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Copyright, 1993, all rights reserved, Charlie Ebner (Dept.of Physics,
The Ohio State University, Columbus OH 43210) and Mark Jarrell (Dept.of
Physics, The University of Cincinnati, Cincinnati, OH 45221-0011). This
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\title{Introduction to Electrostatics}
\author{Charles Augustin de Coulomb\\(1736 - 1806)}
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We shall follow the approach of Jackson, which is more or less historical.
Thus we start with classical electrostatics, pass on to magnetostatics, add
time dependence, and wind up with Maxwell's equations. These
are then expressed within the framework of special relativity. The
remainder of the course is devoted to a broad range of interesting and
important applications.
This development may be contrasted with the more formal and elegant
approach which starts from the Maxwell equations plus special relativity
and then proceeds to work out electrostatics and magnetostatics - as well
as everything else - as special cases. This is the method of \eg Landau and
Lifshitz,~{\em The Classical Theory of Fields}.
The first third of the course, \ie Physics 707, deals with physics which should
be familiar to everyone; what will perhaps not be familiar are the
mathematical techniques and functions that will be introduced in order to
solve certain kinds of problems. These are of considerable usefulness and
therefore will be important to us.
\section{Coulomb's Law}
By performing experiments on small charged bodies (ideally, point charges),
Charles Augustin de Coulomb, working around the time of the American and French
revolutions (1785), was able to empirically infer that the force between
two static
charged particles is proportional to the inverse square of the distance between
them. The following has since become known\footnote{Numerous others, such
as Henry Cavendish, also may legitimately have some claim to the law.} have as
{\em Coulomb's Law}: Given two static charges $q_1$ and $q_2$, there is a force
acting on each of them which is:
\begin{enumerate}
\item Proportional to the product of the magnitudes of the charges, being
attractive for unlike charges and repulsive for like charges
\item Inversely proportional to the square of the
distance between the charges.
\item Directed along the line between the
charges.
\end{enumerate}
In the form of an equation, the law states that
\beq
\F_{21}=k\frac{q_1q_2}{|\x_2-\x_1|^2}\frac{\x_2-\x_1}{|\x_2-\x_1|}
\eeq
where charge $q_i$ is located at $\x_i$, $\F_{21}$ is the force on
charge 2 produced by charge 1, and $k$ is a positive constant; vectors
are denoted by boldface type.
In addition, the force satisfies a superposition law (or principle) in that
the force $\F$ on a charge $q$ in the presence of a number of other charges
$q_i$ at $\x_i$, $i=1$,...,$n$, is simply the sum of the forces arising from
each of the latter as though it were the only other charge present\footnote{
As we shall see, the principle of superposition follows from the
linearity of Maxwell's Equations}. Thus,
\beq
\F=kq\sum_{i=1}^n\frac{q_i(\x-\x_i)}{|\x-\x_i|^3},
\eeq
given that $q$ is at $\x$.
The constant $k$ has units and magnitude which depend on the system of
units employed. We shall adopt cgs Gaussian units. The units of mass
length and time are, respectively grams ($g$), centimeters ($cm$), and
seconds ($sec$). The unit of charge is the statcoulomb ($statcoul$) which is
defined by the statement that the force between two charges, each of one
$statcoul$, one $cm$ apart is one dyne ($dyn$).
Then $k=1\,dyn-cm^2/(statcoul)^2$. In practice one may treat $k$ as
having dimension unity while charge has dimension of $M^{1/2}L^{3/2}/T$.
\section{Electric Field}
It is customary and useful to introduce the concept of the electric field
at this point. This is a vector field, \ie~a vector function of $\x$. It
is written as $\E(\x)$ and is defined as the force that would be
experienced by a charge $q$ at $\x$, divided by $q$\footnote{This definition
is not complete. The field has other attributes as well since it carries
momentum and energy: i.e. photons}. Thus, for a
distribution of charges $q_i$ at $\x_i$, $i$=1,2,...,$n$,
\beq
\E(\x)=\sum_{i=1}^n\frac{q_i(\x-\x_i)}{|\x-\x_i|^3}
\eeq
The electric field has the property of being independent of the `test'
charge $q$; it is a function of the charge distribution which gives rise to
the force on the test charge, and, of course, of the test charge's
position. This object has dimension $Q/L^2$ or $M^{1/2}/L^{1/2}T$.
At this point let us introduce the charge density $\rh(\x)$ which is the
charge per unit volume at, or very close to, $\x$. This object is needed
if we would like to integrate over a source distribution instead of summing
over its constituent charges. Thus a sum is replaced by an equivalent
integral,
\beq
\sum_{i=1}^nq_i\rightarrow \iniv\rh(\x).
\eeq
The charge density has dimension $Q/L^3$. In terms
of $\rh$, the expression for the electric field can be written as
\beq
\E(\x)=\inivp\rh(\xp)\frac{\x-\xp}{|\x-\xp|^3}
\eeq
In the particular case of a distribution of discrete point charges, it is
possible to recover the sum in \eq{3} by writing the charge density in an
appropriate way. To do so we introduce the Dirac delta function $\de(x-a)$.
It is defined by the integral
\beq
f(a)=\int dx f(x)\de(x-a)
\eeq
where $f(x)$ is an arbitrary continuous function of $x$, and the range of
integration includes the point $x=a$. A special case is $f(x)=1$ which
leads to
\beq
\int dx \de(x-a)=1,
\eeq
demonstrating the normalization of the delta function.
From the arbitrariness of $f(x)$, we may conclude that $\de(x-a)$ is zero
when $x$ is not $a$ and sufficiently singular at $x=a$ to give the
normalization property. In other words, it is in essence the charge
density of a point charge (in one dimension) located at $x=a$.
Some important relations involving delta functions are as follows:
\beq
\int_{a_1}^{a_2} f(x) \der{\de(x-a)}{x}dx=-\left.\der{f(x)}{x}\ril_{x=a}
\eeq
and
\beq
\int_{a_1}^{a_2}\de[f(x)]dx=\sum_{i=1}^N\left[ 1/\left|\der{f(x)}{x}
\right|_{x_i}\right]
\eeq
In the final expression the $x_i$ are the 0's of $f(x)$ between $a_1$ and
$a_2$.
A delta function in three dimensions may be built as a product of three
one-dimensional delta functions. In Cartesian coordinates,
\beq
\de(\x)=\de(x)\de(y)\de(z)
\eeq
This function has the property that
\beq
\iniv f(\x)\de(\x-\x_0)=f(\x_0)
\eeq
Returning to electrostatics, we can see that the charge density of a
collection of point charges can be written as a sum of delta functions:
\beq
\rh(\x)=\sum_{1=i}^nq_i\de(\x-\x_i)
\eeq
Thus
\beqa
\E(\x)=\inivp\rh(\xp)(\x-\xp)/|\x-\xp|^3 \nonumber \\
=\sum_{i=1}^n\inivp q_i\de(\xp-\x_i)(\x-\xp)/|\x-\xp|^3 \nonumber \\
=\sum_{i=1}^nq_i(\x-\x_i)/|\x-\x_i|^3.
\eeqa
\section{Gauss's Law}
Although Coulomb's Law is quite sufficient for finding electric fields and
forces, the integral form in which we have expressed it is not always the
most useful approach to a problem. Another integral form, called Gauss's
Law, is often more useful.
Let us look first at a {\bf{two-dimensional}} version
of this law. Consider a point charge $q$ located within a closed path C. In
two dimensions, the field produced by this charge is $2q(\x-\x_0)/|\x-
\x_0|^2$. Consider now the integral around C of that component of $\E$
which is normal to the path. This normal component is $\E\cdot\nn=
q$cos$\th/r$, where $r$ is the distance from the charge to the integration
point on the loop. However, $dl\cos\th/r$ is just the infinitesimal angle
$d\ph$ subtended by $dl$ at the charge. Hence we just need to integrate
$d\ph$ around the loop.
\centerline{\psfig{figure=gauss2d.ps,height=3.5in,width=7.0in}}
\beq
\inlc \E\cdot\nn=\inlc \frac{q\cth}{|\x-\x_0|}=q\int d\ph
\eeq
Since the charge is inside, the integral is $2\pi$;
if it were outside, the integral would be 0 because over one part of the
path, $\cos\th$ is positive and over another part it is negative with the
two parts cancelling one another when the integration is completed.
Thus one finds that
\beq
\int_C dl [\nn\cdot\E(\x)]=\left\{ \begin{array}{cc}
2\pi q, & \mbox{ $q$ inside of C} \\
0, & \mbox{ $q$ outside of C} \end{array} \right.
\eeq
The {\bf{three-dimensional}} case works out much the same way. The field
varies as
$1/r^2$ and so one finds that $d^2x\cos\th/r^2$ is the solid angle element
$d\Om$ subtended by the infinitesimal area element $d^2x$ of S at the position
of the charge. Integration over the surface thus reduces to integration over
the solid angle subtended by the surface at the charge, and this is $4\pi$
if the charge in inside of the surface and 0 otherwise,
\beq
\ina[\E(\x)\cdot\nn]=\left\{ \barr{cc} 4\pi q, & \hbox{$q$ inside of S} \\
0, & \hbox{$q$ outside of S} \ear \right.
\eeq
Next, the superposition principle allows us to add up the fields arising
from an arbitrary collection of charges, with Gauss's Law holding for each
bit of charge. As a consequence, we may say that
\beq
\ina[\E(\x)\cdot\nn]=4\pi Q
\eeq
where $Q$ is the total charge contained inside of the surface,
\beq
Q=\inv\rh(\x).
\eeq
\section{Differential Form of Gauss's Law}
A differential form of this law may be found by applying the divergence
theorem which states that, for a general vector field $C(\x)$,
\beq
\ina[C(\x)\cdot\nn]=\inv[\div C(\x)].
\eeq
Let us apply this equation to Gauss's Law:
\beq
4\pi\inv\rh(\x)=\ina[\E(\x)\cdot\nn]=\inv[\div\E(\x)]
\eeq
or
\beq
\inv[\div\E(\x)-4\pi\rh(\x)]=0
\eeq
Because $V$ is completely arbitrary, we may equate the integrand to zero
and find
\beq
\div\E(\x)=4\pi\rh(\x)
\eeq
which is the differential form of Gauss's Law.
In the process of obtaining this equation from Coulomb's Law, we have lost
some of the information contained in it. Merely specifying the divergence
of a vector field is not sufficient to determine the field. Hence we need
an additional equation to supplement Gauss's Law.
\section{An Equation for $\curl\E$; the Scalar Potential}
Let us start once again from Coulomb's Law:
\beq
\E(\x)=\inivp\rh(\xp)\frac{\x-\xp}{|\x-\xp|^3}
\eeq
But one may write part of the integrand as a gradient,
\beq
\frac{\x-\xp}{|\x-\xp|^3}=-\grad\left(\frac{1}{|\x-\xp|}\right),
\eeq
where the gradient is taken with respect to the variable $\x$. Hence
\beq
\E(\x)=-\grad\inivp\frac{\rh(\xp)}{|\x-\xp|},
\eeq
which is to say, $\E$ can be written as the (negative) gradient of a scalar
function of $\x$. This function we shall call the scalar potential and
denote by $\Ph(\x)$:
\beq
\Ph(\x)\equiv\inivp\frac{\rh(\xp)}{|\x-\xp|};
\eeq
\beq
\E(\x)=-\grad\Ph(\x).
\eeq
From this statement it follows immediately that $\curl\E(\x)=0$ because the
curl of the gradient of a scalar function is always zero.
To summarize,
\beq
\div\E(\x)=4\pi\rh(\x)
\eeq
and
\beq
\curl\E(\x)=0.
\eeq
\subsection{Conservative Potentials}
From our derivation of the curl equation, we can see that this simple
result follows from the fact that the force (or electric field) is central
and depends only on the distance between charges. Such a force is also
called conservative, and the potential function is related in a simple way
to the energy of a charge in an electric field.
To find this relation,
consider that a set of fixed source charges produce a field $\E$ and that a
charge $q$ is placed at point $\x_a$. Here it experiences an electric field
force $\F=q\E(\x_a)$ and so an equal and opposite force
$\F_{ext}=-\F=-q\E(\x_a)$ must be applied by some external agent to keep it
in position.
\centerline{\psfig{figure=fig2.ps,height=3.0in,width=7.0in}}
\noindent If we now move
the charge along a path C from $\x_a$ to $\x_b$, the work done by the
external agent is found by integrating the force along the path,
\beq
W_{a\rightarrow b}=-\invl\cdot\F(\x),
\eeq
or
\beq
W_{a\rightarrow b}=q\invl\cdot\grad\Ph(\x) \\
=q\int_Cd\Ph=q[\Ph(\x_b)-\Ph(\x_a)]
\eeq
This result shows that $q\Ph(\x)$ can be interpreted as the potential
energy of charge $q$ in the electrostatic field at point $\x$, aside from a
constant defining the zero of potential energy. In going from $\x_a$ to
$\x_b$, work $q[\Ph(\x_b)-\Ph(\x_a)]$ is done on the charge, and so the
change in the energy of the system (composed of the charge $q$ and the
sources of the field) is just this work.
Notice especially that the work
does not depend on the path C except through the endpoints. This statement
can always be made of conservative systems. In particular, the integral
of the work done around a closed path is 0,
\beq
\invl\cdot\E(\x)=0
\eeq
It is instructive to apply Stokes' Theorem to this relation. His theorem
states that, for an arbitrary vector field \A, and a closed path C
with a surface S ``linking'' the path (which means that S is an open
surface with edges coinciding with C),
\beq
\invl\cdot\A=\ina[\curl\Ax]\cdot\nn
\eeq
where $\nn$ is a unit normal to the surface in the right-hand sense relative
to the direction in which the path is traversed. As applied to the electric
field, we have
\beq
0=\invl\cdot\E(\x)=\ina[\curl\E(\x)]\cdot\nn.
\eeq
Because C is arbitrary and can in particular be any infinitesimal closed
loop, this relation implies that the integrand is zero, $\curl\E(\x)=0$.
Thus the statement that $\E$ is a conservative field and
$\curl\E(\x)=0$ are equivalent.
\section{Poisson's and Laplace's Equations}
The differential equations we have determined for $\E$ are sufficient to
find it uniquely, given appropriate boundary conditions and the charge
density, but they do not necessarily provide the simplest approach to the
solution of an electrostatics problem. Often, it is best to solve for $\Ph$
from which $\E$ follows easily. Since $\div\E=4\pi\rh$, and
$\E =-\grad\Ph$, we have
\beq
\div\grad\Ph(\x)\equiv\lap\Ph(\x)=-4\pi\rh(\x)\;\;{\rm{Poisson's\;\;Equation}},
\eeq
which is {\em Poisson's Equation}; the operator $\lap$ is the
{\em Laplacian operator}. In those regions of space where the
charge density vanishes, we find the simpler equation,
\beq
\lap\Ph(\x)=0\;\;{\rm{Laplace's\;\;Equation}},
\eeq
which is {\em Laplace's Equation}.
Consider the effect of operating with $\lap$
on the integral expression for $\Ph$:
\beq
-4\pi\rh(\x)=\lap\Ph(\x)=\inivp\rh(\xp)\lap\left(\frac{1}{|\x-\xp|}
\right),
\eeq
or
\beq
\rh(\x)=-\inivp\rh(\xp)\left(\frac{1}{4\pi}\lap\frac{1}{|\x-\xp|}
\right).
\eeq
However, we defined $\de(\x-\xp)$ as
\beq
f(\x)=\inivp f(\xp)\de(\x-\xp)
\eeq
for general $f(\x)$. Since $\rh(\x)$ can be quite general, the quantity in
large parentheses above satisfies the condition placed on the delta
function; hence we conclude that
\beq
\lap\left(\frac{1}{|\x-\xp|}\right)=-4\pi\de(\x-\xp)
\eeq
which is only appropriate because $1/|\x-\xp|$ is the potential of a unit
point charge, and $\de(\x-\xp)$ is the corresponding charge density. Thus,
\eq{40} expresses Poisson's equation for a unit point charge
located at $\xp$.
As an exercise we may derive this result in a different way. Consider
\beq
\lap\lep \frac{1}{|\x|}\rip=\lap\lep\frac{1}{r}\rip=\frac{1}{r}
\frac{d^2}{dr^2}\lep r\frac1r\rip=0,
\eeq
except possibly at $r=0$ where $r/r$ is undefined. To determine what
happens here, we integrate $\lap (1/r)$ over a small sphere centered on the
origin:
\beqa
\inv\lap (1/r)=\inv \div\grad (1/r)=\nonumber\\
\ina[\grad (1/r)]\cdot\nn=-\ina (1/r^2)=-\int r^2dr\sth d\th d\phi (1/r^2)=-4\pi.
\eeqa
Thus we have shown the following:
\beq
\begin{array}{ccl}
(i) & \lap(1/r)=0, & r\neq 0 \\
(ii) & \inv \lap(1/r)(-1/4\pi) =1, & r=0 \in V.
\end{array}
\eeq
These results tell us that $\lap (1/r)=-4\pi\de(\x)$.
\section{Energy in the Electric Field; Capacitance; Forces}
The energy of the static electric field, or of a static charge
distribution,
is of some importance. Let us start our investigation by constructing the
energy of interaction of $n$ point charges $q_i$ located at $\x_i$. As we
have seen, the work required to move a charge $q$ from one point to another
in an applied electric field is $q$ times the difference in the electric
potentials at the end points. If we suppose that this potential is produced
by our collection of point charges, then it is given by
\beq
\Ph(\x)=\sum_{i=1}^n[q_i/|\x-\x_i|]
\eeq
and the work done to bring $q$ from infinitely far away, where $\Phx=0$,
to point $\x$ is
\beq
W=q\sum_{i=1}^n[q_i/|\x-\x_i|].
\eeq
This is therefore the increase in the total energy of the system of charges
when a charge is added to it at some particular point.
We may use this
result to calculate the energy of the collection of charges by bringing
them in one at a time from points at infinity where they are assumed to be
widely separated. The first charge is brought in to $\x_1$, and this costs
no energy because $\Ph=0$ when there are no other charges present.
The second charge costs energy
\beq
W_2=\frac{q_1q_2}{|\x_1-\x_2|}.
\eeq
The third then costs
\beq
W_3= q_3\sum_{j=1}^2\frac{q_j}{|\x_3-x_j|},
\eeq
and so on. The amount of work which must be done to bring in the $i^{th}$
particle is
\beq
W_i=q_i\sum_{j=1}^{i-1}\frac{q_j}{|\x_i-\x_j|}.
\eeq
If we add up these energies to find the total work done, it is
\beq
W=\sum_{i=2}^n\sum_{j=1}^{i-1}\frac{q_iq_j}{|\x_i-\x_j|}
\eeq
which is the sum over all pairs, each pair taken once; it may also be written as
\beq
W=\frac12\sum_{i,j}'\frac{q_iq_j}{|\x_i-\x_j|};
\eeq
the prime on the summation sign means that the terms with $i=j$ are
omitted. In this sum, we include each pair $i,j$ with $i\neq j$ twice and
so have to multiply by a factor of 1/2.
Given a continuous charge distribution, the same argument can be applied
using as the elementary charges infinitesimal charge elements located in
infinitesimal volume elements. The result must be
\beq
W=\frac12\invv\frac{\rh(\x)\rh(\xp)}{|\x-\xp|}
\eeq
where the integrations are unrestricted and include the points
$\x=\xp$ because the interaction energy of an infinitesimal continuously
distributed charge element with itself vanishes in the
limit of zero extent. However, if the charge distribution contains finite
point charges, represented by delta functions in $\rh(\x)$, then one has to
omit the interaction of each of these charges with itself, as in the original
sum, \eq{50}, in order to obtain a finite result.
The expression for $W$ can be cast into a number of other useful forms.
Recall that
\beq
\Ph(\x)=\inivp\frac{\rh(\xp)}{|\x-\xp|};
\eeq
substitution into the expression for $W$ gives
\beq
W=\frac12\iniv\rh(\x)\Ph(\x).
\eeq
Further, $\rh(\x)=-\lap\Ph(\x)/4\pi$, so
\beq
W=-\frac{1}{8\pi}\iniv\Phx\lap\Phx.
\eeq
Let us now do an integration by parts in three dimensions. This operation is
easy to achieve by making use of the divergence theorem; for a vector
field $f(\x)\A(\x)$ consider the integral
\beqa
\inv\div[f(\x)\A(\x)]=\ina f(\x)[\Ax\cdot\nn] \nonumber \\
=\inv[\grad f(\x)\cdot\A(\x)]+\inv f(\x) [\div\A(\x)]
\eeqa
or
\beq
\inv f(\x)[\div\A(\x)]=-\inv[\grad f(\x)]\cdot \A(\x)
+\ina f(\x)[\A(\x)\cdot\nn],
\eeq
where V and S are related in the usual way.
As applied to the integral for $W$, this useful formula gives, letting V be
all space,
\beqa
W=-\frac{1}{8\pi}\iniv\Ph(\x)\div\grad\Ph(\x) \\
=\frac{1}{8\pi}\iniv\grad\Ph(\x)\cdot\grad\Ph(\x)-\inia\Ph(\x)[
\grad\Ph(\x)\cdot\nn]
\eeqa
or
\beq
W=\frac{1}{8\pi}\iniv[\E(\x)\cdot\E(\x)].
\eeq
The surface integral has vanished because $\Ph$ falls off at distances $r$
which are large compared to the extent of the charge distribution at least
as fast as $1/r$. Hence the integrand in the surface integral falls off at
least as fast as $1/r^3$ while the area of the surface at distance $r$
varies as $r^2$. This integral therefore falls off at least as fast as $1/r$
and so vanishes when the surface is at infinity.
An interesting and plausible interpretation of the final expression is that
the integrand is the energy density $u(\x)$ of the electric field,
\beq
u(\x)=\frac{1}{8\pi}\Ex\cdot\Ex.
\eeq
This is only an interpretation, however. All we really know is that the
total energy is the integral of this quantity over all space. The idea is
plausible because $u(\x)$ so defined is everywhere positive or zero (a
negative energy density would be disturbing). Note, too, that our other
expression for the energy as an integral over a single position variable
has an integrand that can be both positive and negative which makes it
unreasonable to interpret that integrand, $\rh(\x)\Ph(\x)/2$, as the energy
density.
Given \eq{59} for $W$, we can see that the energy will be positive
definite. Yet the energy of, \eg~a pair of point charges $q$ and $-q$
at $\x$ and $\xp$ is negative, $-q^2/|\x-\xp|$. The reason is that the
expression we have for the energy of a set of point charges does not
include the (infinite) energy required to assemble each of the point charges
in the first place, but \eq{59} would include this (positive) energy.
A more concrete example involves two oppositely charged masses.
The energy required to bring them together from infinity is negative,
\centerline{\psfig{figure=charged_masses1.ps,height=1.0in,width=4.0in}}
\noindent whereas the energy required to assemble the entire charge
distribution
\centerline{\psfig{figure=charged_masses2.ps,height=1.0in,width=4.0in}}
\noindent at its final location is positive.
\subsection{Conductors}
Consider now the special case that our electrostatic system consists of a
collection of $n$ electrically isolated conductors; for our present purposes, a
conductor may be defined as an object which cannot support an electric
field (because it contains ``free'' charges which move under the influence
of a field until there is no field). Thus the interior of a conductor is an
equipotential. Using \eq{53}, we see that for such a system,
\beq
W=\frac12\iniv\rh(\x)\Ph(\x)=\frac12\sum_{i=1}^nQ_iV_i
\eeq
where $Q_i$ and $V_i$ are, respectively, the charge and potential on the
$i^{th}$ conductor. Now, because the potential is a linear function of
charge (superposition theorem), it is true that
\beq
V_i=\sum_{j=1}^np_{ij}Q_j.
\eeq
The coefficients $p_{ij}$ are independent of the charges; they depend
only on the distribution and shapes of the conductors and are called the
{\em coefficients of potential}.
To see that this relation is valid, one
need only think of the potentials produced on each conductor given charge
$Q_j$ on the $j^{th}$ conductor and zero charge on all others; then superpose
the solutions to each of the problems of this kind. Inversion of \eq{62}
yields the charges $Q_i$ as linear combinations of the potentials
$V_j$,
\beq
Q_i=\sum_{i=1}^nC_{ij}V_j.
\eeq
The coefficients $C_{ij}$ are called {\em coefficients of capacitance}; the
diagonal elements $C_{ii}$ are more commonly referred to simply as {\em
capacitances} while the off-diagonal ones $C_{ij}$ are called {\em
coefficients of electrostatic induction} and are not to be confused with
the inductances introduced in connection with Faraday's Law.
The capacitance of a single conductor, $C_{ii}$, is thus the total charge on
that conductor when it is maintained at unit potential while all other
conductors are held at zero potential.
As an example the capacitance of a pair
of conductors with equal and opposite charge is defined as the ratio of the
charge on one conductor to the potential difference between them when all
other conductors are maintained at zero potential.
\centerline{\psfig{figure=fig3.ps,height=2.0in,width=7.0in}}
\beq
\lep\begin{array}{c} Q\\-Q \end{array}\rip
=\lep\begin{array}{cc} C_{11} & C_{12}\\ C_{21} & C_{22} \end{array}\rip
\lep\begin{array}{c} V_1\\V_2 \end{array}\rip
\eeq
\beq
\lep\begin{array}{c} V_1\\V_2 \end{array}\rip
=\frac{\lep\begin{array}{cc} C_{22} & -C_{21}\\ -C_{12} & C_{11} \end{array}\rip
\lep\begin{array}{c} Q \\-Q \end{array}\rip}{C_{11}C_{22}-C_{12}C_{21}}
\eeq
\noindent The capacitance $C(1,2)=Q/|V_1-V_2|$ turns out to be
\beq
C(1,2)=(C_{11}C_{22}-C_{12}^2)/(C_{11}+C_{22}+2C_{12}).
\eeq
The energy of the system of conductors may be written in terms of
potentials and the coefficients $C_{ij}$ as
\beq
W=\frac12\sum_{i=1}^nQ_iV_i=\frac12\sum_{i,j=1}^nC_{ij}V_iV_j.
\eeq
\subsection{Forces on Charged Conductors}
Another useful application of the expressions for the energy is in the
calculation of forces on charged conductors. Consider the
surface of a conductor. The field at the surface can be inferred from $\div
\Ex=4\pi\rhx$ and $\curl\Ex=0$. Consider an integral of the first of
these equations over a ``pillbox'' or short right circular cylinder
oriented with the ``drumhead'' parallel to the surface of a conductor and
situated half inside and half outside of the conductor.
\centerline{\psfig{figure=pillbox.ps,height=2.5in,width=7.0in}}
\beq
\inv\div\E=\ina \E\cdot\nn=\E_n(\x) \pi \, a^2=
4\pi \inv \rhx= \pi a^2 4\pi \sigma(\x)
\eeq
\noindent Using the
divergence theorem, we may convert to a surface integral. Given that the
height $h$ of the cylinder is much smaller than its radius, $h<_S+\invp G_N(\x,\xp)\rhxp+\frac{1}{4\pi}\inap\pde{\Phxp}{n'}
G_N(\x,\xp)
\eeq
where $<\Ph>_S$ is the average of the potential over the surface S,
\beq
<\Ph>_S\equiv\frac1S\inap\Phxp
\eeq
One can understand the necessity of the presence of this term from the fact
that the Neumann boundary condition problem can only be solved up to an
arbitrary constant.
The Dirichlet Green's function is the one that we shall use most often
as one more commonly specifies the potential on the boundary than the
normal component of the electric field.
\edo