Chapter 1

ELECTROSTATICS IELECTRIC FIELD AND SCALARPOTENTIAL

1.1 Introduction

Atoms and molecules under normal circumstances contain equal number of protons and electrons to

maintain macroscopic charge neutrality. However, charge neutrality can be disturbed rather easily

as we often experience in daily life. "Static electricity" induced when walking on a carpet and

caressing a cat is a familiar phenomenon, and is known as frictional (or tribo) electricity. Before

the invention of chemical batteries (Volta, 1786), electricity generation had been largely done by

frictional electricity generators. In nature, lightnings are caused by electrical discharges of electric

charges accumulated through friction among cloud particles. In frictional electricity, mechanical

disturbance given to otherwise charge neutral molecules either splits electrons o¤ a material body,

or adds them. For example, if an ebony rod is rubbed with a piece of fur, electrons are transfered

from the fur to the ebony, and the ebony rod becomes negatively charged, while the fur becomes

positively charged.

Charge neutrality can be disturbed by various other means, such as thermal, chemical, optical

and electromagnetic disturbances. It is well known that even a candle �ame is weakly ionized

(cuased by thermal ionization) and responds to an electric �eld. Chemical batteries are capable,

through chemical reactions, of separating charges. Some metals are known to emit electrons when

exposed to ultraviolet light (photoelectric e¤ect discovered by Hertz in 1897 and later given quantum

mechanical explanation by Einstein). Finally, as an example of electromagnetic means of charge

separation, the plasma (ionized gas) state in �uorescent lamps and plasma TV may be added to

the list.

Charged bodies exert electric forces (Coulomb force) on each other. Systematic experiments

on electric forces had been carried out by Cavendish and Coulomb in the 18th century and led

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to the establishment of the Coulomb�s law. Like charges repel each other, while unlike charges

attract each other. Therefore, in any attempt to separate charges out of an initially charge neutral

body, energy must be added. For example, in frictional electricity, mechanical energy is expended

to separate charges. In chemical batteries, chemical energy is converted into electric energy, and

in electric generators, either gravitational (as in hydropower generation) or thermal (as in steam

power plants) energy is converted into electric energy.

The Coulomb force to act between two charges is inversely proportional to r2 where r is the

separation distance between the charges. A hydrogen atom consists of one electron �revolving�

around a proton. The centripetal Coulomb force (attracting) is counterbalanced by the mechanical

centrifugal force, and the electron stays on a circular orbit. (This is a classical electrodynamic

model. For a more satisfactory description of a hydrogen atom, quantum mechanical analysis is

required. However, the concept of force balance is basically correct.) A nucleus of heavier atoms

contains more than one proton. The size of a nucleus is of order 10�15 m (compare this with the

atomic size 10�10 m). Therefore, among the protons packed in a nucleus, a tremendously large

repelling Coulomb force should act, and some other force, which is not of electric nature, must

keep protons together. This is provided by the so-called �strong�nuclear force which acts among

hadrons such as protons and neutrons. Obviously, the nature of nuclear force is beyond the realm of

classical electrodynamics. However, it should be realized that nuclear energy that can be released

when a heavy nucleus (such as U235) splits (nuclear �ssion process) is nothing but electric energy

stored in a nucleus.

Electrostatics is one branch of electrodynamics in which electric charges are either stationary

or moving su¢ ciently slowly so that magnetic �eld induction and electromagnetic radiation can be

ignored entirely. All basic laws in electrostatics (Gauss�law, Maxwell�s equations) follow from the

Coulomb�s law. Therefore, formulating the electric �eld and scalar potential in electrostatics will be

deduced from this fundamental law. It should be pointed out that electrostatics (together with mag-

netostatics) provides us with preparation for more general electrodynamics, electromagnetic wave

phenomena in particular. For example, electromagnetic wave propagation in a medium (including

vacuum) requires that the medium be able to store both electric and magnetic energy. Obviously,

electric energy storage leads us to the concept of capacitance. Calculation of the capacitance of a

given electrode system is one important application of electrostatics, but the concept of capacitance

(and inductance) will also play fundamental roles in dynamic electromagnetic phenomena.

1.2 Coulomb�s Law

In the late 1700�s, Cavendish and, independently, Coulomb carried out extensive research on static

electricity. (At that time, no batteries were available, and electricity generation was mainly done

with frictional electricity machines.) Apparently, Cavendish�s discovery of the inverse square law,

now known as Coulomb�s law, was made before Coulomb. However, Cavendish did not publish

his large amount of work on electricity, while Coulomb wrote several papers on electricity and

magnetism. Of course, Cavendish is best known for his gravitational torsion balance experiments

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which experimentally veri�ed the inverse square law of gravitational force originally postulated by

Newton. Both Cavendish and Coulomb used similar torsion balance apparatus to measure electric

forces acting between charged bodies.

Coulomb�s law is stated as follows. When two charges q1 and q2 are at a distance r from each

other, the electric force exerted on each other is proportional toq1q2r2,

F = const.q1q2r2: (1.1)

In the MKS-Ampere unit system (SI unit system), the charge is measured in Coulombs (C), the

distance in meters (m), and the force in Newtons (N). In these units, the constant experimentally

determined is

constant = 8:99� 109�N �m2

C2

�

Figure 1-1: Repelling Coulomb force between like charges.

The force is a vector quantity. In the case of the Coulomb force, the force is directed along the

separation distance vector r, and a more formal expression is given by

F = const.q1q2r2er (1.2)

where

er =r

r(1.3)

is the unit vector along the position vector r. Of course, both charges q1 and q2 experience aforce of the same magnitude, but oppositely directed. The sign of the product q1q2 can be either

positive or negative. When q1q2 > 0, the force is repelling, and when q1q2 < 0, the force is

attractive. In the 18th century when Cavendish and Coulomb conducted experiments, the presence

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of two kinds of charges, positive and negative, was known although their origin, namely charged

elementary particles, was clari�ed much later. The electron was discovered by Thomson in 1897.

The electronic charge presently established is �1:6� 10�19 Coulomb.The proportional constant in the Coulomb�s law corresponds to the force to act between two

equal charges, 1 C each, separated by a distance of 1 m. By arranging such an experimental

situation, the constant could be measured as done by Cavendish and Coulomb. (In more modern

methods, the velocity of light in vacuum provides an indirect measurement of the constant. This

will become clear later in Chapter 7.) In the MKS-Ampere unit system, the connection between

mechanical force (Newtons) and basic electromagnetic unit is actually made in terms of magnetic

force, rather than the Coulomb force, as will be explained in Chapter 7. When two long parallel

currents of equal magnitude separated 1 m exert a force per unit length of 2 � 10�7 N/m, themagnitude of the current is de�ned to be 1 Ampere. The electric charge, 1 C, is then deduced

from,

1 C = 1 Ampere� 1 sec

In other unit systems, the Coulomb�s law itself is employed to de�ne the unit of electric charges. For

example, in the CGS-ESU (ESU for ElectroStatic Unit) system, when two equal charges separated

by 1 cm exert a force of 1 dyne on each other, the charge is de�ned to be 1 ESU. Since 1 N = 107

dyne, and 1 m = 102 cm, we can readily see that

1 C =1 ESUp8:99� 1018

' 1 ESU3:0� 109

For example, the electronic charge in ESU is

1:6� 10�19 � 3:0� 109 = 4:8� 10�10 ESU

Although in engineering, the MKS-Ampere (SI) unit system is universally accepted, the CGS-ESU

and associated Gaussian unit system is still popular in physics. Both have merits and demerits,

and it is di¢ cult to judge one unit system superior to the other.

In the MKS-Ampere unit system, the Coulomb�s law

F = 8:98� 109 q1q2r2er (N) (1.4)

is rewritten as

F =1

4��0

q1q2r2er (N) (1.5)

where

�0 = 8:85� 10�12C2

N �m2 (1.6)

is called the vacuum permittivity. The appearance of the numerical factor 4� may be uncomfortable,

but cannot be entirely avoided. If we do not introduce the factor 4� in the Coulomb�s law, it will

pop up in the corresponding Maxwell�s equation. In fact, the Maxwell�s equation for the static

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Figure 1-2: In MKS unit system, I = 1 Ampere current is de�ned if the force per unit length betweenin�nite parallel currents 1 m apart is 2 � 10�7 N/m. The magnetic permeability �0 = 4� � 10�7H/m is an assigned constant to de�ne 1 Ampere current.

electric �eld in the CGS-ESU system has to be written as

r�E = 4�� (1.7)

because the factor 4� is avoided in the corresponding Coulomb�s law,

F =q1q2r2er (dynes)

The newly introduced constant �0 is one of the fundamental constants in electrodynamics. Note

that �0 is a measured constant, since it is derived from the original measured constant, 8:99� 109

N�m2/C in the Coulomb�s law. The permittivity of air (room temperature, 1 atmospheric pressure)is about 1:0004 �0 due to the polarizability of air molecules. For practical applications, the air

permittivity can be very well approximated by the vacuum permittivity.

1.3 Electric Field E

Interpretation of Coulomb�s law was a matter of debate before the concept of electric �eld was

well established by Faraday and Maxwell. Before Faraday and Maxwell, the so-called �theory of

action at distance� once prevailed. According to this theory, electric e¤ects (such as Coulomb

force) appear through some sort of direct interaction between charges, and space (or vacuum)

has nothing to do with the interaction. In the ��eld� theory, a single charge �disturbs� space

surrounding it by creating an electric �eld. The Coulomb force to act between two charges is due

5

to interaction between one charge and the electric �eld produced by the other. The �eld theory

is now well accepted, and modern electrodynamics is almost entirely described by �eld quantities

such as electric and magnetic �elds.

Let us write down the Coulomb force again,

F =1

4��0

q1q2r2er

This can be written either

F =1

4��0

q1r2er � q2 (1.8)

or

F =1

4��0

q2r2er � q1 (1.9)

We may interpret Eq. (1.8) as the force experienced by a charge q2 placed in an electric �eld,

E1 =1

4��0

q1r2er (1.10)

produced by a charge q1 while Eq. (1.9) as the force experienced by a charge q1 placed in an electric

�eld,

E2 =1

4��0

q2r2er (1.11)

produced by a charge q2. A single charge q in the space with a permittivity thus produces an

electric �eld given by

E =1

4��0

q

r2er (N/C) (1.12)

regardless of the presence of a second charge. Numerically, the electric �eld is equivalent to a

force to act on a unit change. Since the force is a vector, so is the electric �eld, and complete

determination of an electric �eld requires three spatial components.

In general, if a charge q is placed in an electric �eld E, the force to act on the charge is givenby

F = qE (N) (1.13)

This may alternatively be used for de�nition of an electric �eld, namely, if a stationary charge q

experiences a force proportional to the amount of the charge, then we de�ne that an electric �eld is

present. (Elementary particles such as protons and electrons do have masses, and they experience

gravitational forces as well. However, gravitational forces are orders of magnitude smaller than

electromagnetic forces, and in most practical applications, gravitational forces can be ignored.)

Although Eq. (1.12) has been deduced from Coulomb�s law, the inverse is not always true, that is,

electric �elds are not necessarily due directly to charges. A time varying magnetic �eld produces

an electric �eld (Faraday�s law). Also, an object travelling across a magnetic �eld experiences an

electric �eld (motional electromotive force), as we will study in Chapter 9. However, regardless of

the origin of the electric �eld, the force equation, Eq. (1.12), holds.

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q

q

1

2

E

E

1

2Et

P

Figure 1-3: Vectorial superposition of electric �eld.

1.4 Principle of Superposition and Electric Field due to a Distrib-uted Charge

The expression for the electric �eld due to a point charge,

E =1

4��0

q

r2er (1.14)

can be applied repeatedly to �nd an electric �eld due to a system of charges. For example, if there

are two charges q1 at r1 and q2 at r2, the total �eld can be found from,

E =1

4��0

r� r1jr� r1j3

q1 +1

4��0

r� r2jr� r2j3

q2 (1.15)

Note that the quantities,r� r1jr� r1j

;r� r2jr� r2j

are the unit vectors in the directions r� r1 and r� r2, respectively. For larger number of charges,the total electric �eld can still be found as vector sum of electric �elds due to individual charges.

This is known as the principle of superposition. For N discrete charges, the electric �eld can be

written down as,

E =1

4��0

NXi=1

r� rijr� rij3

qi (1.16)

As the number of charges increases, the summation becomes rather awkward. In most practical

applications, the charge distribution can be regarded continuous, rather than discrete. As one

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would expect, the summation in the case of discrete charges can be replaced by an integral in the

case of a continuous charge distribution.,

A continuous charge distribution can be described by a local charge density,

�(r) (C/m3)

which may vary as a function of position r. Since the size of elementary particles is so small, a

collection of those particles (e.g. electrons) can be well approximated by a continuous function

�(r), just as we treat water as a �uid although, microscopically, water consists of discrete water

molecules.

Let us pick up a small volume dV 0 located at a distance r0 from a reference point O as shown

in Fig. 1.4. The charge contained in the volume dV 0 located at r0 is

dq = �(r0)dV 0

By choosing the volume element su¢ ciently small, we may regard the charge dq a point charge.

We already know how to express the electric �eld due to a point charge,

dE =1

4��0

dq(r� r0)jr� r0j3

=1

4��0

r� r0

jr� r0j3�(r0)dV 0 (1.17)

Therefore, the electric �eld at r can be calculated from the following integral,

E(r) =

ZdE =

1

4��0

Zr� r0

jr� r0j3�(r0)dV 0 (1.18)

This is a general formula for calculating an electric �eld due to an arbitrary distribution of charges.

Remember that we have derived it from the Coulomb�s law.

Although the formula we just derived is a complete solution for static electric �elds due to a

charge distribution, it is seldom used in practical applications except for simple (or often trivial)

cases. There are several reasons for this. First, Eq. (1.18) is a vector equation. In the cartesian

coordinates, for example, three components, Ex; Ey and Ez will have to be evaluated separately.

That is, we have to carry out integrations three times for a complete vector solution. Second, in

many potential boundary value problems, the charge distribution, �(r), is not known a priori. For

example, in evaluation of a capacitance of a given electrode system, the problem is reversed, that

is, we calculate the electric �eld (from the scalar potential) �rst, and then evaluate the charge

distribution on the surface of electrodes. As we will study in Section 2, the method based on

the scalar potential is more convenient in practical applications than evaluating the electric �eld

using Eq. (1.18). However, if the charge distribution is known, we can certainly use Eq. (1.18) to

evaluate the electric �eld at an arbitrary point. Let us work on some simple examples.

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dq =ρdV'

r'

r

r ­ r'

dE

O

Figure 1-4: Di¤erential charge dq = �dV 0 produces a di¤erential electric �eld dE:

Charge SheetLet a large, thin insulating sheet carry a uniform surface charge density � (C/m2) (= constant).

If the sheet is large enough, or the point at which we wish to evaluate the electric �eld is close

enough to the sheet, the electric �eld should be perpendicular to the sheet because of cancellation

by an element located opposite with respect to the origin O. Choosing the cartesian coordinates

x; y; z, we then have to evaluate only the z component of the electric �eld at a distance z from the

sheet.

Let us pick up a surface element dx0dy0 located at (x0; y0; z0 = 0) on the sheet. The element

carries a �point charge�dq = �dx0 dy0. Then, the magnitude of the electric �eld on the z axis is,

dE =1

4��0

�dx0dy0

x02 + y02 + z2(1.19)

and its z component is,

dEz =1

4��0

�z

(x02 + y02 + z2)3=2dx0dy0 (1.20)

Therefore, the electric �eld on the axis is given by the following integral,

Ez =�z

4��0

Z 1

�1

Z 1

�1

dx0dy0

(x02 + y02 + z2)3=2(1.21)

However, the double surface integral can be replaced by a single integral over the radius r0, where

r02 = x02 + y02 so that

Ez =�z

4��0

Z 1

0

2�r0

(r02 + z2)3=2dr0 (1.22)

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The integral is elementary, and we �nd

Ez =�z

2�0

�� 1p

r02 + z2

�r0=1r0=0

=�

2�0

z

jzj (1.23)

wherepz2 = jzj has been substituted.

The solution indicates that the magnitude of the electric �eld is independent of the distance

from the sheet. The direction of the electric �eld is negative in the region z < 0, and positive for

z > 0. When there are two oppositely charged sheets, � and �� (C/m2), the electric �eld outsidethe sheets if zero, but the �eld in between the sheets is given by,

Ez =�

�0(1.24)

as can be readily seen from the principle of superposition. This con�guration corresponds to a

parallel plate capacitor provided the edge e¤ects are ignored.

Line ChargeWe assume a long line charge with a line charge density � (C/m). At a distance r from the line

charge, the di¤erential electric �eld due to a charge dq = �dz located at z is

dE =1

4��0

�dz

r2 + z2(cos �er + sin �ez) (1.25)

where,

cos � =rp

r2 + z2; sin � =

zpr2 + z2

The z component is an odd function of z. Therefore, the z component of the electric �eld vanishes

after integration from z = �1 to +1. The radial (r) component remains �nite, and is given by,

Er =�

4��0

Z 1

�1

r

(r2 + z2)3=2dz =

�

2��0r(1.26)

Remember that this result is valid only if the line charge is long, or the point of observation is

su¢ ciently close to a line charge of �nite length.

1.5 Gauss�Law for Static Electric Field

The electric �eld at a distance r from a point charge q is,

E =1

4��0

q

r2er;

10

chargesheet σ(C/m2)

E =σ/2ε0E =σ/2ε0

σ −σ

E = 0E = 0

E =σ/ε0

Figure 1-5: Upper �gure: Single charge sheet. The electric �eld on both sides is Ez = �=2"0: Lower�gure: Equal, opposite charge sheets. The �eld in-between is Ez = �="0: Outside, Ez = 0:

r

zdq=λ dz

dE

dE

dE

z

r

zline charge

Figure 1-6: Electric �eld due to a long line charge (line charge density � C/m). Note that Ez = 0

and the radial electric �eld Er =�

2�"0rcan also be found using Gauss�law.

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b

ar

θ

γ

q

dEn

Figure 1-7: Gauss�law applied to a sphere that is not concentric with the charge. Note r cos +b cos � = a:

On a spherical surface with radius r, the magnitude of the electric �eld is constant, and the quantity,

4�r2Er

is equal to q=�0. This strongly suggests that the closed surface integral of the electric �eld,ISE�dS (1.27)

is equal to the amount of charge enclosed by the closed surface S divided by �0,ISE�dS = q

"0

Noting ISE�dS =

Zr �EdV

and

q =

Z� (r) dV

we �nd the Maxwell�s equation

r �E = �"0

Let us see if this is the case for a spherical surface which is not concentric with the point charge

q. We denote the distance between the charge and the spherical center by b, and the radius of the

sphere by a.

The electric �eld on the surface is no more uniform. It is not normal to the surface, either. Noting

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the relationship,

r2 = a2 + b2 � 2ab cos �

in Fig. ??, we �nd the magnitude of the electric �eld at angle �,

E =q

4��0

1

a2 + b2 � 2ab cos � (1.28)

The component normal to the spherical surface is E cos where the angle is related to � through

r cos + b cos � = a:

Therefore, the entire surface integral reduces to,ISE � dS = q

4��0

Z �

0

a� b cos �(a2 + b2 � 2ab cos �)3=2

2�a2 sin �d� (1.29)

where

dS = 2�a2 sin �d�

is the area of the ring having radius asin� and width ad�. The relevant integral is

Z �

0

(a� b cos �) sin �(a2 + b2 � 2ab cos �)3=2

d� =

8<:2

a2; a > b

0; a < b(1.30)

(In integration, it is convenient to change the variable from � to � through � = cos �. Then the

integral reduces to

R 1�1

a� b�(a2 + b2 � 2ab�)3=2

d�

=1

bpa2 + b2 � 2ab�

�����1

�1

� 1

a2b

a2 + b2 � ab�pa2 + b2 � 2ab�

�����1

�1

Evaluation of the de�nite integral is left for a mathematical exercise. Note thatp(a� b)2 = ja�bj.)

Therefore, the surface integral of the electric �eld becomes,

IE�dS =

8<:q

�0; a > b

0; a < b(1.31)

Obviously, the case a > b corresponds to a sphere enclosing the charge, and a < b corresponds to

the case in which the charge is outside the closed spherical surface. The result may be generalized

to a closed surface of an arbitrary shape. The following formula is called Gauss�law,

�0

ISE�dS = q

Total charge

enclosed by S

!(1.32)

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Again, remember that Gauss� law is equivalent to Coulomb�s law, since we have �deduced� the

former from the latter. In fact, Gauss�law can be mathematically �proven�if we adopt Coulomb�s

law. A formal proof will be given after the Maxwell�s equation is introduced.

Gauss�law can be convenient for simple cases in which a system has a high degree of symmetry,

such as spherical, cylindrical, and planar symmetries. Let us work on a few simple examples.

Uniformly-Charged Insulating SphereLet an insulating sphere of radius a carry a uniform charge density � (C/m3) and a total charge

q = 4�3 a

3� (C). The system has complete spherical symmetry, and the only nonvanishing component

of the electric �eld is the radial component, Er. Outside the sphere (r > a), Gauss�law yields,

4�r2Er =q

�0

or

Er =q

4��0=

�a3

3�0r2(1.33)

which is identical to the �eld due to a point charge q concentrated at the center. Inside the sphere

(r < a), Gauss�law reads

4�r2Er =1

�0

4�

3r3� (1.34)

since in the RHS, only the amount of charge enclosed by the spherical surface (called Gaussian

surface) having a radius r (< a) enters. Solving for Er, we obtain,

Er =�

3�0r (1.35)

Therefore, in the sphere, the �eld linearly increases with the radius r up to the surface, r = a, where

the interior �eld connects to the exterior �eld without dicontinuity. The electric �eld is maximum

at the surface as shown in Fig. 1-8.

The fact that the electric �eld should vanish at the center of a charged (uniformly!) sphere is

understandable from spherical symmetry. At the center, the contribution from a charge dq to the

electric �eld can always be cancelled by another located opposite with respect to the center. By

analogy, the gravitational �eld due to the earth mass itself at the earth�s center should be zero.

Charged Conducting SphereA charge given to a conductor must reside entirely on the conductor surface so that the electric

�eld inside a conductor body should be identically zero in static condition. If an electric �eld were

not zero in a conductor, a large electric current would �ow according to Ohm�s law,

J = �E (1.36)

where � (Siemens/m) is the conductivity. � of metals is large. (For example, copper has � '5:9 � 107 S/m). Therefore, unless E = 0 in a conductor, a large current should �ow, and this

violates the assumption of static electricity. In other words, an electric �eld can exist in a conductor

14

r

Er

a 2a 3a 4a

Emax

aρ

Figure 1-8: Electric �eld of a unifrmly charged sphere. The �eld is 0 at the center. The maximum�eld, Emax =

�a

3"0; occurs at the surface.

only in dynamic conditions in which current �ow is allowed.

If an excess charge q is given to a conducting sphere, the charge uniformly resides on the surface

as a surface charge, with a surface charge density �s = q=4�a2 (C/m2) where a is the sphere radius.

The electric �eld inside (r < a) the sphere is zero, while outside (r > a), it is given by

Er =1

4��0

q

r2(r > a) (1.37)

Note that there is a sudden jump in the �eld at the surface (r = a) where the surface charge density

exists. In general, wherever an in�nitesimaly thin surface charge layer exists, the electric �eld there

becomes discontinuous. We will come back to this subject in Chapter 3 where the concept of the

displacement vector D is introduced.

Experimental Veri�cation of Coulomb�s Inverse Square LawAs stated earlier, Gauss�law and Coulomb�s law are physically identical, in the sense that the

former is derivable from the latter. Therefore, if Gauss�law can be veri�ed experimentally, then

Coulomb�s law is indirectly veri�ed also. Here, we are particularly concerned with a question: How

valid is the inverse square law? Or put alternatively, when the Coulomb�s law is written as

F =const:

rner

how close is the power n to 2.0? Is it exactly 2.0? Experiments in the past have shown that n is

15

r

Er

a 2a 3a 4a

Emax

a

σ

conducting spherecarrying a surfacecharge

Figure 1-9: Electric �eld in a charged conducting sphere is 0. The charge q given to the spheremust reside on the surface as a surface charge � = q=4�a2 C/m2:

indeed very close to 2.0 with an uncertainty of order 10�16. The power n is probably exactly equal

to 2.

The experiment performed by Plimpton and Lawton in 1936 was to measure the intensity of

electric �eld inside a charged conducting shell. As we have just seen in the preceding example, an

excess charge given to a conductor must reside entirely on its outer surface, and the electric �eld in

the conductor must be zero. This holds for a conducting shell, too, and no charge can exist on the

inner surface of the shell. Plimpton and Lawton established n = 2�2�10�9. Further improvementn = 2� 2� 10�16 was achieved later in 1971 by Williams et al.

1.6 Di¤erential Equations for Static Electric Fields-Maxwell�s Equa-tions

For a given charge distribution �(r), the electric �eld can be uniquely evaluated from

E(r) =1

4��0

Zr� r0

jr� r0j3�(r0)dV 0 (1.38)

as we have seen in Sec. 1.4. The electric �eld also satis�es Gauss�law,ISE�dS = 1

�0

ZV�(r)dV (1.39)

16

Wemay use either formulation in evaluating the electric �eld due to a prescribed charge distribution.

The electric �eld is a vector quantity. A vector can be uniquely de�ned if its divergence and

curl are both speci�ed. This mathematical theorem is known as Kirchho¤�s theorem. (Kirchho¤

is a familiar name in electric circuit theory. However, his contributions are not limited to circuit

theory, but quite diversi�ed. Kirchho¤�s scalar di¤raction formula for electromagnetic waves is

another important contribution, and had been used extensively until it was replaced by a more

accurate vector di¤raction formula rather recently. See Chapter 13.) A derivation of vector di¤er-

ential equations for electric and magnetic �elds was formulated by Maxwell in the celebrated book

�Treatise on Electricity and Magnetism� published in 1873. Each �eld (electric and magnetic)

requires both divergence and curl. Therefore, for complete description of electric and magnetic

�elds, four Maxwell�s equations emerge.

In electrostatics, the magnetic �eld is absent (or ignored). Let us start with the divergence of

static electric �elds. In the example of a charged sphere in Sec. 2.5, we have seen that the electric

�eld inside a uniformly charged spherical body can be found from Gauss�law,

4�r2Er =1

�0

4�

3r3� (1.40)

with the result,

Er =�

3�0r (1.41)

Eq. (1.40) holds no matter how small r is chosen. The divergence of the electric �eld is de�ned by

div E =r �E = lim�V!0

H�S E � dS�V

(1.42)

In Eq. (1.40), the LHS is a special case of symmetric surface integration, and the quantity 4�r3=3

in the RHS is of course the volume. Therefore, if we take the limit of r ! 0 in the ratio,

limr!0

4�r2Er4�3 r

3

it reduces to the de�nition of divergence. Consequently, the divergence of the electric �eld satis�es,

r �E = �

�0(1.43)

This is one of Maxwell�s four equations.,

Outside the charged sphere, the surface integration,I�SE � dS

vanishes unless �S intersects the sphere. Therefore, outside the sphere, r � E = 0, as required

because � = 0 outside the sphere.

The divergence equation alternatively follows from Gauss� law, although physical meaning is

17

less clear. The surface integral can be rewritten in terms of a volume integral as follows,ISE�dS =

ZVr �E dV (Gauss�theorem) (1.44)

This is a mathematical theorem, and has nothing to do with the physical Gauss�law, although they

are intimately related. Using this transformation, we can rewrite Gauss�law as,Zr �E dV = 1

�0

Z� dV (1.45)

Therefore, r �E = �=�0 immediately follows.The third method to derive the divergence equation for the electric �eld is to directly take the

divergence of Eq. (1.18),

rr �E(r) =1

4��0rr �

Zr� r0

jr� r0j3�(r0)dV 0 (1.46)

where the subscript r indicates di¤erentiation with respect to the observer�s coordinates r. The

function,

rr ��r� r0

jr� r0j3

�(1.47)

has a peculiar property. It vanishes wherever r 6= r0 but is not properly de�ned where r = r0. (Infact, it diverges at r = r0.) Introducing jr� r0j = R, we indeed see that,

rr ��r� r0jr� r0j3

�=

1

R2d

dR

�R2

1

R2

�= 0 (R 6= 0)

However, its volume integral is well de�ned and remains �nite since

Rrr �

�r� r0

jr� r0j3

�dV

=

Ir� r0

jr� r0j3� dS

=

I1

R2R2d =

Id = 4� (1.48)

where d is the di¤erential solid angle, and 4� is the total solid angle. In other words, the function,

rr ��r� r0

jr� r0j3

�(1.49)

has the property of the delta function,

rr ��r� r0

jr� r0j3

�= 4��(r� r0) (1.50)

18

where

�(r� r0) = �(x� x0)�(y � y0)�(z � z0) (1.51)

is the three-dimensional delta function having the dimensions of m�3. Therefore, the integral in

Eq. (1.46) can be readily performed as

4�

4��0

Z�(r0)�(r� r0)dV 0 = 1

�0�(r) (1.52)

and Eq. (1.46) is equivalent to

r �E = 1

�0� (1.53)

We now evaluate the curl of a static electric �eld given in Eq. (1.18). Taking the curl of both

sides, we �nd,

rr �E(r) =1

4��0

Zrr �

�r� r0

jr� r0j3

��(r0)dV 0 (1.54)

However, in the RHS, the vector,

rr ��r� r0

jr� r0j3

�=rr

�1

jr� r0j3

�� (r� r0) + 1

jr� r0j3rr � (r� r0) (1.55)

identically vanishes, because the vector,

rr

�1

jr� r0j3

�= �3 r� r

0

jr� r0j4(1.56)

is parallel to the vector r� r0, andrr � (r� r0) � 0 (1.57)

identically. In contrast to the divergence in Eq. (1.46),

rr ��r� r0

jr� r0j3

�� 0 (1.58)

holds even at r� r0 = 0. Therefore, for static electric �elds due to charge distributions, we conclude,

r�E = 0 (1.59)

We thus have speci�ed both divergence and curl of static electric �elds in Eqs. (1.53) and (1.59),

respectively, which determine static electric �elds (vector) uniquely. The divergence equation holds

for any electric �elds (electrostatic and non-electrostatic), but the curl equation is the special case

of the more general Maxwell�s equation,

r�E = �@B@t

(1.60)

In static phenomena, @=@t = 0, and r�E = 0 holds only for static electric �elds.

19

What are the physical meanings of the Maxwell�s equations? The nonvanishing divergence of

the static electric �eld indicates that the charge density � is the source (or sink) of the �eld. This

may be illustrated by the electric lines of force. A positive charge �emits�electric �eld lines, while

a negative charge �absorbs�electric lines of force, just like stream lines in water �ow. If two equal,

but opposite charges are near by, the �eld lines �emitted�by the positive charge are all �absorbed�

by the negative charge. If the magnitude of the charges are not equal, some �eld lines are not

absorbed by the negative charge.

The vanishing curl of static electric �elds indicates that an electric �eld line does not close on

itself, that is, a �eld line does not form a loop. (This is in contrast to non-electrostatic �eld for

which the curl is nonvanishing.) In other words, static electric �eld lines always have heads or tails

as clearly seen in the above examples. Static �eld lines start at a positive charge and end at a

negative charge.

1.7 Scalar Potential �(r)

The vanishing curl of static electric �elds has an important implication. Since the curl of a gradient

of an arbitrary scalar function is identically zero,

r�rF � 0 (1.61)

a static electric �eld (vector!) can be deduced from a gradient of some scalar function. We denote

this scalar by �(r) and call it a scalar potential. The electric �eld is given by the gradient of the

scalar potential,

E(r) = �r�(r) (1.62)

The negative sign is introduced so that the electric �eld is directed from higher to lower potential

regions, just like in the familiar gravitational �eld and gravitational potential.

The dimensions of the scalar potential are

Newton �meterCoulomb

=Joule

Coulomb

This is rede�ned as Volts (after Volta). Therefore, an alternative unit for the electric �eld is

Volts/meter.

The physical meaning of the electric �eld is (as discussed before) the force to act on a unit

charge. Then, the scalar potential can be interpreted as the work required to move a unit charge

from one point to another, since

�(r) = �(r1)�Z r

r1

E � dr (1.63)

where �(r1) is the potential at r1 . Clearly, the potential is a relative quantity and is referred

to the potential at a speci�ed reference point. (Again, the analogy to the gravitational potential

20

V

d

E

x0 d

Φ (x)V

Figure 1-10: Potential � (x) =V

dx and electric �eld Ex = �V=d in a parallel plate capacitor.

should be recalled. When we measure the height of a mountain, usually the sea level is chosen as

the reference point.)

As an example, let us consider a point charge q placed at the origin r = 0. The electric �eld is

given by,

Er(r) =1

4��0

q

r2; V/m

If the reference potential � = 0 is chosen on a surface with a radius r1, the potential at an arbitrary

point r becomes

�(r) = �Z r

r1

q

4��0

1

r2dr

=q

4��0

�1

r� 1

r1

�(1.64)

For a charge system of a �nite spatial extent, it is convenient to choose � = 0 at r1 =1, so that

�(r) =1

4��0

q

r; (V) (1.65)

relative to the zero potential at r =1.Similarly, the potential due to a charged conducting sphere with a total charge q and radius a

is given by,

�(r) =1

4��0

q

r(1.66)

21

relative to � = 0 at r =1. The sphere potential can be found by equating the radius to r = a,

�sphere =1

4��0

q

a(1.67)

Note that this result is independent of whether the conducting sphere is solid or shell, since the

electric �eld in the conductor must vanish identically. In general, the potentials at any points on

and in a conductor must be equal because E = 0 in a conductor. In particular, the surface of a

conductor, no matter how complicated is its shape, is an equipotential surface. This fact will play

an important role in potential boundary problems in Chapter III.

In the following, we will work on a few examples in which the potentials can be calculated from

known electric �elds.

Potential of a Charged Insulating SphereThe electric �eld for this problem has been worked out in Sec. 1.5, and given by

Er =

8>>><>>>:�

3�0

a3

r2; (r > a)

�

3�0r; (r < a)

(1.68)

where � is the charge density and a is the sphere radius. We choose � = 0 at r = 1. Then, theexterior potential becomes

�(r) = �Z r

1

�

3�0

a3

r2dr

=�

3�0

a3

r; (r > a) (1.69)

On the surface of the sphere, the potential takes the value,

�(a) =�

3�a2 (1.70)

Therefore, the interior potential becomes,

�(r) = �(a)�Z r

a

�

3�0rdr

=�a2

3�0� �

6�0

�r2 � a2

�=

�a2

2�0� �

6�0r2; (r < a) (1.71)

The potential at the sphere center is ,

�(r = 0) =�a2

2�0(1.72)

which is the maximum (if q > 0) potential. Note that the integration should be carried out starting

22

at the reference point (r =1 in this case), and radially inward.

Potential due to a Long Line ChargeThe electric �eld due to a long line charge has been found in Section 1.5, and given by

E� =�

2��0

1

�(1.73)

where � (C/m) is the line charge density. Let us choose a reference, zero potential surface at a

radius � = a. (a cannot be in�nity, in contrast to the case of spheres, because the potential due to

an in�nitely long line charge does not vanish at in�nity, but diverges. Such a line charge requires

in�nitely large amount of energy, and thus is of mathematical interest only.) Then, the potential

at an arbitrary radial position � can be found as

�(�) = �Z �

aE�d�

=�

2��0

Z a

�

1

�d�

=�

2��0ln

�a

�

�(1.74)

The zero potential surface can be chosen arbitrarily, since the potential is a relative quantity.

However, the potential di¤erence between two radial positions is independent of the choice of the

reference. Indeed, the potential di¤erence between two points at �1 and �2 becomes

�(�1)� �(�2) =�

2��0

�ln

�a

�1

�� ln

�a

�2

��=

�

2��0ln

��2�1

�(1.75)

in which the particular radius a has disappeared.

In practical applications, the potential we just found can be used to calculate the capacitance

of a coaxial cable. Let us consider a coaxial cable having inner and outer conductor radii a and b,

respectively, as shown in Fig. 1-11. We connect a dc power supply of a voltage V between the two

conductors. If we can calculate the charges �q to appear on respective conductors, the capacitancecan be calculated by de�nition from,

C =q

V(F) (1.76)

The charge per unit length is,

� =q

l(1.77)

Therefore, the potential di¤erence between the two conductors becomes,

V = �(a)� �(b) = �

2��0ln

�b

a

�

23

V2a

b

l

Figure 1-11: Coaxial cable.

or

V =q

2��0lln

�b

a

�(1.78)

Hence,

C =q

V=2��0l

ln�ba

� (F) (1.79)

or the capacitance per unit length is given by,

C

l=2��0

ln�ba

� (F/m) (1.80)

If the space between the conductors is �lled with a dielectric having a permittivity �, this should

be modi�ed asC

l=

2��

ln�ba

� (1.81)

1.8 Potential due to a Prescribed Charge Distribution

In the preceding Section, we have (brie�y) learned how to calculate the potential �(r) from a known

electric �eld. However, in most applications, it is more common to reverse the procedure, namely,

we �rst �nd the potential �, and then calculate the electric �eld from,

E = �r� (1.82)

There are several reasons for inverting the procedure. The basic Maxwell�s equations for static

electric �elds are,

r�E = �

�0(1.83)

r�E = 0 (1.84)

24

Obviously, these are vector di¤erential equations, and for a complete solution, we must solve three

di¤erential equations for each component of the �eld E. However, if we substitute

E = �r�

into r�E = �

�0, we �nd a single scalar di¤erential equation for �,

r2� = � ��0

(1.85)

This is known as Poisson�s equation, and mathematically speaking, it is an inhomogeneous version

of the Laplace equation,

r2� = 0 (1.86)

for which exhaustive studies have been made since the 18th century. The Laplace and Poisson�s

equations most frequently appear in physical science and engineering. Analytic solutions to those

equations can be found in a relatively limited number of cases. However, with powerful computers

becoming more easily accessible these days, numerical solutions to Poisson and Laplace equations

are no more prohibitively expensive even for complicated geometries for which analytic solutions are

extremely di¢ cult, if not impossible. We will return to the problem of solving Laplace equations in

the following Chapter. Here, we derive a formal solution to the Poisson�s equation, which enables

us to calculate the potential for a prescribed charge distribution.

There are several methods to �nd the solution to

r2� = � ��0

and we start with a method that is physically most transparent. We have already seen that the

potential due to a point charge q is given by

�(r� r0) = q

4��0

1

jr� r0j (1.87)

where r0 is the location of the point charge.

For a distributed charge, we pick up a small volume dV 0 in the region where the charge density,

�, exists. The charge contained in the volume dV 0 is

dq = � dV 0 (1.88)

By choosing the volume dV 0 su¢ ciently small, the di¤erential charge dq approaches a point charge.

Therefore, the potential due to the charge dq located at r0 becomes

d� =1

4��0

�(r0)

jr� r0j dV0 (1.89)

25

r'

r

r ­ r'

dq

O

dΦ

provided the reference potential � = 0 is at r = 1. By integrating Eq. (??) over the dummyvariable r0, we �nd

�(r) =1

4��0

Z�(r0)

jr� r0j dV0 (1.90)

which is the desired solution to the Poisson�s equation.

The same result can be deduced from the electric �eld due to a prescribed charge distribution

found earlier,

E(r) =1

4��0

Zr� r0jr� r0j3 �(r

0)dV 0:

Sincer� r0jr� r0j3 = �rr

�1

jr� r0j

�where the subscript r indicates di¤erentiation with respect to the observing point r, we �nd

E(r) = � 1

4��0rr

Z�(r0)

jr� r0j dV0 (1.91)

Note that the di¤erential operator rr can be taken out of the integral, because it operates on r

only. Comparing with the basic relationship,

E = �r�

we readily obtain

�(r) =1

4��0

Z�(r0)

jr� r0j dV0

which is identical to Eq. (1.90). Below, we will apply this formula to a few simple problems.

Potential due to a Line Charge of Finite LengthConsider a line charge of length 2a carrying a uniform line charge density � (C=m). A di¤erential

26

z

z'dq

dΦ

a

­a

ρ

Figure 1-12: Line charge (� C/m)of �nite length 2a: dq = �dz0:

charge dq = �dz0 located at z0 creates a potential at the coordinates (�; z),

d� =1

4��0

�dz0p�2 + (z � z0)2

(1.92)

Integrating this over z0 from z0 = �a to +a, we �nd

�(�; z) =�

4��0

Z a

�a

dz0p(z0 � z)2 + �2

=�

4��0

hln�p

(z0 � z)2 + �2 + z0 � z�ia�a

=�

4��0ln

p(z � a)2 + �2 + a� zp(z + a)2 + �2 � a� z

!(1.93)

Let us examine special cases of the potential. At a point very close to the line charge, so that

z ! 0, and �� a, the potential reduces to

�(�) ' �

2��0ln

�2a

�

�(�� a) (1.94)

The potential of the form

�(�) = � �

2��0ln �+ const. (1.95)

has been worked out in Sec. 2.6 for a long coaxial cable, and the result we just obtained is therefore

of an expected form.

Far away from the line charge so that �; z � a, we expect to recover the potential due to a

27

point charge, q = 2a�. On the plane z = 0, the logarithmic function approaches

ln

p�2 + a2 + ap�2 + a2 � a

!' ln

�1 +

2a

�

�' 2a

�(�� a) (1.96)

Therefore, the potential indeed approaches

�(�) ' q

4��0

1

�(�� a) (1.97)

where q = 2a� is the total charge carried by the line. On the z axis, � ! 0, and at jzj � a, the

potential also approaches,

�(z) =�

4��0lim�!0

ln

p(z � a)2 + �2 + a� zp(z + a)2 + �2 � a� z

!

=�

4��0ln

�1 +

2a

z

�' q

4��0

1

z(1.98)

Capacitance of a Thin Linear ConductorThe expression for the potential in the vicinity of a line charge, Eq. (1.94), can be directly used

to �nd the capacitance of a thin linear conductor having a radius b and length 2a with a� b. On

the surface of the conductor, � = b. Therefore, the potential of the conductor can be approximated

by

�(� = b) =�

2��0ln

�2a

b

�' q

4��0aln

�2a

b

�and the approximate capacitance is given by

C = 4��0a= ln

�2a

b

�=

2�"0L

ln(L=b)(1.99)

where L = 2a is the total length of the conductor.

Capacitance of Parallel Wire Transmission LineAs an application of the potential due to a line charge, we consider a long, two-parallel-wire

transmission line, which is often encountered in high voltage power transmission and old-fashioned

telephone lines. The geometry is shown in Fig. ??. We assume two thin parallel conductor wires

28

of an equal radius a separated a distance d. By �thin conductor wires�, we mean d� a. Since the

capacitance we are concerned here is the mutual capacitance between the two wires, we let the two

wires carry equal, but opposite line charge densities, +� and �� (C/m). Then, the potential atarbitrary point can be written down as

�(r+; r�) =�

2��0[� ln r+ + ln r�]

=�

2��0ln

�r�r+

�where r+ and r� are the distance from the observing point to the positive and negative wires,

respectively. Note that the reference, zero potential surface is chosen at the midplane on which

r+ = r�.

When the wire radius a is small compared with the separation distance d, the potential at the

surface of the positive wire can be found by letting r+ = a and r� = d,

�+ =�

2��0ln

�d

a

�(1.100)

Similarly, the potential at the negative wire is,

�� =�

2��0

�ad

�= � �

2��0ln

�d

a

�(1.101)

Therefore, the potential di¤erence between the wires is

V = �+ � �� =�

��0ln

�d

a

�(1.102)

and the capacitance per unit length of the transmission line is given by,

C

l=�

V= ��0= ln

�d

a

�(F/m) (1.103)

The capacitance (per unit length) of a single wire transmission line placed above the ground

at a height h can be found in a similar manner. The ground potential can be chosen to be zero.

Therefore, the potential di¤erence between the wire and the ground is given by

V =�

2��0ln

�2h

a

�(1.104)

where 2h is the distance between the wire and an image line charge located at the mirror point,

z = �h, in the ground. Then, the capacitance becomes

C

l= 2��0= ln

�2h

a

�(1.105)

29

+λ −λr

r+

­

d

Figure 1-13: Parallel wire transmission line. Separation distance d and wire radius a:

Note that this is twice as large compared with the capacitance of the two-wire transmission line.

The latter can be considered to be a series connection of two capacitors, one between the positive

wire and the midplane, and the other between the negative wire and the midplane. The midplane,

which is chosen at zero potential, can be replaced by a grounded large conducting plate without

a¤ecting the potential and electric �eld. The method of images will be discussed more fully in

Chapter 5.

Both formulae in Eqs. (1.103) and (1.105) are subject to the assumption of thin wire radius,

a� d; h. As the wire radius becomes large, they become inaccurate, but only logarithmically. For

example, the exact capacitance of the parallel wire transmission line is given by

C

l= ��0

1

ln

"d+

pd2 � 4a22a

# (1.106)

Even when a = 0:2d (thick conductors indeed), the error caused by using the approximate formula

in Eq. (1.103) is less than 3%.

Electric DipolesTwo charges of opposite signs, but equal magnitude, +q and �q separated by a small distance

a constitute an electric dipole. (A single charge q is called a monopole. Higher order multipoles,

quadrupole, octapoles, etc., can be similarly constructed.) Dipoles are the basic elements in dielec-

tric materials as we will study in detail in Chapter 4.

jsin �j

The potential due to two charges, +q and �q, can be written down as

�(r+; r�) =q

4��0

�1

r+� 1r

�(1.107)

30

Φ

+q

­qa θ

r

r

+

−

Figure 1-14: Electric dipole consists of two charges q and �q separated by a small distance a:

where r+ and r� are the distances to the respective charges from the observation point. They are

related through

r2+ = r2� + a

2 � 2ar� cos � (1.108)

where � is the angle between the z axis and the vector r�. So far, we have not made any approxi-

mations, and Eq. (??) is exact.At a point far away from the dipole such that r+; r� � a, r+ may be approximated by

r+ ' r� � a cos � (1.109)

Then, the potential at r � a becomes

�(r; �) =q

4��0

�1

r � a cos � �1

r

�' q

4��0

a cos �

r2(1.110)

where we have replaced r� (' r+) by r. In contrast to the monopole potential,

�(r) =1

4��0

q

r

the dipole potential is proportional to 1=r2, and thus of higher order in the series expansion in

powers of 1=r. The potentials due to the equal but opposite charges cancel each other in the lowest

(monopole) order. Also, the dipole potential has the angular dependence, cos �. It is convenient to

introduce the dipole moment (vector)

p = qa (1.111)

31

­0.6 ­0.4 ­0.2 0.2 0.4 0.6

­1.0

­0.8

­0.6

­0.4

­0.2

0.2

0.4

0.6

0.8

1.0

x

y

Figure 1-15: Equipotential surfaces of electric dipole pz:

where a is the vector directed from the negative charge �q to the positive charge +q. Since � isthe angle between a and r, we may rewrite Eq. (1.110) as

� =1

4��0

p � rr3

(1.112)

where

p � r = pr cos � = aqr cos �

The equipotential surface of the dipole is shown in Fig. 1.17.pjsin �j

The electric �eld associated with the dipole can be calculated by taking the gradient of the

potential,

E = �r�(r; �)

= ��er@

@r+e�r

@

@�

�1

4��0

aq

r2cos �

=aq

4��0

�2

r3cos �er +

sin �

r3e�

�(1.113)

32

The electric �eld lines are described by the following di¤erential equation,

dr

Er=rd�

E�(1.114)

since the electric �eld is tangent to the �eld lines. The components, Er and E� are

Er =aq

4��0

2

r3cos � (1.115)

E� =aq

4��0

1

r3sin �

Substituting these into Eq. (??), we �nd the following di¤erential equation to describe the �eldline,

dr = 2r cot �d�

ordr

r= 2 cot �d� (1.116)

Integrating both sides, we obtain

ln r = ln(c sin2 �) (1.117)

or

r = c sin2 �

where c is a constant. The electric �eld lines of the diople are shown in Fig. 1.18. Note that the

electric �eld lines are normal to the equi potential surfaces.

cos2 �

1.9 Linear Quadrupole

Charges +q; �2q; and +q placed along the z axis at z = a; 0; �a constitute a linear quadrupole.The potential at r � a can be found as superposition of 3 potemtials,

� (r; �) =q

4�"0

�1p

r2 + a2 � 2ar cos �� 2r+

1pr2 + a2 + 2ar cos �

�' qa2

4�"0

3 cos2 � � 1r3

where use is made of binomial expansion

1p1 + x

= 1� 12x+

3

8x2 � ��

33

­1.0 ­0.8 ­0.6 ­0.4 ­0.2 0.2 0.4 0.6 0.8 1.0

­0.3

­0.2

­0.1

0.1

0.2

0.3

x

y

Figure 1-16: Electrif �eld lines of the diople pz = qa:

Note that the dipole potemtial vanishes. This is because the quadrupole consists of two dipoles

oppositely directed. The equipotebtial surfaces of the linear quadrupole is shown in Fig. 1.

1� 3 sin2 �

­1.0 ­0.5 0.5 1.0

­2

­1

1

2

x

y

The function

P2 (x) =1

2

�3x2 � 1

�is the Legendre polymonial of order l = 2: The binomial expansion of the potential due to the

34

charge q at z = a is

�1 =q

4�"0

1pr2 + a2 � 2ar cos �

=q

4�"0

�1

r+a

r2cos � +

a2

r33 cos2 �

2+ � � �

�=

q

4�"0

1Xl=0

al

rl+1Pl (cos �)

Likewise

�2 =q

4�"0

1pr2 + a2 + 2ar cos �

=q

4�"0

1Xl=0

(�1)l al

rl+1Pl (cos �)

35