37

Selection Problems - Answers To a...Answ ers T o a Selection of Problems from Classical Electro dynamics John David Jackson b y Kasp er v an Wijk Cen ter for W a v e Phenomena Departmen

Embed Size (px)

Citation preview

Page 1: Selection Problems - Answers To a...Answ ers T o a Selection of Problems from Classical Electro dynamics John David Jackson b y Kasp er v an Wijk Cen ter for W a v e Phenomena Departmen

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

Page 2: Selection Problems - Answers To a...Answ ers T o a Selection of Problems from Classical Electro dynamics John David Jackson b y Kasp er v an Wijk Cen ter for W a v e Phenomena Departmen

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�

Page 3: Selection Problems - Answers To a...Answ ers T o a Selection of Problems from Classical Electro dynamics John David Jackson b y Kasp er v an Wijk Cen ter for W a v e Phenomena Departmen

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 � � � � � � � � � � � � � � � � � � � � � � � � � � � � � �

Page 4: Selection Problems - Answers To a...Answ ers T o a Selection of Problems from Classical Electro dynamics John David Jackson b y Kasp er v an Wijk Cen ter for W a v e Phenomena Departmen

� 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�

Page 5: Selection Problems - Answers To a...Answ ers T o a Selection of Problems from Classical Electro dynamics John David Jackson b y Kasp er v an Wijk Cen ter for W a v e Phenomena Departmen

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�

Page 6: Selection Problems - Answers To a...Answ ers T o a Selection of Problems from Classical Electro dynamics John David Jackson b y Kasp er v an Wijk Cen ter for W a v e Phenomena Departmen

� CHAPTER �� INTRODUCTION TO ELECTROSTATICS

qq

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

Page 7: Selection Problems - Answers To a...Answ ers T o a Selection of Problems from Classical Electro dynamics John David Jackson b y Kasp er v an Wijk Cen ter for W a v e Phenomena Departmen

���� 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� �����

Page 8: Selection Problems - Answers To a...Answ ers T o a Selection of Problems from Classical Electro dynamics John David Jackson b y Kasp er v an Wijk Cen ter for W a v e Phenomena Departmen

� 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��

Page 9: Selection Problems - Answers To a...Answ ers T o a Selection of Problems from Classical Electro dynamics John David Jackson b y Kasp er v an Wijk Cen ter for W a v e Phenomena Departmen

���� 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

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� ������

Page 10: Selection Problems - Answers To a...Answ ers T o a Selection of Problems from Classical Electro dynamics John David Jackson b y Kasp er v an Wijk Cen ter for W a v e Phenomena Departmen

�� 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 ����

Page 11: Selection Problems - Answers To a...Answ ers T o a Selection of Problems from Classical Electro dynamics John David Jackson b y Kasp er v an Wijk Cen ter for W a v e Phenomena Departmen

����� 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 � ������ �����

Page 12: Selection Problems - Answers To a...Answ ers T o a Selection of Problems from Classical Electro dynamics John David Jackson b y Kasp er v an Wijk Cen ter for W a v e Phenomena Departmen

� 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�� ��� ��

Page 13: Selection Problems - Answers To a...Answ ers T o a Selection of Problems from Classical Electro dynamics John David Jackson b y Kasp er v an Wijk Cen ter for W a v e Phenomena Departmen

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����

Page 14: Selection Problems - Answers To a...Answ ers T o a Selection of Problems from Classical Electro dynamics John David Jackson b y Kasp er v an Wijk Cen ter for W a v e Phenomena Departmen

�� 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�

��������

Page 15: Selection Problems - Answers To a...Answ ers T o a Selection of Problems from Classical Electro dynamics John David Jackson b y Kasp er v an Wijk Cen ter for W a v e Phenomena Departmen

���� 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 ����

Page 16: Selection Problems - Answers To a...Answ ers T o a Selection of Problems from Classical Electro dynamics John David Jackson b y Kasp er v an Wijk Cen ter for W a v e Phenomena Departmen

�� CHAPTER �� BOUNDARYVALUE PROBLEMS IN ELECTROSTATICS �

’z

aV

z

x

P

q

φ

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�� �����������

Page 17: Selection Problems - Answers To a...Answ ers T o a Selection of Problems from Classical Electro dynamics John David Jackson b y Kasp er v an Wijk Cen ter for W a v e Phenomena Departmen

���� 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

Page 18: Selection Problems - Answers To a...Answ ers T o a Selection of Problems from Classical Electro dynamics John David Jackson b y Kasp er v an Wijk Cen ter for W a v e Phenomena Departmen

�� CHAPTER �� BOUNDARYVALUE PROBLEMS IN ELECTROSTATICS �

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�

Page 19: Selection Problems - Answers To a...Answ ers T o a Selection of Problems from Classical Electro dynamics John David Jackson b y Kasp er v an Wijk Cen ter for W a v e Phenomena Departmen

���� 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�����

Page 20: Selection Problems - Answers To a...Answ ers T o a Selection of Problems from Classical Electro dynamics John David Jackson b y Kasp er v an Wijk Cen ter for W a v e Phenomena Departmen

� 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� �� �

Page 21: Selection Problems - Answers To a...Answ ers T o a Selection of Problems from Classical Electro dynamics John David Jackson b y Kasp er v an Wijk Cen ter for W a v e Phenomena Departmen

����� 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� �� ��

Page 22: Selection Problems - Answers To a...Answ ers T o a Selection of Problems from Classical Electro dynamics John David Jackson b y Kasp er v an Wijk Cen ter for W a v e Phenomena Departmen

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�

Page 23: Selection Problems - Answers To a...Answ ers T o a Selection of Problems from Classical Electro dynamics John David Jackson b y Kasp er v an Wijk Cen ter for W a v e Phenomena Departmen

����� 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 ���

Page 24: Selection Problems - Answers To a...Answ ers T o a Selection of Problems from Classical Electro dynamics John David Jackson b y Kasp er v an Wijk Cen ter for W a v e Phenomena Departmen

� 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 �� �����

Page 25: Selection Problems - Answers To a...Answ ers T o a Selection of Problems from Classical Electro dynamics John David Jackson b y Kasp er v an Wijk Cen ter for W a v e Phenomena Departmen

����� 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

���

Page 26: Selection Problems - Answers To a...Answ ers T o a Selection of Problems from Classical Electro dynamics John David Jackson b y Kasp er v an Wijk Cen ter for W a v e Phenomena Departmen

� 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 � �� � � �

Page 27: Selection Problems - Answers To a...Answ ers T o a Selection of Problems from Classical Electro dynamics John David Jackson b y Kasp er v an Wijk Cen ter for W a v e Phenomena Departmen

����� 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�����

Page 28: Selection Problems - Answers To a...Answ ers T o a Selection of Problems from Classical Electro dynamics John David Jackson b y Kasp er v an Wijk Cen ter for W a v e Phenomena Departmen

� 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

Page 29: Selection Problems - Answers To a...Answ ers T o a Selection of Problems from Classical Electro dynamics John David Jackson b y Kasp er v an Wijk Cen ter for W a v e Phenomena Departmen

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��� � � �

Page 30: Selection Problems - Answers To a...Answ ers T o a Selection of Problems from Classical Electro dynamics John David Jackson b y Kasp er v an Wijk Cen ter for W a v e Phenomena Departmen

� 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� � ���

Page 31: Selection Problems - Answers To a...Answ ers T o a Selection of Problems from Classical Electro dynamics John David Jackson b y Kasp er v an Wijk Cen ter for W a v e Phenomena Departmen

���� 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�

Page 32: Selection Problems - Answers To a...Answ ers T o a Selection of Problems from Classical Electro dynamics John David Jackson b y Kasp er v an Wijk Cen ter for W a v e Phenomena Departmen

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� � ����

Page 33: Selection Problems - Answers To a...Answ ers T o a Selection of Problems from Classical Electro dynamics John David Jackson b y Kasp er v an Wijk Cen ter for W a v e Phenomena Departmen

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���

Page 34: Selection Problems - Answers To a...Answ ers T o a Selection of Problems from Classical Electro dynamics John David Jackson b y Kasp er v an Wijk Cen ter for W a v e Phenomena Departmen

� 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���

Page 35: Selection Problems - Answers To a...Answ ers T o a Selection of Problems from Classical Electro dynamics John David Jackson b y Kasp er v an Wijk Cen ter for W a v e Phenomena Departmen

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���

Page 36: Selection Problems - Answers To a...Answ ers T o a Selection of Problems from Classical Electro dynamics John David Jackson b y Kasp er v an Wijk Cen ter for W a v e Phenomena Departmen

� APPENDIX A� MATHEMATICAL TOOLS

Page 37: Selection Problems - Answers To a...Answ ers T o a Selection of Problems from Classical Electro dynamics John David Jackson b y Kasp er v an Wijk Cen ter for W a v e Phenomena Departmen

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� �����