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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 
 material may not be reproduced for profit, modified or published in any 
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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, $h<<a$, the
only important contribution to the surface integral must come from the
drumheads. But $\Ex=0$ on the one inside of the conductor, so we pick up
only the contribution from the component of $\E$ normal to the surface of
the conductor on the outside. Given that $a$ is much smaller than distances
over which the field varies appreciably, we get simply $\pi a^2\E_n(\x)$
where $\x$ is a point just outside of the surface of the conductor and the
subscript $n$ designates the outward (from the conductor) normal component of
the field. The
volume integral of $\rhx$, on the other hand, yields the total charge
within the pillbox. Because $h<<a$, we will get a vanishingly small
contribution from any finite volume density of charge. However, we get an
important contribution from a {\em surface} charge density; we shall denote
such a thing by $\six$. It can be written as a volume charge density by
using a $\de$ function:
\beq
\rhx=\six\de(\xi(\x))
\eeq
where $\xi$ is the normal distance of the point $\x$ from the surface of the
conductor. When this surface charge density is integrated over the volume
of the pillbox, it reduces to a surface integral of the surface charge
density over a disc on the conductor surface and having the area of the
crosssection of the pillbox, \ie~$\pi a^2$. Hence one finds $\pi a^2(4\pi\six
)$. Putting the two sides of the equation together, we find the following
relation between the charge density on the surface of a conductor and the
normal component of the electric field just outside of the conductor:
\beq
\E_n(\x)=4\pi\six
\eeq

We may also find an equation for the {\bf{tangential component}} of the electric
field at the surface of a conductor. Consider the line integral of $\dvl
\cdot\Ex$ around a rectangle which straddles the conductor's surface. 

\centerline{\psfig{figure=fig4.ps,height=2.5in,width=7.0in}}
\beq
\ina (\curl\E)\cdot\nn=\invlc\cdot\E=\E_t h =0 \;\Longrightarrow\; \E_t=0
\eeq

\noindent The width $w$ of the rectangle, which is its size in the 
direction normal to
the interface, is much smaller than its height $h$, which is its size
parallel to the interface. The dominant contribution to the line integral,
which is zero, comes from the two sides parallel to the interface. On the
inside, $\Ex$ is zero, so we have only the integral along the side which is
exterior to the conductor. Since the whole integral is zero, the integral
along this single side must be zero, and hence we can conclude that the
tangential component of $\Ex$, or $\E_t(\x)$, just outside of a
conductor must vanish,
\beq
\E_t(\x)=0 \hbox{\hsp{0.5} at the surface of a conductor.}
\eeq

\centerline{\psfig{figure=fig6.ps,height=2.5in,width=3.0in}}

Now we are in a position to consider {\bf{the force on the surface of a
conductor}}. We use the method of virtual work. Imagine moving a small element
$dA$ of the conductor's surface, along with the charge on it, a distance $dx$
from its initial position in the direction normal to the surface. It will
sweep out a volume $dAdx$. 
In this volume, to a first approximation \footnote{This approximation 
improves as the ratio of $dx$ to $\sqrt{A}$ 
goes to zero}, there will no longer be any electric field (since the field
is zero within a conductor) while the electric field elsewhere will be
unchanged. Hence there will be a change in the field energy of
\beq
dW=-dAdx\frac{1}{8\pi}E_n^2=-2\pi\si^2dAdx + {\cal O}(dAd^2x,d^2Adx)
\eeq
where \eq{70} has been used for the normal component of the electric field.
Energy conservation demands that the amount of work done on the system in
making this displacement is equal to $dW$. It is also $dx$ times the negative
of the electric force acting on the area element $dA$. Thus,
\beq
-Fdx=dW \hbox{~or~} F=2\pi\si^2\,dA.
\eeq
Hence the force per unit area on the surface of the conductor is
$2\pi\si^2$; it is directed normally outward from the conductor 
(``negative pressure'').

A second way of looking at this problem is to calculate the force directly.
It must be the charge on $dA$, or $\si dA$, times the electric field which
acts on this charge, \ie~that part of the electric field at the surface
which is produced by charges other than those on $dA$. This field is just $2\pi
\si$ (Why?), so the force comes out at before.

\section{Green's Theorem}

In everything we have discussed thus far, we have assumed that $\rhx$ is
known, and that there is a simple boundary condition on $\phx$ at infinity
(that it must at least as fast as $1/r$).  {\em{This is not generally
true}}!  Usually, we only know $\rhx$ within some finite volume $V$,
and the value of $\phx$ on the corresponding surface $S$.

\centerline{\psfig{figure=fig5.ps,height=2.5in,width=7.0in}}

\noindent For these problems Green's theorem and functions are useful.
The simplest greens function is that for free space
\beq
G_{free}(\x,\x')=\frac{1}{|\x-\x'|}\,,
\eeq
and the corresponding potential is
\beq
\phx=\invp G_{free}(\x,\x') \rhxp \,.
\eeq
Note that $G_{free}(\x,\x')$ gives the response at $\x$, due to a unit 
point charge at $\x'$.  For  the response due
to a collection of charges $\rhx$, superposition yields the integral above.

For the more general problem of a collection of charges and boundary 
conditions on a surface, we might expect that this integral relation 
will become
\beq
\phx=\invp G(\x,\x')\rhxp + \;{\rm{surface\;\;term}}
\eeq
Before we can proceed further, we must develop some formalism.

\subsection{Green's Theorem}

To this end, let us develop Green's Theorem,
starting from the divergence theorem as applied to a vector field $\A$:
\beq
\inv[\div\Ax]=\ina[\Ax\cdot\nn]
\eeq
where S is the surface bounding the domain V.
Consider the special case $\Ax=\ph(\x)\grad\ps(\x)$; then
\beq
\div\Ax=\grad\ph(\x)\cdot\grad\ps(\x)+\ph(\x)\lap\ps(\x)
\eeq
and
\beq
\nn\cdot\Ax=\ph(\x)[\nn\cdot\grad\ps(\x)]=\ph(\x)\pde{\ps(\x)}{n}
\eeq
where $\pde{\ps}{n}$ is the outward normal derivative of $\ps$ at the
surface. Substitution into the divergence theorem produces
\beq
\inv[\grad\phx\cdot\grad\psx+\phx\lap\psx]=\ina\phx\pde{\psx}{n},
\eeq
a result known as {\em Green's first identity}. We may also start from a
vector field $\Ax=\ps(\x)\grad\ph(\x)$ and wind up with
\beq
\inv[\grad\psx\cdot\grad\phx+\psx\lap\phx]=\ina\psx\pde{\phx}{n}.
\eeq
Subtract the second expression from the first and obtain
\beq
\inv[\phx\lap\psx-\psx\lap\phx]=\ina\leb\phx\pde{\psx}{n}-\psx\pde{\phx}{n}
\rib,
\eeq
which is {\em Green's second identity}, also known as
{\em Green's Theorem}.

\subsection{Applying Green's Theorem 1}

We next make a particular choice of the scalar functions and also change
the integration variable to $\xp$:
\beq
\phxp=\Ph(\xp) \mbox{ and } \psxp=1/|\x-\xp|\equiv1/R.
\eeq
In the latter function, $\x$ is to be regarded as a parameter which will
eventually become the point at which we evaluate the potential.
Substitution into Green's Theorem gives
\beqa
\invp[\Phxp\lapp(1/R)-(1/R)\lapp\Phxp] \\
=\inap\leb\Phxp\pde{(1/R)}{n'}-\frac{1}{R}\pde{\Phxp}{n'}\rib,
\eeqa
or
\beq
-4\pi\Phx+4\pi\invp\frac{\rhxp}{|\x-\xp|}=\inap\leb\Phxp\pde{(1/R)}{n'}
-\frac1R\pde{\Phxp}{n'}\rib
\eeq
where we have assumed $\x$ is inside of V. With a little
rearrangement, the final equation can be written as
\beq
\Phx=\invp\frac{\rhxp}{|\x-\xp|}+\frac{1}{4\pi}\inap\leb\frac1R\pde{\Phxp}{n'}-
\Phxp\pde{(1/R)}{n'}\rib
\eeq
The first term on the right is the familiar volume integral over the charge
density, but notice that it no longer is over all space. The charge outside
of V does, of course, contribute to $\Phx$; its contribution is now taken
into account by the integral over the surface surrounding V. Note, too,
that if the point $\x$ is outside of V, then the left-hand side (LHS)
of the equation is zero. Further, if V is all space so that S is at
infinity, then the surface integral will vanish and we recover the volume
integral over all space. Note also that if $\rhxp=0$ for all $\xp$ in V,
then the potential is found simply from the surface integral of $\Ph$ and
its normal derivative.

\subsection{Applying Green's Theorem 2}

This equation is not the best one of its kind. It
is in fact (as we shall see) possible to find $\Phx$ from the charge
density in V  and from {\bf either} $\Phxp$ {\bf or}
$\partial{\Phxp}/\partial{n'}$ on the surface S; that is, it is not necessary
to know both of these things on the surface. When $\Ph$ is specified, that
is called {\em Dirichlet boundary conditions}; when the normal derivative
of $\Ph$ is given, that is called {\em Neumann boundary conditions}. Various
combinations are also possible, such as Dirichlet conditions on part of S
and Neumann conditions on the remainder, a case known as {\em mixed
boundary conditions}.

\begin{center}
\begin{tabular}{||l|l||}\hline
specification			& Boundary Condition	\\ \hline\hline
$\phxp$ specified on $S$	&	Dirichlet \\\hline
$\partial{\Phxp}/\partial{n'}$ specified on $S$ & Neumann \\ \hline
$\phxp$ specified on part of $S$ &	Mixed	\\
$\partial{\Phxp}/\partial{n'}$ specified on remainder & \\\hline\hline
\end{tabular}
\end{center}

Let us demonstrate that it is sufficient to know either $\Phxp$ or its
normal derivative on S in order to obtain a unique solution
to
\beq
\lap\Ph(\x)=-4\pi\rhx
\eeq
with some volume $V$ bounded by $S$. 
Start by supposing that either type of boundary
condition is given and that there are two distinct solutions $\Ph_1$ and
$\Ph_2$. Define $U=\Ph_1-\Ph_2$. This function is such that
\beq
\lap U(\x)=0  \hbox{~inside of V}
\eeq
and
\beq  \barr{ccl}
\hbox{either} &   U(\x)=0,          & \hbox{for $\x$ on S (Dirichlet)} \\
\hbox{or}  &      \pde{U(\x)}{n}=0,  & \hbox{for $\x$ on S (Neumann).}
\ear  \eeq
In Green's first identity \eq{81}, let $\ps=\ph=U$:
\beq
\inv[U(\x)\lap U(\x)+\grad U(\x)\cdot\grad U(\x)]=\ina U(\x)\pde{U(\x)}{n}.
\eeq
Now, since $U$ satisfies the Laplace equation, the first term in brackets
on the left vanishes. From the boundary condition for either the Neumann or
the Dirichlet problem, the surface integral also vanishes. Hence we have
just
\beq
\inv|\grad U(\x)|^2=0
\eeq
from which it follows that $\grad U(\x)=0$ in V.
Therefore $U(\x)$ is a constant in V and so the two solutions $\Ph_1$ and
$\Ph_2$ are the same up to a constant. For Dirichlet conditions, this
constant is zero since the two functions are the same on the boundary. For
Neumann conditions it is arbitrary and amounts only to a choice of the
zero of potential.

The preceding proof is also valid for the case of mixed boundary conditions
because the surface integral vanishes in this case also. Finally, one
cannot in general specify {\em both} $\Phx$ and $\pde{\Ph}{n}$ everywhere
on S ({\em Cauchy} boundary conditions); either one alone is sufficient to
determine a unique solution and the two solutions so determined are not
necessarily the same. However, if Cauchy boundary conditions are given on
just an appropriate part of S, that can be sufficient to give a unique
solution for the potential.

In the light of what we have learned, it is evident that our integral
expression for $\Phx$, \eq{88}, which involves surface integrals of both the
potential and its normal derivative, is not a very effective way to solve
an electrostatic boundary value problem; it requires more input information
than is actually needed to determine a solution and so is an integral
equation as opposed to a solution in the form of an integral. If we had made a
better choice of $\psxp$ at the outset, we could have come up with a better
result. Let's try again, choosing for $\psxp$ a function we shall call
$\gxxp$; it is given by $1/\xxpa$ plus an as yet undetermined function
$\fxxp$ which is to be a solution of the Laplace equation in V,
\beq
\lapp\fxxp=0
\eeq
for $\x$ and $\xp$ in V.
Since
\beq
\gxxp=\xxpi+\fxxp,
\eeq
it is the case that
\beq
\lapp\gxxp=-4\pi\de(\x-\xp)
\eeq
for $\x$ and $\xp$ in V.
Physically, the function $G$, viewed as a function of $\xp$, is a solution of
Poisson's equation in V given a unit point charge at $\x$. The function
$\fxxp$ is a solution of Laplace's equation; we shall presently determine its
properties further by requiring that it satisfy certain conditions on S.
Using $\gxxp$ for $\psxp$ and $\Phx$ for $\phx$ in Green's Theorem, 
we can easily show that
\beq
\Phx=\invp\gxxp\rhxp+\frac{1}{4\pi}\inap\leb\gxxp\pde{\Phxp}{n'}
-\Phxp\pde{\gxxp}{n'}\rib.
\eeq

\subsubsection{Greens Theorem with Dirichlet B.C.}

Now consider in turn two different sets of boundary conditions for $G$.
First, require that $\gxxp=0$ for $\xp$ on S and denote this function by
$G_D$. Then the preceding equation
becomes
\beq
\Phx=\invp G_D(\x,\xp)\rhxp-\frac{1}{4\pi}\inap\pde{G_D(\x,\xp)}{n'}
\Phxp.
\eeq
This is a useful equation when we have a Dirichlet problem with $\Phxp$
specified for $\xp$ on S. Then we have in principle the information we need
to complete the integration and so find $\Phx$ in which case \eq{98} is an
integral solution for $\Phx$ as opposed to an integral equation for the
potential. 

The question naturally
arises, does the function $G_D$ exist? That is, is it possible to find the
Dirichlet Green's function $G_D$ which is specified by the conditions
\beq
\lap G_D(\x,\xp)=-4\pi\de(\x-\xp); \hbox{~$\x,\xp$ in V}
\eeq
and
\beq
G_D(\x,\xp)|_{\xp \hbox{~on S}}=0?
\eeq
The answer is that this function does exist; further, it is unique. The
preceding two conditions are sufficient to determine it completely. We know
this without resorting to a mathematical proof because we can see that
$G_D(\x,\xp)$ is just the scalar potential at $\xp$ given a unit point charge
at $\x$ inside of a cavity with conducting walls coincident with S and held at
zero potential. 

\centerline{\psfig{figure=fig7.ps,height=1.5in,width=2.5in}}

\noindent This is the physical interpretation of the Dirichlet
Green's function. Notice in particular that this is a strongly
geometry-dependent function (it depends on S very much) but it is not
dependent on any other properties of the system. In other words, we can
solve any Dirichlet problem for a given geometry if we can solve the
``point charge with grounded conducting surfaces'' problem for the same
geometry in the sense that we can reduce the solution to a quadrature, \ie
to an integral.

An important property of the Dirichlet Green's function is that it is
invariant under interchange of $\x$ and $\xp$, $\gxxp=G(\xp,\x)$. To
demonstrate this property, let $\ph(\y)
=G(\x,\y)$ and $\ps(\y)=G(\xp,\y)$. Then insert these functions into
Green's Theorem (with $\y$ as the integration variable) 
\beqa
&&\invy[G(\x,\y)\lap G(\x',\y)-G(\x',\y)\lap G(\x,\y)]=\nonumber \\
&&\inay\leb G(\x,\y)\pde{G(\x',\y)}{n}-G(\x',\y)\pde{G(\x,\y)}{n}
\rib,
\eeqa
and make use of the
properties of the Dirichlet Green's function that $\lap_{\y}G(\x,\y)=
-4\pi\de(\x-\y)$ and $G(\x,\y)=0$ for $\y$ on S. The result is that
$G(\x,\xp)=G(\xp,\x)$.

\subsubsection{Greens Theorem with Neumann B.C.}

The second case of boundary conditions we consider on $G$ is
$\partial\gxxp/\partial n'=0$ for $\xp$ on S. Then application of Green's
Theorem (\eq{97}) leads to
\beq
\Phx=\invp\gxxp\rhxp+\frac{1}{4\pi}\inap\pde{\Phxp}{n'}\gxxp.
\eeq
Unfortunately, this $G$ does not exist as we may show by applying
Gauss's Law,
\beq
0=\inap\pde{\gxxp}{n'}=\inap[\nnp\cdot\gradp\gxxp]=\invp\lapp\gxxp=-4\pi.
\eeq
Clearly, we cannot have a $G$ with zero normal derivative everywhere on S.
The next simplest possibility is that
\beq
\pde{G_N(\x,\xp)}{n'}|_{\xp \hbox{~on S}}=-\frac{4\pi}{S},
\eeq
where $S$ is the area of the surface. Given such a function, we can use it
in Green's Theorem and will be led to the following integral expression for
the scalar potential:
\beq
\Phx=<\Ph>_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


