.
(11.16)
c
1 − β2
The twin in the rocket ship feels a force but does not move in its reference frame: x = 0, so
dt = γdt .
(11.17)
I wrote this in diﬀerentials because γ is function of velocity and the velocity of the space ship as
seen from earth increases with time. We’ll ﬁnd this velocity via the acceleration a. We know a = g
and that
3/2
a
1 − v2
3/2
c2
v2
a = a ˆ
x =
ˆ
x = g 1
ˆ
x
vu
3
− c2
⇔
1 + c2
dv
v2 3/2
= g 1
ˆ
x
dt
− c2
⇔
v2 −3/2
gdt =
1 −
dv
c2
⇔
v
gt =
.
(11.18)
1 − v2c2
Rewriting this result as a function of velocity leads to
gt
v(t) =
(11.19)
2
1 + gt
c
Plugging this result into equation 11.16 gives
dt
dt = dt/γ =
.
(11.20)
2
1 + gt
c
This we integrate to
c
τ =
sinh−1
gt
.
(11.21)
g
c
The time on the clock in the space ship records τ = 5 years for the ﬁrst leg. Equation 11.21 written
as a function of time on earth t is
c
gτ
t =
sinh
= 83.7612 years.
(11.22)
g
c
Since all four stages are symmetric the total time spent away from earth is 4 · 83.7612 = 335.05
years, while 20 years passed on the rocket. The space ship left in 2100, so the year of return is
2435.05 AD.

11.6. THE ROCKET SHIP
51
b. How far did the rocket ship get?
From Earth dx = vdt, where the velocity for leg one is calculated in equation 11.19. Therefore
the distance traveled is merely an integration over time from zero to the t1 seconds from part a
(83.7612 years):
t1
t1
gt
 gt 2 
d
1
1 =
v(t)dt =
dt = f racc2g
− 1
0
0
2
c
1 + gt
 1+
=7.83×1017m=82.77lightyears.
c
(11.23)
The total is distance is just twice the ﬁrst leg distance: d = 2d1 = 165.55 light years.

52
CHAPTER 11. SPECIAL THEORY OF RELATIVITY

Chapter 12
Practice Problems
12.1
Angle between Two Coplanar Dipoles
p
p
1
2
Er
θ
Eθ
r
θ’
Figure 12.1: Coplanar dipoles separated by a distance r.
Two dipoles are separated by a distance r. Dipole p1 is ﬁxed at an angle θ as deﬁned in ﬁgure
12.1, while p2 is free to rotate. The orientation of the latter is deﬁned by the angle θ as deﬁned
in ﬁgure 12.1. Let us calculate the angular dependence between the two dipoles in equilibrium.
The electric ﬁeld due to a dipole can be decomposed into a radial and a tangential component
(D3.39):
2pcosθ
psinθ
E(r, θ) =
ˆ
r +
ˆ
θ,
(12.1)
4π 0r3
4π 0r3
where 0 is the dielectric permittivity. Writing p2 in terms of r and θ:
p2(r, θ) = (p2 · ˆr)ˆr+ (p2 · ˆθ = p2cosθ ˆr+ p2sinθ ˆθ
(12.2)
The potential energy of the second dipole in the electric ﬁeld due to the ﬁrst dipole is
p
U
1p2
2(r, θ, θ ) = −E1 · p2 = −
(2cosθcosθ
4π
− sinθsinθ )
(12.3)
0r3
53

54
CHAPTER 12. PRACTICE PROBLEMS
The second dipole will rotate to minimize its potential energy, deﬁning the angular dependence
between the two dipoles:
∂U2 = 0
∂θ
⇔
p1p2 (2cosθsinθ + sinθcosθ ) = 0
4π
⇔
0r3
2cosθsinθ = −sinθcosθ ⇔
tanθ = −(tanθ)/2.
(12.4)
12.2
The Potential in Multipole Moments
z
b
a
y
+q
-q
x
Figure 12.2: Concentric rings of radii a and b. Their charge is q and −q, respectively.
This exercise is how to ﬁnd the potential due to two charged concentric rings (see ﬁgure 12.2) in
terms of the monopole dipole and quadrupole moments. Discarding higher order moments (J4.10):
q
p
1
Φ(r) =
+
· r +
x
4π
i xj Qij + ....
(12.5)
0r
4π 0r3
8π 0r5 i,j
The monopole moment is zero since there is no net charge. The dipole moment is (J4.8):
p =
rρ(r)dV,
(12.6)
V

12.3. POTENTIAL BY TAYLOR EXPANSION
55
where the volume charge density for our case can be written in cylindrical coordinates:
q
q
ρ(r) =
δ(r
δ(r
2πa
− a)δ(z) − 2πb − b)δ(z)
(12.7)
After the x-component of the dipole moment is also written in cylindrical coordinates, the integral
can be easily evaluated:
∞
2π
∞
px =
xρ(r, z)dV =
cosφρ(r, z)r2drdφdz = 0,
(12.8)
V
−∞ 0
0
because the cosφ integrated over one period is zero. The y-component is zero, because the inte-
gration involves a sinusoid over one period. Finally, the z-component is zero, because
pz ∝ zδ(z)dz = 0
(12.9)
if the value 0 is within the integration limits.
The quadrupole moment is a tensor:
Qij =
(3xixj − r2)ρ(r)dV
(12.10)
V
We use the following properties of the δ-function:
δ(z)dz = 1 and
rδ(r − a)dr = a,
(12.11)
where 0 and a are within the respective integral limits. Knowing this, the calculations for each
element of Q is left to the reader. After some algebra, the quadrupole moment turns out to be
 1 0 0 
Qij =  0 1 0 a2−b2
(12.12)
0 0 −2  q2
This means the potential of the two charged concentric rings is approximately
(a2
Φ(r) ≈
− b2)(x2 + y2
16π
− 2z2)
(12.13)
0r5
12.3
Potential by Taylor Expansion
Here, I will show that the electrostatic potential Φ(x, y, z) can be approximated by the average of
the potentials at the positions perturbed by a small quantity +/ − a by doing a Taylor expansion.

56
CHAPTER 12. PRACTICE PROBLEMS
This expansion is correct to the third order. First, the Taylor expansion around the Φ(x, y, z)
perturbed in the positive x-component
∂Φ(x, y, z)
∂2Φ(x, y, z)
Φ(x + a, y, z) = Φ(x, y, z) +
a +
a2 +
∂x
∂x2
∂3Φ(x, y, z) a3 + O(a4)
(12.14)
∂x3
Next, the same expansion around Φ(x − a,y,z):
∂Φ(x, y, z)
∂2Φ(x, y, z)
Φ(x − a,y,z) = Φ(x,y,z) −
a +
a2
∂x
∂x2
−
∂3Φ(x, y, z) a3 + O(a4)
(12.15)
∂x3
When we add up these two equations, the odd powers of a cancel. This is the same for the y- and
z-component. The a2-term adds up to the Laplacian
2. Assuming there is no charge within the
radius a of (x, y, z), Laplace’s equation holds:
2Φ = 0
(12.16)
and thus the a2 term is zero, too. Therefore, the potential at (x, y, z) can be given by:
Φ(x, y, z) = 1/6(Φ(x + a, y, z) + Φ(x − a,y,z) + Φ(x,y + a,z)
+Φ(x, y − a,z) + Φ(x,y,z + a) + Φ(x,y,z − a)) + O(a4)
(12.17)

Appendix A
Mathematical Tools
A.1
Partial integration
b
b
f (x)g (x)dx = [f (x)g(x)]b
f (x)g(x)dx
(A.1)
a −
a
a
A.2
Vector analysis
Stokes’ theorem
M · dl = curlM · dS
(A.2)
L
S
Gauss’ theorem
M · dS =
divM · dV
(A.3)
S
V
Computation of the curl
h ˆ
ˆ
ˆ
ii
h2j
h3k
1
curlM =
∂
∂
∂
(A.4)
h
∂x
∂x
∂x
1h2h3
1
2
2
h1M1 h2M2 h3M3
57

58
APPENDIX A. MATHEMATICAL TOOLS
hi are the geometrical components that depend on the coordinate system. For a Cartesian coordi-
nate system they are one. For the cylindrical system:
h1 = 1
h2 = r
h3 = 1
For the spherical coordinate system:
h1 = 1
h2 = r
h3 = r sin θ
Computation of the divergence
1
∂
∂
∂
divM =
(h
(h
(h
h
2h3M1) +
1h3M2) +
1h2M3)
(A.5)
1h2h3
∂x1
∂x2
∂x3
With the same factors hi depending on the coordinate system.
a × (b × c) = (a · c)b − (a · b)c.
(A.6)
Relations between grad and div
× × M =
· M − 2M
(A.7)
M =
Φ +
× A
(A.8)
A.3
Expansions
Taylor series
(x
xn
f (x) = f (a) + (x − a)f (a) + − a)2f (a) + .... + (x
2!
n!
− a)nf(n)(a)
(A.9)

A.4. EULER FORMULA
59
A.4
Euler Formula
eiφ = cos φ + i sin φ
(A.10)
eıθ
sin θ
=
− e−ıθ
(A.11)
2i
eıθ + e−ıθ
cos θ
=
(A.12)
2
eθ
sinh θ
=
− e−θ
(A.13)
2
eθ + e−θ
cosh θ
=
(A.14)
2
A.5
Trigonometry
1
sin2 θ =
− cos(2θ)
(A.15)
2
cos(φ − θ) = Re eı(φ−θ) = Re eıφe−ıθ
= Re [cos φ cos θ + sin φ sin θ + i(..)]
= cos φ cos θ + sin φ sin θ
(A.16)

60
APPENDIX A. MATHEMATICAL TOOLS

Bibliography
[1] R. Bracewell. The Fourier Transform and Its Applications. McGraw-Hill Book Company, ﬁrst
edition, 1965.
[2] W.J. Duﬃn. Electricity and Magnetism. McGraw-Hill Book Company, fourth edition, 1990.
[3] ?.?. Griﬃth. Electricity and Magnetism. John Wiley & Sons, Inc., third edition, 1999.
[4] J.D. Jackson. Classical Electrodynamics. John Wiley & Sons, Inc., third edition, 1999.
61