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Answers To a Selection of Problems from
Classical Electrodynamics
John David Jackson
by Kasper van Wijk
Center for Wave PhenomenaDepartment of GeophysicsColorado School of Mines
Golden� CO �����
SamizdatPress
Published by the Samizdat Press
Center for Wave PhenomenaDepartment of GeophysicsColorado School of MinesGolden� Colorado �����
andNew England Research
�� Olcott DriveWhite River Junction� Vermont ����
c� Samizdat Press� ����
Release ��� January ����
Samizdat Press publications are available via FTP or WWW fromsamizdat�mines�edu
Permission is given to freely copy these documents�
Contents
� Introduction to Electrostatics �
��� Electric Fields for a Hollow Conductor � � � � � � � � � � � � � � � � � � � � � � � � �
��� Charged Spheres � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � �
�� Charge Density for a Hydrogen Atom � � � � � � � � � � � � � � � � � � � � � � � � � �
��� Charged Cylindrical Conductors � � � � � � � � � � � � � � � � � � � � � � � � � � � � �
��� Green�s Reciprocity Theorem � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � ��
� Boundary�Value Problems in Electrostatics� � ��
� The Method of Image Charges � � � � � � � � � � � � � � � � � � � � � � � � � � � � � �
�� An Exercise in Green�s Theorem � � � � � � � � � � � � � � � � � � � � � � � � � � � � �
�� Two Halves of a Conducting Spherical Shell � � � � � � � � � � � � � � � � � � � � � � ��
��� A Conducting Plate with a Boss � � � � � � � � � � � � � � � � � � � � � � � � � � � � �
��� Line Charges and the Method of Images � � � � � � � � � � � � � � � � � � � � � � � �
�� Two Cylinder Halves at Constant Potentials � � � � � � � � � � � � � � � � � � � � � � �
� A Hollow Cubical Conductor � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � �
� Practice Problems ��
�� Angle between Two Coplanar Dipoles � � � � � � � � � � � � � � � � � � � � � � � � � �
� The Potential in Multipole Moments � � � � � � � � � � � � � � � � � � � � � � � � � � �
� Potential by Taylor Expansion � � � � � � � � � � � � � � � � � � � � � � � � � � � � � �
� CONTENTS
A Mathematical Tools ��
A�� Partial integration � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � �
A� Vector analysis � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � �
A� Expansions � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � �
A�� Euler Formula � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � �
A� Trigonometry � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � �
Introduction
This is a collection of my answers to problems from a graduate course in electrodynamics� Theseproblems are mainly from the book by Jackson � �� but appended are some practice problems� Myanswers are by no means guaranteed to be perfect� but I hope they will provide the reader with aguideline to understand the problems�
Throughout these notes I will refer to equations and pages of Jackson and Du�n ��� The latter isa textbook in electricity and magnetism that I used as an undergraduate student� References toequations starting with a �D� are from the book by Du�n� Accordingly� equations starting withthe letter �J� refer to Jackson�
In general� primed variables denote vectors or components of vectors related to the distance betweensource 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�
Chapter �
Introduction to Electrostatics
��� Electric Fields for a Hollow Conductor
a� The Location of Free Charges in the Conductor
Gauss� law states��
��� r �E� �����
where � is the volume charge density and �� is the permittivity of free space� We know thatconductors allow charges free to move within� So� when placed in an external static electric �eldcharges move to the surface of the conductor� canceling the external �eld inside the conductor�Therefore� a conductor carrying only static charge can have no electric �eld 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 �gure���a�� Gauss� law simpli�es to Laplace�s equation in the cavity� The conducting material forms avolume of equipotential� because the electric �eld in the conductor is zero and
E � �r� ����
Since the potential is a continuous function across a charged boundary� the potential on the innersurface of the conductor has to be constant� This is now a problem satisfying Laplace�s equationwith Dirichlet boundary conditions� In section ��� of Jackson� it is shown that the solution for thisproblem is unique� The constant value of the potential on the outer surface of the cavity satis�esLaplace�s equation and is therefore the solution� In other words� the hollow conductor acts like aelectric �eld shield for the cavity�
� CHAPTER �� INTRODUCTION TO ELECTROSTATICS
q
a b
Figure ���� 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 �gure ���b�� we use the following representation ofGauss� law� I
E � dS �q
����� �
Therefore� the electric �eld inside the hollow conductor is non�zero� Note� the electric �eld outsidethe conductor due to a point source inside is in�uenced by the shape of the conductor� as you canin the next problem�
c� The Direction of the Electric Field outside a Conductor
An electrostatic �eld is conservative� Therefore� the circulation of E around any closed path iszero I
E � dl � � �����
This is called the circuital law for E �E���� or J����� I have drawn a closed path in four legs
12
3
4
Figure ��� Electric �eld 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 ��gure ���� Sections � and � can be chosennegligible small� Also� we have seen earlier that the �eld in the conductor �section � is zero� For
���� CHARGED SPHERES �
the total integral around the closed path to be zero� the tangential component �section � has tobe zero� Therefore� the electric �eld is described by
E ��
���r ����
where � is the surface charge density� since � as shown earlier � free charge in a conductor is locatedon the surface�
��� Charged Spheres
Here we have a conducting� a homogeneously charged and an in�homogeneously charged sphere�Their total charge is Q� Finding the electric �eld for each case in� and outside the sphere is anexercise in using Gauss� law I
E � dS �q
�������
For all cases�
� Problem ���c showed that the electric �eld is directed radially outward from the center ofthe spheres�
� For r � a� E behaves as if caused by a point charge of magnitude of the total charge Q ofthe sphere� at the origin�
E �Q
����r��r
As we have seen earlier� for a solid spherical conductor the electric �eld inside is zero �see �gure�� �� For a sphere with a homogeneous charge distribution the electric �eld at points inside thesphere increases with r� As the surface S increases� the amount of charge surrounded increases�see equation �����
E �Qr
����a��r �����
For points inside a sphere with an inhomogeneous charge distribution� we use Gauss� law �onceagain�
E��r� � ����
Z ��
�
Z �
�
Z r
�
��r��r��sin��dr�d�d�� �����
Implementing the volume charge distribution
��r�� � ��r�n�
the integration over r� for n � � is straightforward�
E ���r
n��
���n� ��r� �����
� CHAPTER �� INTRODUCTION TO ELECTROSTATICS
adistance from the origin
0
elec
tric
fiel
d st
reng
th
homogeneously chargedinhomogeneously charged: n=2inhomogeneously charged: n=−2conductor
Figure �� � Electric �eld for di�erently charged spheres of radius a� The electric �eld outside the
spheres is the same for all� since the total charge is Q in all cases�
where
Q �
Z a
�
����rn��dr �
����an��
n� �
�� �Q�n� �
��an��������
It can easily be veri�ed that for n � �� we have the case of the homogeneously charged sphere�equation ����� The electric �eld as a function of distance are plotted in �gure �� for the conductor�the homogeneously charged sphere and in�homogeneously charged spheres with n � �� �
��� Charge Density for a Hydrogen Atom
The potential of a neutral hydrogen atom is
��r� �q
����
e��r
r
�� �
r
�������
where equals divided by the Bohr radius� If we calculate the Laplacian� we obtain the volumecharge density �� through Poisson�s equation
�
��� r��
���� CHARGED CYLINDRICAL CONDUCTORS �
Using the Laplacian for spherical coordinates �see back�cover of Jackson�� the result for r � � is
��r� � � �q
��r�e��r �����
For the case of r � �limr��
��r� � limr��
q
����r���� �
From section ��� in Jackson we have �J�� ���
r����r� � �����r� ������
Combining ���� �� ������ and Poisson�s equation� we get for r � �
��r� � q��r� �����
We can multiply the right side of equation ����� by e��r without consequences� This allows fora more elegant way of writing the discrete and the continuous parts together
��r� �
���r� � �
��r�
�qe��r ������
The discrete part represents the stationary proton with charge q� Around the proton orbits anelectron with charge �q� The continuous part of the charge density function is more a statisticaldistribution of the location of the electron�
��� Charged Cylindrical Conductors
Two very long cylindrical conductors� separated by a distance d� form a capacitor� Cylinder � hassurface charge density � and radius a�� and number has surface charge density �� and radius a��see �gure ����� The electric �eld for each of the cylinders is radially directed outward
-λ
0 d
aa
1
2
P
λ
Figure ���� Top view of two cylindrical conductors� The point P is located in the plane connecting
the axes of the cylinders�
E ��
���jrj �r� ������
�� CHAPTER �� INTRODUCTION TO ELECTROSTATICS
where �r is the radially directed outward unit vector� Taking a point P on the plane connectingthe axes of the cylinders� the electric �eld is constructed by superposition�
E ��
���
��
r�
�
d� r
�������
The potential di�erence between the two cylinders is
jVa� � Va� j ��
���
Z a�
d�a�
��
r�
�
d� r
�dr
��
����lnr � ln�d� r��
a�d�a�
��
����lna� � ln�d� a�� � ln�d� a��� lna�� ������
If we average the radii of the cylinders to a� � a� � a and assume d � a� then the potentialdi�erence is
jVa� � Va� j ��
���
�lnd
a
������
The capacitance per unit length of the system of cylinders is given by
C ��
jVa� � Va� j� ���
ln�d�a������
From here� we can obtain the diameter � of wire necessary to have a certain capacitance C at adistance d�
� � a � d � e����
C � ����
where the permittivity in free space �� is � �� � �����F�m� If C � � � �����F�m and
� d � � cm� the diameter of the wire is � � cm�
� d � � cm� the diameter of the wire is � cm�
� d � cm� the diameter of the wire is � cm�
���� Green�s Reciprocity Theorem
Two in�nite grounded parallel conducting plates are separated by a distance d� What is the inducedcharge 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 �rst is the situation as sketched byJackson! two in�nitely large grounded conducting plates� one at x � � and one at x � d �see theright side of �gure ���� In between the plates there is a point charge q at x � x�� We will applyGreen�s reciprocity theorem using a �mirror� set�up� In this geometry there is no point charge butthe plates have a �xed potential �� and ��� respectively �see the left side of �gure ����
����� GREEN�S RECIPROCITY THEOREM ��
| | |
q
Plate 1Plate 2
φ1φ
2
x00 d
S2S1
S3
S4| |
ψ2ψ1
Plate 1Plate 2
0 d
S2S1
S3
S4
Figure ��� 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 on the plates�
Green�s theorem �J�� � statesZV
�r�� � �r�
�dV �
IS
���
�n� �
�
�n
�dS ��� �
The volume V is the space between the plates bounded by the surface S� S� and S� bound theplates and S� and S run from plate � to plate at � and �� respectively� The normal derivative��n at the surface S is directed outward from inside the volume V�
When the plates are grounded� the potential in the plates is zero� The potential is continuousacross the boundary� so on S� and S � �� Note that if the potential was not continuous theelectric �eld �E � �r� would go to in�nity� At in�nite distance from the point source� thepotential is also zero� I
S
��
�ndS � � �����
The remaining part of the surface integral can be modi�ed according to Jackson� page ��
�
�n� r � n � �E � n ����
The electric �eld across a boundary with surface charge density � is �Jackson� equation ����
n � �Econductor �Evoid� � ���� �����
However� inside the conductor E � �� therefore
n �Evoid � ������ �����
� CHAPTER �� INTRODUCTION TO ELECTROSTATICS
for each of the plates� The total surface integral in equation �� is thenIS
��
�ndS �
ZS�
������dS�
ZS�
������dS �����
In case of the plates of �xed potential �� the legs S� and S have opposite potential and thuscancel� Using Z
S
�dS � Q� �����
the surface integral of Greens theorem isIS
��
�ndS � �������QS� � ��QS�� ��� ��
In the volume integral in equation ��� �� the mirror case of the charged boundaries includes nofree charges�
r�� � ��total charge���� � � ��� ��
Applying Gauss� law to the case with the point charge gives us
r� � ��total charge���� � �q�����x� x��� ��� �
where x� is the x�coordinate of the point source location� The volume integral will beZV
�q�����x� x����x�dV � �q�����x�� ��� �
For two plates with �xed potentials� the potential in between is a linear function
��x�� � �� � ��� � ��
d���� x��d � x��� � ����� x�� ��� ��
Green�s theorem is now reduced to equation ��� �� and equation ��� �� in ��� ��
�q�x��� � ��� x����� � QS��� �QS��� ��� �
Since this equality must hold for all potentials� the charges on the plates must be
QS� � �qx� and QS� � �q��� x�� ��� ��
Chapter �
Boundary�Value Problems in
Electrostatics� �
� The Method of Image Charges
a� The Potential Inside the Sphere
This problem is similar to the example shown on pages �� � and �� of Jackson� The electric �elddue to a point charge q inside a grounded conducting spherical shell can also be created by thepoint charge and an image charge q� only� For reasons of symmetry it is evident that q� is locatedon the line connecting the origin and q� The goal is to �nd the location and the magnitude of theimage charge� The electric �eld can then be described by superposition of point charges�
��x� �q
����jx� yj �q�
����jx� y�j ����
In �gure �� you can see that x is the vector connecting origin and observation point� y connectsthe origin and the unit charge q� Finally� y� is the connection between the origin and the imagecharge q�� Next� we write the vectors in terms of a scalar times their unit vector and factor thescalars y� and x out of the denominators�
��x� �q�����
xjn� yxn�j �
q������y�jn� � x
y�nj ���
The potential for x � a is zero� for all possible combinations of n �n�� The magnitude of the imagecharge is
q� � �ayq �� �
at distance
y� �a�
y����
�
�� CHAPTER �� BOUNDARYVALUE PROBLEMS IN ELECTROSTATICS �
q q’y y’θ
a
0
Px
Figure ��� 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 �likein the Jackson example�� After implementing the amount of charge �� � and the location of theimage ���� in ����� potential in polar coordinates is
��r� �� �q
����
�BB �p
y� � r� � yrcos��
�ay
�r�
a�
y
��� r� �
�a�
y
�rcos�
CCA � ���
where � is the angle between the line connecting the origin and the charges and the line connectingthe origin and the point P �see �gure ���� r is the length of the vector connecting the origin andobservation point P �
b� The Induced Surface Charge Density
The surface charge density on the sphere is
� � ��
���
�r
�r�a
����
Di�erentiating equation ��� is left to the reader� but the result is
� � � q
��a�
�a
y
� ���ay
���� �
�ay
��� ay cos�
��������
���� AN EXERCISE IN GREEN�S THEOREM �
c� The Force on the Point Charge q
The force on the point charge q by the �eld of the induced charges on the conductor is equal tothe force on q due to the �eld of the image charge�
F � qE� ����
The electric �eld at y due to the image charge at y� is directed towards the origin and of magnitude
jE�j � q�
�����y� � y������
We already computed the values for y� and q� in equation ���� and �� �� respectively� The forceis also directed towards the origin of magnitude
jFj � �
����
q�
a�
�a
y
����
�a
y
�����
�����
d� What If the Conductor Is Charged�
Keeping the sphere at a �xed non�zero potential requires net charge on the conducting shell� Thiscan be imaged as an extra image charge at the center of the spherical shell� If we now computethe force on the conductor by means of the images� the result will di�er from section c�
�� An Exercise in Green�s Theorem
a� The Green Function
The Green function for a half�space �z � �� with Dirichlet boundary conditions can be found by themethod of images� The potential �eld of a point source of unit magnitude at z� from an in�nitelylarge grounded plate in the x�y plane can be replaced by an image geometry with the unit chargeq and an additional �image� charge at z � �z� of magnitude q� � �q� The situation is sketched in�gure �� The potential due to the two charges is the Green function GD�
�p�x � x��� � �y � y��� � �z � z���
� �p�x � x��� � �y � y��� � �z � z���
�����
b� The Potential
The Green function as de�ned in equation ����� can serve as the �mirror set�up� required inGreen�s theorem� Z
V
�r�� � �r�
�dV �
IS
���
�n� �
�
�n
�dS ����
�� CHAPTER �� BOUNDARYVALUE PROBLEMS IN ELECTROSTATICS �
’z
aV
z
x
P
q
φ
rθ
q’
yz
Figure �� A very large grounded surface in which a circular shape is cut out and replaced by aconducting material of potential V �
with GD � � and � � � There are no charges the volume z � �� so Laplace equation holdsthroughout the half�space V�
r�� � � ��� �
Chapter one in Jackson �J�� �� showed
r�GD � ������� ��� �����
leaving ����� 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� andthe other parts stretch to in�nity� ThereforeI
S
GDdS � � ����
The potential � � V in the circular area with radius a� but everywhere else � � �� Also�IS
rGD � �ndS � �IS
�GD
�zdS� �����
since the normal �n is in the negative z�direction ��k� Thus we are left with the following remainingterms in Green�s theorem �in cylindrical coordinates��
��r�� � ��V
Z a
�
Z ��
�
�GD
�z�d�d �����
From here on we will exchange the primed and unprimed coordinates� This is OK� since thereciprocity theorem applies� Some algebra left to the reader leads to
��r� �V z
�
Z a
�
Z ��
�
��d��d�
��� � ��� � z� � ���cos�� �����������
���� TWO HALVES OF A CONDUCTING SPHERICAL SHELL ��
c� The Potential on the z�axis
For � � �� general solution ����� simpli�es to
���� � �V z
�
Z a
�
Z ��
�
��d��d�
���� � z�����
� �V z
�p��� � z�
�a�
� V
��� zp
a� � z�
������
d� An Approximation
Slightly rewriting equation ������
��r� �V z
� ��� � z�����
Z a
�
Z ��
�
��d��d��� � ��������cos������
���z�
���� ����
The denominator in the integral can be approximated by a binomial expansion� The �rst threeterms of the approximation give
��r� � V a�z
��� � z�����
��� a�
� ��� � z���
a
� ��� � z��� �
�a���
� ��� � z���
�����
Along the axis �� � �� the expression simpli�es to
��� z� � V a�
z�
��� a�
�z��
a
�z
����
This is the same result when we expand expression ������
���� � � V
��� zp
a� � z�
�
� V
���
�� �
a�
z�
������
� V a�
z�
��� a�
�z��
a
�z
��� �
� Two Halves of a Conducting Spherical Shell
A conducting spherical shell consists of two halves� The cut plane is perpendicular to the homoge�neous �eld �see �gure � �� The goal is to investigate the force between the two halves introduced
�� CHAPTER �� BOUNDARYVALUE PROBLEMS IN ELECTROSTATICS �
r̂
zE0
a
E0
Figure � � A spherical conducting shell in a homogeneous electric �eld directed in the z�direction�
by the induced charges�
The electric �eld due to the induced charges on the shell is �see ��� p� ���
Eind ��
���r ����
You can see this as the resulting �eld in a capacitor with one of the plates at in�nity� The electric�eld inside the conducting shell is zero� Therefore the external �eld has to be of the same magnitude�see �gure ����
The force of the external �eld on an elementary surface dS of the conductor is�
dF � Eextdq ���dS
��� ���
directed radially outward from the sphere�s center� From the symmetry we can see that all forcescancel� except the component in the direction of the external �eld��
a� An Uncharged Shell
The derivation of the induced charge density on a conducting spherical shell in a homogeneouselectrical �eld E� is given �� �� p� ���� The homogeneous �eld is portrayed by point charges ofopposite magnitude at � and � in�nity� Next� the location and magnitude of the image chargesare computed� The result is
���� � ��E�cos� ����
�The external �eld is a superposition of the homogeneous electric �eld plus the electric �eld due to the inducedcharges excluding dq� This external �eld is perpendicular to the surface of the conductor with magnitude ��������This is not addressed in Jackson�
���� TWO HALVES OF A CONDUCTING SPHERICAL SHELL ��
cavity
E
E E
E
ind ind
extext
Figure ��� Zooming in on that small part of the conductor with induced charges� where theexternal �eld is at normal incidence�
When we plug this result into equation ���� we get for the horizontal component of the force�dFz� on an elementary surface�
dFz � dFcos� ��
��E
��cos
��dS ����
Now we can integrate to get the total force on the sphere halves� From symmetry we can also seethat the force on the left half is opposite of that on the right half �see �gure � �� So we integrateover the right half and multiply by two to get the total net force�
Fz �
Z ��
�
Z ���
�
�
��E
��cos
��a�sin�d�d �z
� ��a���E��
Z ���
�
cos��sin�d� �z
� ��a���E��
�����cos������
�z
��
��a���E
���z ����
b� A Shell with Total Charge Q
When the shell has a total charge Q it changes the charge density of equation ���� to
���� � ��E�cos� �Q
��a�����
� CHAPTER �� BOUNDARYVALUE PROBLEMS IN ELECTROSTATICS �
When we plug this expression into equation ��� and compute again the net �horizontal� com�ponent of the force� we �nd that
Fz �
��
��a���E
�� �
Q�
���a��E�Q
��z �� ��
The total force is bigger then for the uncharged case� This makes sense when we look at equation���! when the shell is charged there is more charge per unit volume to� hence the force is bigger�
��� A Conducting Plate with a Boss
a� � On the Boss
By inspection it can be seen that the system of images as proposed in �gure ��� �ts the geometryand the boundary conditions of our problem� We can write the potential as a function of thesefour point charges� This is done in Jackson �p� � �� It has to be noted that R has to be chosen atin�nity to apply to the homogeneous character of the �eld� The potential can then be describedby expanding �the radicals after factoring out the R���
��r� �� �Q
����
��
R�rcos� �
a�
R�r�cos�
��
� �E�
�r � a�
r�
�cos� �� ��
The surface charge density on the boss �r � a� is
� � � ����
�r
����r�a
� ��E�cos� �� �
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�
Q � ��E�
Z ��
�
Z ���
�
cos�a�sin�d�d
� ��E��a�
��
sin��
�����
� ��E��a� �� �
����� A CONDUCTING PLATE WITH A BOSS �
E
E
0
0V0
0
az
rP
θ
D
0
a
qq’z
-q’
rP
θ
R
-q
Figure �� 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 �eld between the two plates
is E�� On the right is the set of charges that image the �eld due to the conducting plates� In part
a and b� R� to image a homogeneous �eld� In c� R � d�
c� The Charge On the Boss Due To a Point Charge
Now we do not have a homogeneous �eld to image� but the result of a point charge on a groundedconducting plate with the boss� Again we use the method of images to replace the system withthe plate by one entirely consisting of point charges� Checking the boundary conditions leads tothe same set of four charges as drawn in �gure �� The only di�erence is that R is not chosenat in�nity to mimic the homogeneous �eld� but R � d� The potential is the superposition of thepoint charge q at distance d and its three image charges�
��r� �� �q
����
��
�r� � d� � rdcos������ �
�r� � d� � rdcos�����
� a
d�r� � a�
d� � �a�rd cos�
���� �a
d�r� � a�
d� � �a�rd cos�
����A �� ��
The charge density on the boss is
� � ��� ��
�r
����r�a
�� �
The total amount of charge is the surface charge density integrated over the surface of the boss�
Q � �a�Z ���
�
�sin�d� �� ��
CHAPTER �� BOUNDARYVALUE PROBLEMS IN ELECTROSTATICS �
The di�erentiation of equation �� �� to obtain the the surface charge density and the followingintegration in equation �� �� are left to the reader� The resulting total charge is�
Q � �q��� d� � a�
dpd� � a�
��� ��
��� 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 b�
R �see �gure ��� satis�es the conditions the boundary conditions
limr��
��r� � � � and
��b� � � V�
b
V0
P
τ τ-
r rr2
1
φ
R
Figure ��� A cross sectional view of a long cylinder at potential V� and a line charge � at distance
R� parallel to the axis of the cylinder� The image line charge �� is placed at b��R from the axis
of the cylinder to realize a constant potential V� at radius r � b�
�I have chosen to keep Q as the symbol for the total charge� Jackson calls it q�� I �nd this confusing since theprimed q has been used for the image of q�
����� LINE CHARGES AND THE METHOD OF IMAGES
b� The Potential
The potential in polar coordinates is simply a superposition of the line charge � and the image linecharge � � with the conditions as proposed in section a� The result is
��r� � ��
����ln
��R�r� � b � rRb�cos�
R��r� �R� � Rrcos�
��� ��
For the far �eld case �r �� R� we can factor out �Rr��� The b and R� in equation �� �� can beneglected�
��r� � � �
����ln
��Rr����� �b�
Rr cos�
�Rr����� �Rr cos�
��� ��
The �rst order Taylor expansion is
��r� � � �
����ln
���� b�
Rrcos��� �
R
rcos�
�
� �
����ln
��� �
�R
rcos� b�
Rrcos�
��
� �
���
�R� � b��
Rrcos �using ln�� � x� � x� �����
c� The Induced Surface Charge Density
��� � ������r
����r�b
�����
Di�erentiation of equation �� �� and substituting r � b�
��� � � �
��
�bR� � Rb�cos
R�b� �R� � b � b�Rcos� R��b� Rcos
R��b� �R� � Rbcos
�
� � �
�
��R�b�� � �
�R�b�� � �� �R�b�cos
�����
When R�b � � the induced charge as a function of is
���jR��b � � �
�b
�
� �cos
���� �
When the position of the line charge � is four radii from the center of the cylinder� the surfacecharge density is
���jR�b � � �
�b
��
��� �cos
������
The graphs for either case are drawn in �gure ���
� CHAPTER �� BOUNDARYVALUE PROBLEMS IN ELECTROSTATICS �
1 2 3 4 5 6
-3
-2.5
-2
-1.5
-1
-0.5φ(rad)
τ /2πb( )σ
R/b=2
R/b=4
Figure ��� The behavior of the surface charge density � with angle for two di�erent ratios betweenthe 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�
F � �E�R� �� � � ��
���
�
�R� � b���� per meter ����
where �� is the directed from the the axis of the cylinder to the line charge� perpendicular to theline charge and the cylinder axis�
��� Two Cylinder Halves at Constant Potentials
a� The Potential inside the Cylinder
In this case �see �gure ��� there are no free charges in the area of interest� Therefore the potential� inside the cylinder obeys Laplace�s equation�
r�� � � �����
We can write the potential in cylindrical coordinates and separate the variables�
���� � � R�r�F �� �����
����� TWO CYLINDER HALVES AT CONSTANT POTENTIALS
The general solution is �see J�����
���� � � a� � b�ln��
�Xn��
�an�
nsin�n� n� � bn��ncos�n� n�
������
From this geometry it is obvious that at the center � � � the solution may not blow up� so�
bn � b� � � �����
This results in a potential
���� � � a� �
�Xn��
an�nsin�n� n� ����
The next step is to implement the boundary conditions
V1
b
P
r
V2
φ
Figure ��� Cross�section of two cylinder halves with radius b at constant potentials V� and V��
��b� � � V� � a� �
�Xn��
anbnsin�n� n� for ���� � � ���
��b� � � V� � a� ��Xn��
anbnsin�n� n� for ��� � � ��� ����
b� The Surface Charge Density
� � � ����
�r
����r�b
���
� CHAPTER �� BOUNDARYVALUE PROBLEMS IN ELECTROSTATICS �
�� A Hollow Cubical Conductor
a� The Potential inside the Cube
z
y
x
a
a
a
V
Vz
z
Figure ��� A hollow cube� with all sides but z�� and z�a grounded�
r�� � � �� �
Separating the variables��
X
d�X
dx��
�
Y
d�Y
dy��
�
Z
d�Z
dz�� � ����
x and y can vary independently so each term must be equal to a constant ����
X
d�X
dx�� � � � X � Acosx�Bsinx ���
�
Y
d�Y
dy�� �� � � Y � Ccos�y �Dsin�y ����
�
Z
d�Z
dz�� �� � � Z � Esinh��z� � F cosh��z�� ����
where �� � � � ��� The boundary conditions determine the constants�
���� y� z� � � A � �
��a� y� z� � � n � n��a �n � �� � � �
����� A HOLLOW CUBICAL CONDUCTOR �
��x� �� z� � � C � �
��x� a� z� � � �m � m��a �m � �� � � �
�nm � �pn� �m�
The solution is thus reduced to
��x� y� z� �
�Xn�m��
sin�n�x
a
�sin
�m�y
a
� hAnmsinh
��nmza
��Bnmcosh
��nmza
�i����
Now� let�s use the last boundary conditions to �nd the coe�cients Anm and Bnm� The top andbottom of the cube are held at a constant potential Vz � so
��x� y� �� � Vz �
�Xn�m��
Bnmsin�n�x
a
�sin
�m�y
a
�����
This means that Bnm are merely the coe�cients of a double Fourier series �see for instance ��� onFourier series��
Bnm ��Vza�
Z a
�
Z a
�
sin�n�x
a
�sin
�m�y
a
�dxdy �����
It can be easily shown that the individual integrals in equation ����� are zero for even integervalues and �a
n� for n is odd� Thus Bnm is
Bnm ���Vz��nm
for odd �n�m� �����
The top of the cube is also at constant potential Vz� so
��x� y� �� � Vz � ��x� y� a��Bnm � Anmsinh��nm� �Bnmcosh��nm��Anm � Bnm
�� cosh��nm�
sinh��nm�����
Substituting the expressions for Anm and Bnm into equation ����� gives us
��x� y� z� ���Vz��
�Xn�m odd
�
nmsin
�n�xa
�sin
�m�y
a
���� cosh��nm�
sinh ��nm�sinh
��nmza
�
� cosh��nmz
a
�i� ��� �
where �nm � �pn� �m��
b� The Potential at the Center of the Cube
The potential at the center of the cube is
��a
�a
�a
� �
��Vz��
�Xn�m odd
�
nmsin
�n�
�sin
�m�
���� cosh��nm�
sinh ��nm�sinh
��nm
�
� cosh��nm
�i�����
� CHAPTER �� BOUNDARYVALUE PROBLEMS IN ELECTROSTATICS �
With just n�m � �� the potential at the center is
��Vz��
�� cosh�
p��
sinh�p��
sinh
��p
�� cosh
��p
��� � ����Vz ����
When we add the two terms �n � � m � �� and �n � �� m � �� the potential is � ���Vz�
c� The Surface Charge Density
The surface charge density on the top surface of the cube is given by
� � ������z
����z�a
�����
In the appendix it is shown that the di�erentiation of the hyperbolic sine is the hyperbolic cosine�Furthermore
dcosh�az�
dz� asinh�az� �����
Using this equality in di�erentiating the expression for the potential in equation ��� �� we get
��
�z�
��Vz��
�Xn�m odd
�nmnma
sin�n�x
a
�sin
�m�y
a
���� cosh��nm�
sinh ��nm�cosh
��nmza
�
� sinh��nmz
a
�i� �����
where �nm � �pn� �m�� Now we evaluate this expression for z � a�
� � ������z
����z�a
�
�����Vz��
�Xn�m odd
�nmnma
sin�n�x
a
�sin
�m�y
a
���� cosh��nm�
sinh ��nm�cosh ��nm� � sinh ��nm�
��
�����Vz��
�Xn�m odd
�nmnma
sin�n�x
a
�sin
�m�y
a
����� cosh��nm��coth��nm� � sinh ��nm�� �����
Further simpli�cation
Chapter �
Practice Problems
��� Angle between Two Coplanar Dipoles
p2
p1
ErEθ
θ’
θ
r
Figure ��� Coplanar dipoles separated by a distance r�
Two dipoles are separated by a distance r� Dipole p� is �xed at an angle � as de�ned in �gure ��� while p� is free to rotate� The orientation of the latter is de�ned by the angle �� as de�ned in�gure ��� Let us calculate the angular dependence between the two dipoles in equilibrium�
The electric �eld due to a dipole can be decomposed into a radial and a tangential component�D � ���
E�r� �� �pcos�
����r��r�
psin�
����r���� � ���
where �� is the dielectric permittivity� Writing p� in terms of r and ��
p��r� �� � �p� � �r��r� �p� � �� � p�cos���r� p�sin�
��� � ��
The potential energy of the second dipole in the electric �eld due to the �rst dipole is
U��r� �� ��� � �E� � p� � � p�p�
����r��cos�cos�� � sin�sin��� � � �
�
� CHAPTER �� PRACTICE PROBLEMS
The second dipole will rotate to minimize its potential energy� de�ning the angular dependencebetween the two dipoles�
�U�
���� � �
p�p�����r�
�cos�sin�� � sin�cos��� � � �cos�sin�� � �sin�cos�� �
tan�� � ��tan��� � ���
�� The Potential in Multipole Moments
ba
+q
-q
x
y
z
Figure �� Concentric rings of radii a and b� Their charge is q and �q� respectively�
This exercise is how to �nd the potential due to two charged concentric rings �see �gure �� interms of the monopole dipole and quadrupole moments� Discarding higher order moments �J������
��r� �q
����r�
p � r����r�
��
����r
Xi�j
xixjQij � � ��
The monopole moment is zero since there is no net charge� The dipole moment is �J�����
p �
ZV
r��r�dV� � ���
���� POTENTIAL BY TAYLOR EXPANSION �
where the volume charge density for our case can be written in cylindrical coordinates�
��r� �q
�a��r � a���z�� q
�b��r � b���z� � ���
After the x�component of the dipole moment is also written in cylindrical coordinates� the integralcan be easily evaluated�
px �
ZV
x��r� z�dV �
Z�
��
Z ��
�
Z�
�
cos��r� z�r�drddz � �� � ���
because the cos 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 �Zz��z�dz � � � ���
if the value � is within the integration limits�
The quadrupole moment is a tensor�
Qij �
ZV
� xixj � r����r�dV � ����
We use the following properties of the ��function�Z��z�dz � � andZr��r � a�dr � a� � ����
where � and a are within the respective integral limits� Knowing this� the calculations for eachelement of Q is left to the reader� After some algebra� the quadrupole moment turns out to be
Qij �
� � � �
� � �� � �
A q
�a� � b�
�� ���
This means the potential of the two charged concentric rings is approximately
��r� � �a� � b��
�����r�x� � y� � z�� � �� �
��� Potential by Taylor Expansion
Here� I will show that the electrostatic potential ��x� y� z� can be approximated by the average ofthe potentials at the positions perturbed by a small quantity ��� a by doing a Taylor expansion�
CHAPTER �� PRACTICE PROBLEMS
This expansion is correct to the third order� First� the Taylor expansion around the ��x� y� z�perturbed in the positive x�component
��x� a� y� z� � ��x� y� z� ����x� y� z�
�xa�
����x� y� z�
�x�a� �
����x� y� z�
�x�a� �O�a� � ����
Next� the same expansion around ��x� a� y� z��
��x� a� y� z� � ��x� y� z�� ���x� y� z�
�xa�
����x� y� z�
�x�a� �
����x� y� z�
�x�a� �O�a� � ���
When we add up these two equations� the odd powers of a cancel� This is the same for the y� andz�component� The a��term adds up to the Laplacian r�� Assuming there is no charge within theradius a of �x� y� z�� Laplace�s equation holds�
r�� � � � ����
and thus the a� term is zero� too� Therefore� the potential at �x� y� z� can be given by�
��x� y� z� � ������x� a� y� z� � ��x� a� y� z� � ��x� y � a� z�
���x� y � a� z� � ��x� y� z � a� � ��x� y� z � a�� �O�a� � ����
Appendix A
Mathematical Tools
A�� Partial integration
Z b
a
f�x�g��x�dx � �f�x�g�x��ba �
Z b
a
f ��x�g�x�dx �A���
A� Vector analysis
Stokes� theorem
IL
M � dl �ZS
curlM � dS �A��
Gauss� theorem
IS
M � dS �
ZV
divM � dV �A� �
Computation of the curl
curlM ��
h�h�h�
���������
hi�i h��j h��k
��x�
��x�
��x�
h�M� h�M� h�M�
����������A���
� APPENDIX A� MATHEMATICAL TOOLS
hi are the geometrical components that depend on the coordinate system� For a Carthesiancoordinate system they are one� For the cylindrical sytem�
h� � �
h� � r
h� � �
For the spherical coordinate system�
h� � �
h� � r
h� � rsin�
Computation of the divergence
divM ��
h�h�h�
��
�x��h�h�M�� �
�
�x��h�h�M�� �
�
�x��h�h�M��
��A��
With the same factors hi depending on the coordinate system�
Relations between grad and div
r�r�M � rr �M�r�M �A���
Any vector can be written in terms of a vector potential and a scalar potential�
M � r��r�A �A���
A�� Expansions
Taylor series
f�x� � f�a� � �x� a�f ��a� ��x � a��f ���a�
"� �
xn
n"�x� a�nf �n��a� �A���
A�� Euler Formula
ei� � cos� isin �A���
A� � TRIGONOMETRY
sinx �eix � e�ix
i�A����
cosx �eix � e�ix
�A����
sinhx �ex � e�x
�A���
coshx �ex � e�x
�A�� �
A�� Trigonometry
sin�x ��� cos�x�
�A����
cos�� �� � Rehei�����
i� Re
�ei�e�i�
�� Re �coscos� � sinsin� � i� ��
� coscos� � sinsin� �A���
� APPENDIX A� MATHEMATICAL TOOLS
Bibliography
��� R� Bracewell� The Fourier Transform and Its Applications� McGraw�Hill Book Company� �rstedition� ����
�� W�J� Du�n� Electricity and Magnetism� McGraw�Hill Book Company� fourth edition� �����
� � J�D� Jackson� Classical Electrodynamics� John Wiley # Sons� Inc�� third edition� �����
�