electrostatics i electric field and scalar potentialphysics.usask.ca/~hirose/ep464/ch1-09.pdf ·...
TRANSCRIPT
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.
8
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.
11
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 � ��
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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
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