\documentstyle[12pt,psfig]{article}
\message{
 	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 
 form (this includes electronic redistribution) without the prior written 
 permission of the authors listed above.	
}
\input{../defs.tex}
\def\baselinestretch{1.4}
\title{Multipoles; Macroscopic Media; Dielectrics}
\author{Henry Cavendish\\(1731 - 1810)}
\bdo
\maketitle

\tableofcontents

\pagebreak

\large

In this chapter, we shall first develop the multipole expansion for the
electrostatic potential and field. This is useful not only for
expressing the field produced by a localized distribution of charge but
is also a helpful preliminary investigation for the business of describing the
electrostatics of materials containing a large number of charges and
which are not conductors.
These are called {\em dielectrics}. After developing a means of describing
their electrostatic properties, we shall turn to boundary value
problems in systems comprising dielectrics and conductors.

\section{Multipole Expansion: An Alternate Approach}
In this section we will develop the multipole expansion for a charge 
distribution by an alternate means to that used in Jackson (the method
used in Jackson is discussed in the appendix to this chapter).

We begin by writing the general expression for the potential due to 
a finite charge distribution $\rh(\x)$,
\beq
\Phx=\inivp\rhxp\xxpi
\eeq
\medskip
Let us consider the case where the origin is within the charge distribution 
and where $|\x|=r$ is large compared to the size of the charge distribution.
Then we may expand the denominator in the integrand.
\beq
\xxpi=\frac{1}{\sqrt{r^2-2\x\cdot\xp+r'^2}}=
\frac{1}{r\sqrt{1-2\x\cdot\xp/r^2 + r'^2/r^2}}.
\eeq
Using the Taylor series expansion
\beq
\frac{1}{\sqrt{1-x}}=1 + \frac{1}{2}x + \frac{3}{8}x^2 + \cdots
\eeq
We get
\beq
\xxpi=\frac{1}{r}\lep 1 + \frac{1}{2}\lep 2\x\cdot\xp/r^2 - r'^2/r^2\rip
+\frac{3}{8}\lep 4(\x\cdot\xp)^2/r^4 \rip \rip + 
{\rm{order}}\lep (r'/r)^3\rip
\eeq
\beq
\xxpi=\frac{1}{r} + \frac{\x\cdot\xp}{r^3} + \frac{1}{2r^5}\lep
3(\x\cdot\xp)^2 - r'^2r^2 \rip + \cdots
\eeq
With this expansion, we can rewrite $\Ph(\x)$ as
\beqa
\Phx &= &\inivp \frac{\rhxp}{r}  + \nonumber\\
&&\inivp \frac{(\x\cdot\xp)\rhxp}{r^3}  +\nonumber\\
&&\inivp \frac{\rhxp}{2r^5} \lep
3(\x\cdot\xp)^2 - r'^2r^2 \rip + \cdots \nonumber
\eeqa
\beqa
\Phx &= &\frac{1}{r}\inivp \rhxp \;\;\;{\rm{monopole}} + \nonumber\\
&& \frac{\x}{r^3} \cdot \inivp \xp\rhxp \;\;\; {\rm{ dipole}} +\nonumber\\
&&\frac{1}{2r^5} \inivp \rhxp \lep
3(\x\cdot\xp)^2 - r'^2r^2 \rip \;\;\; {\rm{quadrupole}} + \cdots \nonumber 
\eeqa
or
\beqa
\Phx &= &\frac{1}{r}\inivp \rhxp  + \nonumber\\
&& \frac{\x}{r^3} \cdot \inivp \xp\rhxp +\nonumber\\
&& \frac{1}{2}\sum_{i,j} \frac{x_ix_j}{2r^5} \inivp \rhxp \lep
3x'_ix'_j-r'^2\delta_{ij} \rip  + \cdots 
\eeqa
where the sum is over the 3 coordinates of space.  If we follow the 
conventional designation of these terms, then
\beq
\Phx= q/r + \frac{\x\cdot\p}{r^3} + 
\frac{1}{2}\sum_{ij}Q_{ij}\frac{x_ix_j}{r^5}
+ \cdots
\eeq
where
\beq
q=\inivp \rhxp \;\;\;\; {\rm{Monopole\; Moment}}
\eeq
\beq
\p=\inivp \rhxp \xp \;\;\;\; {\rm{Dipole\; Moment}}
\eeq
\beq
Q_{ij} =\inivp \rhxp \lep 3x'_ix'_j-r'^2\delta_{ij} \rip \;\;\;\; 
{\rm{Quadrupole\; Moment}}
\eeq

Note that the matrix $Q_{ij}$ is real an symmetric ($Q_{ij}=Q_{ji}$).
Thus only six of its elements are independent.  In fact, only 5 are,
since there is an additional constraint that ${\rm{Tr}}(Q)=0$.
\beq
{\rm{Tr}}(Q)=\sum_i Q_{ii}=\inivp \rhxp \sum_i  3x'_ix'_i-r'^2\delta_{ii}
\eeq
then as
\beq
\sum_i  3x'_ix'_i= 3r'^2
\eeq
\beq
\sum_i  r'^2\delta_{ii} =3r'^2
\eeq
\beq
{\rm{Tr}}(Q)=\inivp \rhxp  \lep 3r'^2- 3r'^2 \rip =0
\eeq
Thus it must be that $Q_{33}=-Q_{11}-Q_{22}$ and only two of the 
diagonal components are independent.  This is important,
since we will relate $Q$ to the set of five $Y_2^m$.

\subsection{Interpretation of the Moments}
What is the interpretation of these terms?  The {\em{monopole moment}}
is just the total charge of the distribution.  Thus the monopole term
gives the potential due to the charge as a whole.  Since the monopole
term in the potential falls off like $1/r$ at large $r$, it will
dominate the far field potential whenever $q$ is finite.  The
{\em{dipole moment}} is the first moment of the charge distribution;
and refers to how the charge is distributed in space.  Similarly,
the quadrupole moment is a second moment of the distribution.

	Let's consider the dipole in some detail, with the model
shown below. 

\centerline{\psfig{figure=fig1.ps,height=1.8in,width=7.0in}}

\beq
\Phx= \frac{q}{|\x-\a/2|} + \frac{-q}{|\x+\a/2|}
\eeq
For $|\x|=r\gg a$ we can expand
\beq
\frac{1}{|\x\mp\a/2|}=\frac{1}{r}\lep 1 \pm \frac{\x\cdot\a}{2r^2}\rip +
{\rm{order}}(a^2/r^2)
\eeq
\beqa
\Phx&\approx &
\frac{q}{r}\lep 1 + \frac{\x\cdot\a}{2r^2}\rip -
\frac{q}{r}\lep 1 - \frac{\x\cdot\a}{2r^2}\rip \nonumber \\
&\approx& \frac{q}{r^3} \x\cdot\a \nonumber \\
\eeqa
If $\a\to 0$ in the diagram (such that $q\a=\p=$ constant), then
the higher order terms vanish, and this result becomes exact.
In such a limit ($\a\to 0$, $q\a=\p$) we obtain a point dipole.

Now consider the field due to a dipole $\p=p\zh$
\beq
\Phx=\frac{p\cth}{r^2}
\eeq
\beq
E_r=-\pde{\Ph}{r}=\frac{2 p \cth}{r^3}
\eeq
\beq
E_{\th}=-\frac{1}{r}\pde{\Ph}{\th}=\frac{p\sth}{r^3},
\eeq

\centerline{\psfig{figure=fig2.ps,height=1.8in,width=3.0in}}

\noindent or more formally (in Cartesian coordinates).
\beqa
E_{\p}&=&-\grad\lep\frac{\p\cdot\x}{r^3}\rip=-\sum_i {\bf{e}}_i \pde{}{x_i}
\sum_j\frac{p_j x_j}{r^3} \nonumber \\
&=&-\sum_i {\bf{e}}_i \lep \frac{p_i}{r^3}- 3\sum_j\frac{p_j x_j}{r^4}
\frac{x_i}{r}\rip \nonumber\\
\eeqa
where we have used the fact that $\pde{r}{x_i}=\frac{x_i}{r}$.
\beq
E_{\p}=
\frac{3\x(\p\cdot\x)}{r^5} - \frac{\p}{r^3}
\eeq

In a similar fashion the potential term involving the quadrupole moment
may be interpreted as due to an assembly of four charges (hence the
name).  

\vspace{0.1in}
\centerline{\psfig{figure=fig3.ps,height=1.8in,width=1.8in}}

\noindent Higher-order moments (octapole, hexadecapole, etc.) may be
generated in a like fashion.

	It is important to note that the interpretation of the moments
depends strongly upon the origin.  For example consider a point charge
located at the origin.

\vspace{0.1in}
\centerline{\psfig{figure=fig4.ps,height=1.8in,width=1.8in}}

\beq
\Phx=q/r
\eeq
It has only a monopole term.  Now displace the charge by a vector
$\a$.

\vspace{0.1in}
\centerline{\psfig{figure=fig5.ps,height=1.8in,width=1.8in}}

\beq
\Phx=\frac{q}{|\x-\a|}=q/r + q\frac{\a\cdot\x}{r^3} +
\frac{q}{2} \frac{3(\a\cdot\x)^2-a^2r^2}{r^5} + \cdots
\eeq
This has moments to all orders!  {\em{watch your origin}}!

\subsection{Dipole Field}
It is interesting to ask what is going on at the origin, where our expansion
fails. Let's look at the particular case of the dipole field, assuming a
point dipole at $r=0$. For any $r>0$, we know that the potential is as
given in \eq{17}. Once before we found such a potential when we solved the
problem of a
conducting sphere of radius $a$ in a uniform external applied field $\E_0$.
What we found was that the potential outside of the sphere is
\beq
\Phx=-E_0r\cth+E_0a^3\frac\cth{r^2}
\eeq
while inside of the sphere the potential is a constant and the field is zero.

\vspace{0.1in}
\centerline{\psfig{figure=fig6.ps,height=2.0in,width=5.0in}}

If one removes the applied field but retains the field produced by the
charges on the surface of the sphere, then the potential for $r>a$ is simply
$E_0a^3\cth/r^2$ which may
also be written as $\edm\cdot\x/r^3$ with $\edm=E_0a^3\zh$. By the
superposition principle, the field at $r<a$ is now $-\E_0=-\edm/a^3$. 
What
this means is that the surface charge on the sphere has a dipole moment
and, remarkably, no other multipole moments. Now let us fix $\edm$ while
letting $a\rightarrow0$. The region $r<a$ shrinks,
while the field inside gets bigger. As the region shrinks to zero, the
field strength at the origin (\ie inside the sphere) diverges. The integral of
the field over the spherical domain $r<a$ is, however, a constant and equal to
$-(4\pi/3)\edm$. Consequently, in
the limit of vanishing $a$, this field may be represented by a delta
function, $-(4\pi/3)\edm\de(\x)$. The total field of a point dipole of
moment $\edm$ is thus the dipolar
field, \eq{21}, for $r>0$ plus a delta-function piece at the origin,
\beq
\Ex=\frac{3\nn(\edm\cdot\nn)-\edm}{r^3}-\frac{4\pi}{3}\edm\de(\x).
\eeq

Our derivation of this result is not completely general since it is based
on the limiting form of the solution to one particular problem involving a
sphere; the result is, however, quite correct for any point dipole. See
Jackson, Chapter 4, Section 1, for a more complete discussion of this
point.
\section{Energy of the Charge Distribution}
In this section we consider the energy of a localized charge distribution
$\rhx$ in an external applied electric field $\Ex$ which may be described
through its potential $\Phx$. This energy is, as we know from Chapter 1,
\beq
W=\iniv\rhx\Phx.
\eeq

\vspace{0.1in}
\centerline{\psfig{figure=fig7.ps,height=1.5in,width=4.0in}}

\noindent Notice that there is no factor of 1/2; that is because we are 
finding the
interaction energy of a charge distribution with a field which is {\em not}
produced by that same charge distribution and so we do not double count the
energy in \eq{24} by omitting this factor.

Now if we assume that $\Ph$ changes slowly over the region where
$\rho$ is appreciable, then we can expand the potential $\Ph$ around 
the origin of coordinates using a
Taylor series:
\beqa
\Phx=\Ph(0)+\x\cdot\grad\Phx|_{\x=0}+\frac12\sum_{i,j=1}^3x_ix_j\left.\frac
{\partial^2\Phx}{\partial x_i\partial x_j}\ril_{\x=0}+...\nonumber\\
=\Ph(0)-\x\cdot\E(0)-\frac12\sum_{i,j=1}^3x_ix_j\left.\pde{E_j(\x)}{x_i}
\ril_{\x=0}+...
\eeqa
where $\Ex$ is the external applied field. Now, this field is such that
its sources are far away, or at least are zero in the region where the
charge distribution $\rhx$ is located. Therefore $\div\Ex=0$ in this
region and so we can add a term proportional to $\div\Ex$ to the potential
$\Phx$ without changing the result of the integral in \eq{24}. We choose
this term to be
\beq
\frac16r^2\div\Ex|_{\x=0}=\frac16\sum_{i,j=1}^3r^2\de_{ij}\left.\pde{E_j(\x)}
{x_i}\ril_{\x=0}.
\eeq
Hence we have
\beq
\Phx=\Ph(0)-\x\cdot\E(0)-\frac16\sum_{i,j=1}^3(3x_ix_j-r^2\de_{ij})
\left.\pde{E_j(\x)}{x_i}\ril_{\x=0},
\eeq
plus higher-order terms. If we substitute this expansion into the
expression for the energy, we find
\beq
W=q\Ph(0)-\edm\cdot\E(0)-\frac16\sum_{i,j=1}^3Q_{ij}\left.\pde{E_j(\x)
}{x_i}\ril_{\x=0}+... .
\eeq

\subsection{Example: Dipole Energies}
As an example making use of this result, suppose that we have a dipole of
moment $\edm_1$ at point $\x_1$ in the presence of a second dipole of
moment $\edm_2$ at $\x_2$. Then the energy of interaction is

\vspace{0.1in}
\centerline{\psfig{figure=fig8.ps,height=1.5in,width=4.0in}}

\beq
W=-\edm_1\cdot\E_2(\x_1)=\frac{-3(\edm_2\cdot\nn)(\edm_1\cdot\nn)+
\edm_1\cdot\edm_2}{|\x_1-\x_2|^3}
\eeq
where $\nn=(\x_1-\x_2)/|\x_1-\x_2|$ is a unit vector pointing from the
second dipole to the first (or vice versa).

\subsection{Example: Quadrupole Energies}
A second example has to do with the coupling of a nucleus' electric
quadrupole moment to an external field (such as that from the electrons).
By choosing the origin in an appropriate fashion, one can guarantee that
any nucleus (any object with a non-zero net charge, in fact) has no dipole
moment. Hence the first interesting term in the nucleus' interaction with
an external field is the electric quadrupole interaction. Further, a nucleus
in an angular momentum eigenstate $|J,M>$ will have a charge density which
is invariant under rotation around the $z$-axis, 
\[
\rho\propto Y_J^M Y_J^{M*} \propto e^{i M\phi}e^{-i M\phi}
\]
leading to a diagonal
electric quadrupole moment tensor (the matrix of $Q_{lm}$'s) which is such
that $Q_{xx}=Q_{yy}$. Since the trace of this tensor (or matrix) is zero,
this means that\footnote{Or maybe that means this: $Q_{11}=Q_{22}=-Q_{33}/
2$.} $Q_{xx}=Q_{yy}=-Q_{zz}/2$. The upshot is that the interaction of the
nuclear quadrupole with the applied field is
\beq
W=-\frac14Q_{zz}\left.\pde{E_z(\x)}{z}\ril_{\x=0}.
\eeq
Bear in mind that the moment $Q_{zz}$ is a function of the internal state
of the nucleus and in particular of its angular momentum states. The
quadrupolar coupling thus provides a way to lift the degeneracy associated
with the different quantum numbers $M$ for the $z$-component of angular
momentum.

\section{Dipoles in Nature: Permanent and Induced}

Why are dipoles so interesting?? The reason is that many atoms and 
molecules have dipole moments which affects their chemical and electrical
properties.

\subsection{Permanent Dipoles}
	An example of a molecule with a permanent electric dipole moment
is water H$_2$O.
\[
|\p|=1.86x10^{-18} {\rm{esu-cm}}=1.86 {\rm{Debyes}}
\]

\vspace{0.1in}
\centerline{\psfig{figure=fig9.ps,height=1.5in,width=2.0in}}

This dipole moment corresponds to approximately one electron charge
separated across the size of the molecule.  Other polar molecules
have similar dipole moments (NH$_3$:  1.47 debyes, HCl:  1.03 debyes).

	Atoms or Nuclei cannot have permanent dipole moments, since
they are in states of good angular momentum $l$:  the dipole moment
of an electron in such a state vanishes.  In contrast the molecules 
mentioned above are in sp hybrid orbitals, so that $l$ is not a good 
quantum number, and thus a dipole moment is allowed.

\subsection{Induced Dipoles}
	Atoms and molecules that lack permanent dipoles {\em{can}} 
have induced dipole moments when placed in an external electric
field.  

\subsubsection{Static Models}
We have already seen the effect of an external field
inducing a dipole moment in a metallic sphere.

\vspace{0.1in}
\centerline{\psfig{figure=fig10.ps,height=1.8in,width=3.0in}}

\beq
\Phx=\Phi_{external}(\x) + \Phi_{induced}(\x)
\eeq
\beq
\Phi_{induced}(\x)=\frac{\p\cdot\x}{r^3}\;\;\;\p=E_o a^3\zh
\eeq
The induced dipole moment is proportional to the external
electric field.  If we define $\alpha$ to be the polarizability
of the body, then
\beq
\p=\alpha\E
\eeq
where, in this case, $\alpha=a^3$.  We see that in general the polarizability
of the order of magnitude of the volume of the body.  Thus for an
atom
\beq
\alpha_{atom}\approx {\rm{atomic\; volume}}\approx 10^{-24} {\rm{cm}}^3
\eeq
This is consistent with experiment.

	To see what this means in realistic terms, consider an atom
placed in a relatively large electric field $E= 100\;\;{\rm{statvolts/cm}}$.

\centerline{\psfig{figure=fig10a.ps,height=1.0in,width=1.0in}}

\noindent Lets assume that the induced dipole moment is 
\beq
\p=\alpha\E =e\a
\eeq
where $e$ is an electronic charge and $|\a|$ is the distance which
separates the charge.  Then
\beq
a=\frac{\alpha E}{e} \approx 
\frac{ (10^{-24} {\rm{cm}}^3)(100 {\rm{statvolts/cm}})}
{4.8\times10^{-10} {\rm{esu}}} \approx 2\times10^{-13} {\rm{cm}}
\eeq
This is of nuclear dimensions ( 2 fermi's).  Thus the atom is quite
rigid to polarization. To have a distortion of the order of
and Angstrom, we need a field of order $E= 5\times10^{6} {\rm{statvolts/cm}}$.
This type of field strength is only available with a laser.

\subsubsection{Dynamic Model}
	We may also calculate $\alpha$ for an atom using a simple dynamical
model.  Suppose that the electron is bound to the ion by a spring,
so that, if displaced from equilibrium, it feels a restoring
force.
\beq
\F_{restore}=-m\wo^2\x
\eeq
where $m$ is the electronic mass, and $\wo$ the frequency of oscillation.
If we apply an external electric field $\E$, the displacement $\x$ of
the electron from equilibrium will grow until $\F_{restore}$ is 
equal and opposite to the electronic force on the electron.
\beq
-(-e\E)=-m\wo^2\x \;\;\; {\rm{so}}\;\; \x=\frac{-e\E}{m\wo^2}
\eeq
The induced dipole moment is then
\beq
\p=-e\x=\frac{e^2}{m\wo^2} \E= \alpha\E
\eeq
So that the atomic polarizability is 
\beq
\alpha=\frac{e^2}{m\wo^2}
\eeq
Now, we expect that $\wo\approx$ angular frequency of oscillation,
which is approximately the frequency of the light which is emitted by
atoms.  For a wavelength of $3000$ Angstroms $\wo\approx 6\times10^{15} s^{-1}$,
giving 
\[
\alpha\approx 6\times10^{-24} {\rm{cm}}^3.
\]
This is in accord with our previous estimate.


\section{Dielectric Materials}

The electrostatic properties of some insulating materials may be modeled 
by a collection of dipole molecules, each with a dipole moment.  Higher
order moments are usually neglected.  Our main interest here is not in 
the dipole moments of individual
atoms or molecules, but rather the dipole moments of atoms or molecules
in a solid.  In such a medium, we expect that there will be no net
permanent dipole moment.  This is for two reasons:
\begin{tabbing}
xxxxxxx\=  \kill\\
(1)	\>If there is a permanent dipole moment in the atoms or \\
	\>molecules which make up the system, then the orientation\\
	\>of them will be random.  Thus the average dipole moment\\
	\>$<\p>$ will be zero. \\
	\> \\
(2)	\>If there is no permanent dipole moment of the component\\
	\>atoms or molecules, then in the absence of an external\\
	\>field, each will have no dipole moment, and thus the average\\
	\>dipole moment $<\p>$ will also be zero\\
\end{tabbing}
	
\subsection{Statistical Mechanics}

Now suppose that we do apply an external electric field, what will
$<\p>$ be then?  What effect will thermal fluctuations have?
We must again consider the ensemble of molecules for
two different cases.

\subsubsection{Induced dipoles}

If each molecule has an induced dipole moment, then the Hamiltonian for 
each molecule is
\beq
U=-\edm\cdot\E+\frac12m\om_0^2r^2.
\eeq
We can thus find the thermal average value of $\x$ by averaging it over the
distribution $\exp (-U/kT)$, where $k$ is Boltzmann's constant and $T$ is
the temperature:
\beq
<\x>=\frac{\iniv  e^{-U/kT}\x}{\iniv  e^{-U/kT}}
\eeq
{\bf{Let $\E$ define the $z$-direction}} and have
\beqa
<\x>=\frac{\iniv e^{(eEr\cos\th-\frac12m\om_0^2r^2)/kT}r\cos\th\zh}
{\iniv e^{(eEr\cos\th-\frac12m\om_0^2r^2)/kT}} \nonumber \\
=\frac{\int dze^{(eEz-\frac12m\om_0^2z^2)/kT}z\zh}{\int dze^{(eEz-\frac12m
\om_0^2z^2)/kT}}\nonumber\\
=\frac{\int du e^{-\frac12m\om_0^2u^2/kT}(u+eE/m\om_0^2)\zh}{\int du e^{
-\frac12m\om_0^2u^2/kT}}
\eeqa
where $u\equiv z-eE/m\om_0^2$. The remaining integrals cancel nicely and we
find that 
\beq
<\x>=(eE/m\om_0^2)\zh\;\;\;\p=(e^2E/m\om_0^2)\zh\;\;\;\alpha=(e^2/m\om_0^2)
\eeq
the same as before we introduced thermal
fluctuations in the separation. Thus thermal effects vanish.

\subsubsection{Permanent Dipoles}

 For permanent dipoles (remember H$_2$O) we may do 
something similar using a Boltzmann distribution
$\exp(-U/kT)$, as in the previous example but this time $U$ is simply
\beq
U=-\edm\cdot\E
\eeq
with $\edm$ fixed in magnitude. Thus, letting $\E$ define
the $z$-direction again, we have
\beq
<\edm>=\frac{\inom e^{pE\cos\th/kT}p\cos\th\zh}{\inom
e^{pE\cos\th/kT}}=kT\zh\der{}{E}\ln\leb\int_{-1}^1due^{pEu/kT}\rib.
\eeq
The integral is easy; upon taking the derivative and simplifying the result
insofar as possible, one finds
\beq
<\edm>=p\zh\lep\coth(Ep/kT)-kT/Ep\rip.
\eeq
As $T\rightarrow0$, this becomes $p\zh$, meaning that the dipole is
perfectly aligned with the field. For large $T$, $kT>>pE$, we may expand
the hyperbolic cosine and find the leading term
\beq
<\edm>=\frac13\frac{p^2E}{kT}\zh.
\eeq
This is the most frequently encountered situation at \eg room temperature;
it leads to a polarizability which is
\beq
\al=\frac13\frac{p^2}{kT}.
\eeq

\subsubsection{Both}

Finally, if a molecule has both a permanent moment and the possibly of
being polarized, then the polarizability consists of both a
temperature-independent term and one which varies inversely as the
temperature,
\beq
\al=\frac{e^2}{m\om_0^2}+\frac{p^2}{3kT}.
\eeq

	In each of the model calculations, where the dipoles are induced or 
permanent or both, the mean dipole moment induced in the material by an 
external field is proportional to that field.  This is the basic assumption
which we will use to explore the electrostatics of dielectric materials.

\subsection{Macroscopic Electrostatics; Dielectrics}
Before this section, we have considered only one kind of macroscopic material,
conductors. Within conductors, there is no electric field, we said, because
a conductor is an equipotential. If we had bothered to think a bit about
that statement, we would have realized that it is a statement which applies
only in some average sense. If one looks at the microscopic structure of a
conductor or any other material, one finds electrons and nuclei with very
strong electric fields reflecting the forces that act between these
objects. 

\vspace{0.1in}
\centerline{\psfig{figure=fig11.ps,height=1.5in,width=6.0in}}

\noindent There is no electric field only in some {\em macroscopic} sense,
that is, only if one averages over some region with a size large compared
to an atomic size (which can still be much smaller than the size of a
macroscopic probe whose size is at least of order 1$\mu$).

Now we want to do the same with other materials, \ie nonconductors or
insulators. Such materials are termed {\em dielectrics}. We concern
ourselves again only with the macroscopic electric field, which is the true
electric field averaged over some small domain, but it will no longer be
zero, so that we must work a little harder to understand how to describe
these materials.

Start by supposing that a piece of material is subjected to an externally
applied electric field. This field will alter the multipole moments of the
constituents of the material, which we shall call molecules (They could
also be atoms or ions), yielding a net polarization of the material.

	Now let's calculate the potential do to this polarization.
If we regard it as a sum over the dipoles of each molecule, then
\beq
\Phx=\sum_j \frac{\p_j\cdot(\x-\x_j)}{|\x-\x_j|^3}
\eeq
For now assume the the molecules are neutral, so that there is no
monopole term.  In addition, assume that quadrupole and higher terms in the
series are negligible.  If we define the polarization vector
\beq
\Px=\;{\rm{dipole\; moment\; per\; unit\; volume}},
\eeq
then this becomes,
\beqa
\Phx=\inivp\leb\frac{\Pxp\cdot(\x-\xp)}{|\x-\xp|^3}\rib
\nonumber \\
=\inivp\leb\Pxp\cdot\nabla'\lep\xxpi\rip\rib.
\eeqa
where we have used the expression
\beq
\frac{(\x-\xp)}{|\x-\xp|^3}=-\nabla\frac{1}{|\x-\xp|}
=\nabla'\frac{1}{|\x-\xp|}
\eeq
If we integrate by parts, we get the form
\beq
\Phx=\inivp \nabla'\cdot\lep\frac{\Pxp}{|\x-\xp|}\rip
-\inivp\frac{\lep \nabla'\cdot\Pxp\rip}{|\x-\xp|}
\eeq

	There are two ways to regard this expression.  Assume we
have a volume V with $\Pxp$ finite inside and zero outside.

	{\bf{First case}}.  Let V be bounded by a surface S
just inside the volume.  Then using the divergence theorem, 
the equation above becomes.

\vspace{0.1in}
\centerline{\psfig{figure=fig12.ps,height=1.5in,width=4.0in}}

\beq
\Phx=\inap \frac{\np\cdot\Pxp}{|\x-\xp|}
-\invp\frac{\lep \nabla'\cdot\Pxp\rip}{|\x-\xp|}
\eeq
Thus we can define the surface and volume polarization charge densities:
\beq
\si_p(\x)=\P_n(\x) \hbox{ or } \si_p(\x)=\P(\x)\cdot\nn
\eeq
\beq
\rho_p(\x)=-\div\Px
\eeq
These have simple physical interpretations.  For example, in the
figure below on the left, the material has a constant finite $\P$ throughout 
its volume, so that at the surface, charge congregates since all of the 
dipoles are aligned.  Also, in the figure on the right a a certain location
within a material the dipoles point radially outward (yielding a positive
divergence).  At the center of this region, where the tails of the 
dipoles are concentrated, there is an excess of negative charge (hence
the $-$ sign in $\rho_p(\x)=-\div\Px$.

\centerline{\psfig{figure=fig12a.ps,height=1.7in,width=5.0in}}


{\bf{Second case}}.  Let V be bounded by a surface S just {\em{outside}}
of the region of finite polarization.  Then

\vspace{0.1in}
\centerline{\psfig{figure=fig13.ps,height=1.5in,width=2.5in}}

\beqa
\Phx&=&\inap \frac{\np\cdot\Pxp}{|\x-\xp|}
-\invp\frac{\lep \nabla'\cdot\Pxp\rip}{|\x-\xp|}\\
&=& \invp \frac{\rho_p(\xp)}{|\x-\xp|}
\eeqa

	Although the surface charge does not appear explicitly, it
still must be there.  It may be obtained from the discontinuity in $\Px$ 
at the surface of the polarization region.  To see this consider a small
pill box enclosing  a small section of the volume and surface of the 
polarized material.  From the divergence theorem

\centerline{\psfig{figure=fig15.ps,height=1.5in,width=6.375in}}

\beq
\inv \div\Px =\ina \nn\cdot\Px,
\eeq
or if the region is small enough
\vspace{0.2in}
\beq
\lep-\nn\cdot\P_{in} + \nn\cdot\P_{out}\rip da=\nabla\cdot\P d^3x
\eeq
so that 
\beq
\nn\cdot\P\, da=-\nabla\cdot\P\, d^3x ,
\eeq
Thus $-\nabla\cdot\P$ must have a delta-function at the surface, and we
still have the surface polarization charge.

	To understand the surface polarization charge, consider a uniformly
polarized slab of dielectric.

\centerline{\psfig{figure=fig16.ps,height=3.0in,width=8.5in}}

\beq
\p=\left\{
\begin{array}{ll}
{\rm{constant}} &{\rm{inside}}\\
0 &{\rm{outside}}
\end{array}
\right.
\eeq
We may actually regard this as two overlapping slabs, one of uniform
positive charge $+\rho$, and one of uniform negative charge $-\rho$,
separated by a small distance distance $\a$.  The whole is then 
electrically neutral, with uniform polarization 
\beq
\P=\rho\a= \;{\rm{dipole\; moment\; per\; unit\; volume}}.
\eeq
The charge density is then
\beq
\rho_p=-\nabla\cdot\P=0 \;{\rm{Inside\;the\;slab}}
\eeq
\beq
\sigma_p=\P\cdot\nn=\left\{
\begin{array}{ll}
P>0 &{\rm{on\;rhs}}\\
-P<0 &{\rm{on\;lhs}}
\end{array}
\right.
\eeq

	The potential due to the slab is just that of two oppositely charged
sheets separated by a distance $d$.
The corresponding electric field is just obtained by summing that due 
to each sheet

\centerline{\psfig{figure=fig17.ps,height=1.5in,width=4.25in}}

\beq
\E=
\left\{
\begin{array}{ll}
0 &{\rm{outside\;the\;slab}}\\
-4\pi \P &{\rm{inside\; the\; slab}}
\end{array}
\right.
\eeq
(We used $E_{in}=2\pi\sigma + 2\pi\sigma=4\pi\sigma$ and $\sigma=P$)

\subsubsection{Electric Displacement}

	Thus far we have assumed that the dielectric is neutral.  If
there are free charges present as well, then the total charge density is
\beq
\rho_{tot}=\rho_{free}-\grad\cdot\P
\eeq
Then as $\E$ is generated by {\em{all}} charges, we have
\beq
\grad\cdot\E=4\pi\rho_{tot}=4\pi\lep \rho_{free}-\grad\cdot\P \rip
\eeq
or,
\beq
\grad\cdot\lep\E+4\pi\P\rip=4\pi\rho_{free}
\eeq

	The vector $\lep\E+4\pi\P\rip$ is generated by free charges only.
We will define the {\em{electric displacement}} $\D$ as this field
\beq
\D=\E+4\pi\P\;\;\;\grad\cdot\D=4\pi\rho_{free}
\eeq

As an example, consider the uniformly polarized slab

\centerline{\psfig{figure=fig18.ps,height=1.0in,width=4.25in}}

\noindent, here
\beq
\rho_{free}=0 \;\;\;{\rm{everywhere}}
\eeq
Thus $\E+4\pi\P=0$, and
\beq
\E=
\left\{
\begin{array}{ll}
0 &{\rm{where}}\; \P=0\\
-4\pi \P &{\rm{where\;}}\P\ne 0
\end{array}
\right.
\eeq

Another simple example using $\D$ with the same slab geometry
is the parallel plate capacitor. 
 
\centerline{\psfig{figure=fig19.ps,height=1.5in,width=6.375in}}

\noindent If we have two charged plates in a 
vacuum, then $\P=0$ everywhere, and 
\[
\D=\E=
\left\{
\begin{array}{ll}
0 &{\rm{outside}}\\
-4\pi \sigma_{free} &{\rm{between\;the\;plates}}
\end{array}
\right.
\]
If we now slide a dielectric slab between the plates, then we expect
it to obtain a uniform polarization, giving rise to surface charges.
 
\centerline{\psfig{figure=fig20.ps,height=1.7in,width=6.375in}}

\noindent However, $\D$ responds to only free charges, thus it is unchanged by
the introduction of the dielectric slab.  $\E$ responds to all
charges, so it {\em{is}} changed.  Since $\E=\D-4\pi\P$, we see that
$\E$ decreases in magnitude inside the dielectric, and since
$\D$, $\P$, and $\E$ are parallel:
\[
\frac{E_{\rm{dielectric}}}{E_{\rm{vacuum}}}=
\frac{E_{\rm{dielectric}}}{D}=
\frac{E_{\rm{dielectric}}}{E_{\rm{dielectric}}+4\pi P}
\]

\subsubsection{Summary and Discussion}  At this point a summary of the dielectric
equations will be useful.
\[
\Px=\;{\rm{dipole\; moment\; per\; unit\; volume}},
\]
\[
\Phx= \inivp \frac{\rho_{total}(\xp)}{|\x-\xp|}
\]
\[
\rho_{total}=\rho_{free}-\grad\cdot\P
\]
\[
\E(\x)=-\grad\Phx \Longrightarrow \grad \times \E(\x)=0
\]
\[
\D(\x)=\E(\x)+4\pi\P(\x)
\]
\[
\grad\cdot\D=4\pi\rho_{free}
\]
Several points must be made in relation to these. The first is that they do
not form a sufficient set from which we can solve for $\Ex$ as we have no
way of writing $\E$ in terms of $\D$ or vice versa. The defining relation
does not help as we don't know $\P$. What is needed is a {\em constitutive
relation} which can be of the form $\D=\D(\E)$ or $\P=\P(\E)$. If there is
no nonalalytic behavior entering this relation, then one can expand
components of $\P$ as a power series in components of $\E$. If $\E$ is not
too large, then only the linear term in these expansions need be kept,
\beq
P_i=\sum_{j=1}^3\ch_{ij}E_j.
\eeq
where the nine numbers $\ch_{ij}$ are the components of the {\em electric
susceptibility tensor}. When this is a good approximation, one says that
the dielectric is {\em linear} .In disordered materials as well as highly
ordered ones with a high degree of symmetry (cubic crystals, for example),
this tensor reduces to a single non-zero number,
\beq
\ch_{ij}=\ch_e\de_{ij};
\eeq
$\ch_e$ is called simply the electric susceptibiliity and such materials are
said to be {\em isotropic}. Finally, if a material is uniform in its electrical
properties, $\ch_e$ will be a constant, independent of position; then the
material is said to be {\em homogeneous}. If all of these things are true,
the dielectric material is as simple as it can be.

For a linear, isotropic, homogeneous dielectric, the connection between $\E$
and $\D$ is
\beq
\Dx=\Ex+4\pi\ch_e\Ex\equiv\ep\Ex
\eeq
where
\beq
\ep=1+4\pi\ch_e
\eeq
is the {\em dielectric constant} of the material.

A second point is that the electric displacement is neither fish nor fowl,
that is, neither field (force on a test charge) nor source. Look again at the
integral expression for $\Phx$; from it we see that the negative of the
divergence of $\Px$ must be a (macroscopic) charge density; it is called
the {\em polarization charge density},
\beq
\rh_p(\x)=-\div\Px.
\eeq
To see how this can be so, imagine a polarization which points in the
$z$-direction and decreases in this direction so that its divergence is
negative. Because of the variation of $\Px$, the molecules at smaller $z$
are more polarized than those at slightly larger $z$, meaning that less
positive charge ``sticks out'' on the larger-$z$ side of the former than
negative charge sticks out on the smaller-$z$ side of the latter.
 
\centerline{\psfig{figure=fig21.ps,height=1.5in,width=6.25in}}

\noindent  Hence
there is a net positive charge density in the region between the two sets
of molecules, and this is the polarization charge density. This argument
leads one to believe that the total polarization charge must be zero.
One can easily show by an application of the divergence theorem that it is
indeed zero.

Having understood that the polarization leads to a charge density, how then
may we understand the electric displacement? It is a linear combination of
a macroscopic field (representing the force on a test charge) and of the
polarization, whose divergence is a charge density. The polarization is
itself source, being the dipole moment density of the constituent molecules
of the material. Hence the displacement is neither field ($\E$) nor
source ($\P$). Its usefulness lies in the fact that problems involving
macroscopic electrostatics, and especially boundary value problems, are
conveniently approached by making use of both the electric field and the
electric displacement.

Another point that should be mentioned has to do with the higher multipole
moments. We have seen how it is essential to keep the sources associated
with the electric dipole moments of the molecules. What of the higher
multipole moments? One may show that they contribute negligibly at the
macroscopic level.

Finally, there is the question of solving for the macroscopic electric
field. Given a medium such that \eq{40} is valid, we may use $\D=\ep\E$ and
have the field equations
\beq
\curl\Ex=0\hbox{ and }\div\Ex=4\pi(\rhx/\ep);
\eeq
these are the same as we have been working with right along except that the
charge density is rescaled by a factor of $1/\ep$; hence all of the lore
that we have learned may be applied to solve for the macroscopic field.

\section{Boundary-Value Problems in Dielectrics}
In this section we shall solve a few representative boundary-value problems
involving dielectrics.  Since $\curl\E=0$, $\E=-\grad\Phx$, $\D=\ep\E$,
and $\div\D=4\pi\rho$ in a dielectric, we may write
\[
\lap\Phx=4\pi\rho/\ep\,.
\]
Thus, all the methods we have learned (images, greens functions, series 
expansion etc.) will all work if properly modified.

There is of course the question of {\bf{boundary conditions}}. 
At an interface
between two materials (dielectric-vacuum, dielectric-dielectric,
dielectric-conductor, etc.), we have a choice. We can either learn how to
solve for the field in a system with non-homogeneous properties, or we can
split the system up into pieces in each of which the material properties
are uniform and then solve a boundary value problem. The latter course is
the simpler if the interfaces may be treated as abrupt. 

The appropriate
boundary or continuity conditions may be found from the basic differential
equations for $\D$ and$\E$. Applying the divergence theorem and
Stokes' theorem as we did once before, one can show that the appropriate
boundary conditions are

\centerline{\psfig{figure=fig21a.ps,height=4.0in,width=6.5in}}

\beq
[\D_2(\x)-\D_1(\x)]\cdot\nn=4\pi\si\hbox{ and }[\E_2(\x)-\E_1(\x)]\times
\nn=0
\eeq
which say that the discontinuity in the normal component of $\D$ is equal
to $4\pi$ times the surface charge density ({\em not} including the surface
charge density arising from the polarization) and that the tangential
component of $\E$ is continuous. The unit normal in the equation for $\D$
points into medium 2 from medium 1.

\subsection{Example: Point Charge Near a Boundary}

Consider that we have two dielectric materials; the first, with dielectric
constant $\ep_1$, occupies the half-space $z>0$, and the second, with
$\ep_2$, occupies the half-space $z<0$. Let there be a point charge $q$
inside of the first dielectric at point $\x_0=(0,0,z_0)$.  

Without a boundary, we can solve the problem easily.  Since $\D$ is
unchanged by the dielectric, 
\beq
\D=-\grad\lep\frac{q}{R}\rip=\ep\E\;\;\;{\rm thus}\;
\E=-\frac{1}{\ep}\grad\lep\frac{q}{R}\rip\;\;\;
{\rm and}\; \Phi=\frac{q}{\ep R}
\eeq
where $R$ is the distance between the charge and where the electric
displacement is evaluated

We will try to solve
for the electric field using the method of images.  For the region
$z>0$, following our earlier
success with this approach, let us locate an image charge $q'$ at the image
position $\x_i=(0,0,-z_0)$. 
 
\centerline{\psfig{figure=fig22.ps,height=2.25in,width=6.5in}}

\noindent The potential produced by these two charges,
embedded in a medium which {\bf everywhere} has the properties of the first
medium, is
\beq
\Ph_1(\x)=\frac1{\ep_1}\lep\frac{q}{R_1}+\frac{q'}{R_2}\rip,
\eeq
where $R_1$ and $R_2$ are, respectively, the distances of the field point
from $x_0$ and $x_i$; this becomes our potential in the region $z>0$ for the
real system. 

For the region $z<0$, we imagine in the fictitious system that
there is a charge $q''$ at the location of the real charge, embedded in a
medium whose dielectric constant is everywhere $\ep_2$. The potential of such
a system is
\beq
\Ph_2(\x)=\frac1{\ep_2}\frac{q''}{R_1}.
\eeq
This becomes our potential in the region $z<0$. 

Now we try to pick the
image charges in such a way that the boundary conditions are satisfied.
these conditions involve the following derivatives:
\beq
\pde{}{z}\left.\lep\frac1{R_1}\rip\ril_{z=0}=-\pde{}{z}\left.\lep
\frac1{R_2}\rip\ril_{z=0}=\frac{z_0}{(\rh^2+z_0^2)^{3/2}}
\eeq
and
\beq
\pde{}{\rh}\left.\lep\frac1{R_1}\rip\ril_{z=0}=\pde{}{\rh}\left.\lep\frac
{1}{R_2}\rip\ril_{z=0}=-\frac{\rh}{(\rh^2+z_0^2)^{3/2}}.
\eeq
Using these one finds that the condition of continuous normal component of
$\D$ or,  $D_{1z}=D_{2z}$ leads to (since $\sigma_{free}=0$)
\beq
q-q'=q''
\eeq
and that the condition of continuous tangential component of $\E$ or
$E_{1\rh}=E_{2\rh}$ leads to
\beq
\frac1{\ep_1}(q+q')=\frac1{\ep_2}q''\,.
\eeq
The solution of these two linear equations is
\beqa
q'=\lep\frac{\ep_1-\ep_2}{\ep_1+\ep_2}\rip q \nonumber \\
q''=\lep\frac{2\ep_2}{\ep_1+\ep_2}\rip q.
\eeqa
Hence the potential on the right side, $z>0$, is
\beq
\Ph_1(\rh,z)=\frac{q}{\sqrt{\rh^2+(z-z_0)^2}}+\frac{q(\ep_1-\ep_2)/(\ep_1
+\ep_2)}{\sqrt{\rh^2+(z+z_0)^2}}
\eeq
while that on the left, $z<0$, is
\beq
\Ph_2(\x)=\frac{2q\ep_2/(\ep_1+\ep_2)}{\sqrt{\rh^2+(z-z_0)^2}}.
\eeq

{\bf{consequences}} The constitutive relations yield several interesting
results for this problem.  Since $\grad\D=4\pi\rho$, 
where $\rho$ is the free charge density,  it must be that
\[
\grad \cdot \D=0\;\;\;{\rm{except\;at\;the\;real\;charge}}
\]
Then as $\D=\ep\E$,
\[
\grad \cdot\E=0\;\;\;{\rm{except\;at\;the\;real\;charge}}
\]
Then since $\D=\E+4\pi\P$, it must be that
\[
\grad \cdot \P=0\;\;\;{\rm{except\;at\;the\;real\;charge}}
\]
Thus the polarization charge density is zero except at the real
charge! (this is consistent with the potential for a point
charge retaining $1/r$ behavior).
\vspace{5.0in}

	This line of reasoning breaks down at the surface between
the two dielectrics, since there $\grad\ep\ne 0$ corresponding to
the polarization surface charge $\P\cdot\nn$ discussed earlier. Thus
\beq
\si_p=(\P_2-\P_1)\cdot\nn
\eeq
where $\nn$ is the unit normal outward from medium 2 into medium 1. If we
apply this to our present example, we first need to find the polarizations.
These are given by, for $i=1,2$,
\beq
\P_i=\ch_i\E_i=\frac{\ep_i-1}{4\pi}\E_i.
\eeq
and so
\beq
\si_p=\frac{\ep_2-1}{4\pi}\E_{2z}-\frac{\ep_1-1}{4\pi}\E_{1z},
\eeq
evaluated at $z=0$. Here,
\beq
\E_{1z}=-\frac1{\ep_1}\frac{(q-q')z_0}{(\rh^2+z_0^2)^{3/2}}
\eeq
and
\beq
\E_{2z}=-\frac1{\ep_2}\frac{q''d_0}{(\rh^2+z_0^2)^{3/2}}.
\eeq
The polarization surface-charge density is then
\beqa
\si_p=-P_{1z}+P_{2z}=\frac{z_0}{(\rh^2+z_0^2)^{3/2}4\pi}\lec\frac{\ep_1-
1}{\ep12}(-q+q')-\frac{\ep_2-1}{\ep_2}q''\ric\nonumber \\
=\frac{qz_0}{4\pi(\rh^2+z_0^2)^{3/2}}\lec\frac{\ep_1-1}{\ep_1}\frac{2
\ep_2}{\ep_1+\ep_2}-\frac{\ep_2-1}{\ep_1}\frac{2\ep_1}{\ep_1+\ep_2}\ric
\nonumber \\ =\frac{qz_0(\ep_1-\ep_2)}{2\pi\ep_1(\ep_1+\ep_2)(\rh^2+z_0^2)
^{3/2}}.
\eeqa
An important limiting case is given by $\ep_1=1$ and $\ep_2\rightarrow
\infty$, in which case the material at $z<0$ cannot support an electric
field and behaves like a conductor. Then our system reduces to a point
charge outside of a conductor, for which we already know that the answer is
\beq
\si_p=-\frac{qz_0}{2\pi(\rh^2+z_0^2)^{3/2}}.
\eeq

\subsection{Example: Dielectric Sphere in a Uniform Field}

Our second example is a dielectric sphere placed in a uniform externally
applied field. 
 
\centerline{\psfig{figure=fig23.ps,height=1.5in,width=6.25in}}

\noindent Rather than use the image charge method, this time we shall
make use of an orthogonal function expansion. Because $\curl\Ex=0$
everywhere, we can write the electric field as the gradient of a scalar
potential. Further, for a uniform medium $\Dx=\ep\Ex$ with constant $\ep$,
so, from $\div\Dx=4\pi\rhx$, we know that $\div\Ex=0$ where there is no
macroscopic charge density $\rhx$. Given that the radius of the sphere is
$a$, we have such conditions for $r<a$ and for $r>a$. Hence the scalar
potential satisfies the Laplace equation in these two regimes, and we can
expand it in the usual way in spherical coordinates. The symmetries in the
problem imply that the solution is independent of $\ph$, so we need to use
a Legendre polynomial expansion. Thus, for $r<a$, we have
\beq
\Ph_<(r,\th)=E_0a\sum_{l=0}^\infty A_l\lep\frac ra\rip^lP_l(\cos\th)
\eeq
and, for $r>a$,
\beq
\Ph_>(r,\th)=-E_0r\cos\th+E_0a\sum_{l=0}^\infty B_l\lep\frac ar\rip^{l+1}
P_l(\cos\th).
\eeq
The first term in the second of these expansions is the potential
associated with the applied field; the others come from the sources induced
on the dielectric sphere (polarization charge). The boundary conditions
that must be applied are (i) tangential $E$ and (ii) normal $D$ continuous;
these are
\beq
\left.\pde{\Ph_<}{\th}\ril_{r=a}=\left.\pde{\Ph_>}{\th}\ril_{r=a}
\eeq
and
\beq
\left.\ep\pde{\Ph_<}{r}\ril_{r=a}=\left.\pde{\ph_>}{r}\ril_{r=a},
\eeq
where $\ep$ is the dielectric constant of the sphere.
By proceeding with the solution in the by now familiar way for orthogonal
function expansions, one finds that all $A_l$ and $B_l$ are zero except for
$l=1$. For $l=1$ the conditions are
\beq
A_1=-1+B_1 \hbox{ and }\ep A_1=-1-2B_1.
\eeq
These are easily solved to yield
\beq
A_1=-3/(2+\ep)\hbox{ and }B_1=(\ep-1)/(\ep+2).
\eeq
Hence, the potential is
\beq
\Ph_<(r,\th)=-\frac3{\ep+2}E_0r\cos\th
\eeq
and
\beq
\Ph_>(r,\th)=-E_0r\cos\th+E_0a\lep\frac{\ep-1}{\ep+2}\rip\lep\frac ar
\rip^2\cos\th.
\eeq
Notice that if $\ep\rightarrow\infty$, we recover the result for the
conducting sphere.

From this result, and since $\Phx_{dipole}=\p\cdot\x/r^3$, we can see that 
the sphere has a dipole moment which is
\beq
\edm=E_0a^3\lep\frac{\ep-1}{\ep+2}\rip\zh.
\eeq
The electric field inside of the sphere is a constant, and so is $\D$,
\beq
\D=\ep\E=\ep\lep\frac3{\ep+2}\rip E_0\zh\equiv\E+4\pi\P,
\eeq
so
\beq
4\pi\P=\D-\E=3\lep\frac{\ep-1}{\ep+2}\rip E_0\zh,
\eeq
or
\beq
\P=\frac3{4\pi}\lep\frac{\ep-1}{\ep+2}\rip E_0\zh.
\eeq
 
\centerline{\psfig{figure=fig24.ps,height=2.25in,width=6.25in}}

Although there is no macroscopic charge density anywhere in the system,
there is polarization charge density. There is no volume polarization
charge density because $\P$ has zero divergence. However, there is a
surface charge density; it is given by
\beq
\si_p=\P\cdot\rhh=P_r=\frac3{4\pi}E_0\lep\frac{\ep-1}{\ep+2}\rip\cos\th.
\eeq

	As an application of the polariazble sphere problem, consider
a water drop in air.  For this system, roughly
\beq
\ep_{rm{air}}\approx 1\;\;\;\;\ep_{rm{water}}\approx 81\,.
\eeq
Water is a dielectric composed of permanent dipoles.
The polarizability $\alpha$ ($\p=\alpha\E$) of the water drop is then
\beq
\alpha_{H_2O}=\frac{81-1}{81+2}a^3 \approx a^3
\eeq
Water drops look like metallic spheres to a static E-field.

\subsubsection{The Inverse Problem}

The inverse problem of a dielectric with a spherical cavity is easy to
solve because one has only to change $\ep$ into $1/\ep$ in the results
found here. The reason is that the relative dielectric constant of the
cavity to that of the surrounding medium is $1/\ep$. In this way we find
\beq
\Ph_<(r,\th)=-\frac{3\ep}{2\ep+1}rE_0\cos\th
\eeq
and
\beq
\Ph_>(r,\th)=-E_0r\cos\th-E_0a\frac{\ep-1}{2\ep+1}\lep\frac ar\rip^2\cos
\th.
\eeq
 
\centerline{\psfig{figure=fig25.ps,height=2.0in,width=8.5in}}


\subsection{Clausius-Mossotti equation}

In writing $\al$ above note
that the electric field in
\[
\p=\al\E
\]
is the external field {\em{not including the field of the induced dipole 
itself}}.  However, in 
\[
\P=\chi_e\E
\]
the field $\E$ {\em{does}} include the the field due to the dipoles 
in $\P$.  If we can relate these two, we can calculate the relation
between $\al$ (a microscopic quantity), and the macroscopic quantities 
$\chi_e$ and $\ep$.  If we define two different $\E$s
\[
\p=\al\E_{loc}=\al({\rm{electric\;field\;at\;the\;site\;of\;the\;molecule}})
\]
\[
\P=\chi_e\E_{med}=\chi_e({\rm{electric\;field\;in\;the\;medium}})
\]
If $n$ is the number of molecules per unit volume, then $\P=n\p$ and
\[
\P=n\al\E_{loc}.
\]

	With these equations we can find a relation between $\al$ and
$\chi_e$ if we can relate $\E_{loc}$ and $\E_{med}$.  We note that
$\E_{loc}$ is the field at the site of a molecule if the molecule is
removed.  To calculate this, we will consider a spherical cavity in
a dielectric medium.  If the cavity were filled with dielectric, then
the field at the center would be $\E_{med}$, so
 
\centerline{\psfig{figure=fig26.ps,height=2.25in,width=6.25in}}

\[
\E_{med}=\E_{loc}+\E_{sphere}
\]
\noindent where $\E_{sphere}$ is the field at the center of a 
uniformly polarized dielectric sphere due {\em{only}} to the polarization.  
From \eq{111} it is clear 
that the field inside the sphere (due to {\em{both}} the polarization
and the external field) is
\[
\E_{inside\;sphere}=\frac{3}{2+\ep}\E_0=\E_0+\frac{1-\ep}{2+\ep}\E_0
\]
Thus that due only to the polarization is
\[
\E_{sphere}=\frac{1-\ep}{2+\ep}\E_0
\]
Comparing this to \eq{113} we see that
\[
\E_{sphere}=-\frac{4\pi}{3}\P
\]
Thus
\[
\E_{loc}=\E_{med}-\E_{sphere}=\E_{med}+\frac{4\pi}{3}\P.
\]
Now, if we multiply by $n\al$ and solve for $\P$, we get
\[
\P=\frac{n\al}{1-\frac{4\pi}{3} n\al}\E_{med}
\]
Then since $\P=\chi_e\E_{med}$ and $\ep=1+4\pi\chi_e$ we get
\beq
\frac{4\pi}{3} n\al=\frac{\ep-1}{\ep+2}\;\;\;{\rm{Clasius-Mossotti\;Equation}}
\eeq

\section{Electrostatic Energy in Dielectrics}
In free space we derived the energy of a distribution of charge $\rhx$ by
assembling the distribution little by little, bringing infinitesimal pieces
of charge in from infinity. Following this reasoning we found that
\[
W=\frac{1}{2}\iniv \rhx\Phx
\]
This is in general {\em{not}} true in the presence of dielectrics
(however, as we will see, it may be true in some cases).
In the presence of dielectrics work must also be done to induce 
polarization in the dielectric, and it is not clear if this work is
included in the equation above.

When dielectrics are present we shall use a somewhat different 
argument (which still
corresponds to the same procedure). Suppose that there is initially some
macroscopic charge density $\rhx$, potential $\Phx$, and fields $\Ex$ and
$\Dx$. The imagine that some infinitesimal change in the charge density,
$\de\rhx$, is made. To first order in $\de\rh$, the change in energy of the
system is
\beq
\de W=\inv\Phx\de\rhx
\eeq
where the integration is done over that region of space where the integrand
is non-zero. The point is that this is the interaction energy of $\de\rhx$
with the sources already present (and which produce $\Phx$); the
interaction energy of $\de\rh$ with itself is second-order in small
(infinitesimal) quantities.  (this form is consistent with the fact
that $W$ is a natural thermodynamic function of the charges, not
the potential).

The change in $\D$ which arises as a consequence of the change $\de\rh$ in
the charge density is related to the latter by the equation
$\div(\D+\de\D)=4\pi(\rho +\de\rho)$, or
\beq
\div(\de\Dx)=4\pi\de\rhx,
\eeq
so we may write the change in the energy as
\beq
\de W=\inv\Phx\frac1{4\pi}(\div\de\Dx).
\eeq
We next do an integration by parts in the by now familiar way.
\beqa
\de W=\frac1{4\pi}\inv\div(\Phx\de\Dx)-\frac1{4\pi}\inv\grad\Phx\cdot\de\Dx
\nonumber\\=\frac1{4\pi}\ina\Phx\de\Dx\cdot\nn+\frac1{4\pi}\inv\Ex\cdot\de
\Dx.
\eeqa
The surface integral is zero for a localized charge distribution if
V includes all space. Thus we have simply
\beq
\de W=\frac1{4\pi}\iniv\Ex\cdot\de\Dx.
\eeq
Now we must introduce some statement about the properties of the medium. If
it is {\bf linear} ($\D=\ep\E$), then
\[
\E\cdot\de\D=\E\cdot(\ep\de\E)=\frac{1}{2}\ep\de(\E\cdot\E)=
\frac{1}{2}\de(\E\cdot\D) 
\]
so that
\beq
[\Ex\cdot\de\Dx]=\frac12\de[\Ex\cdot\Dx].
\eeq
and so
\beq
\de W=\frac1{8\pi}\iniv\de[\Ex\cdot\Dx].
\eeq
If we now integrate from zero field up to the final fields (a {\em
functional integration}),
\[
W=\frac1{8\pi}\iniv\int_0^{\D}\de[\Ex\cdot\Dx]
\]
we find
\beq
W=\frac1{8\pi}\iniv\Ex\cdot\Dx.
\eeq
This result is valid {\bf{only}} for linear media. 

	There are several amusing consequences obtainable from this
relation.  First, by writing $\Ex=-\grad\Phx$ and integrating by parts, we
obtain
\[
W=\frac{-1}{8\pi} \iniv \div\lep\Phx\Dx\rip
+\frac{1}{8\pi} \iniv\Phx \div\Dx
\]
Through the divergence theorem, the first term yields a surface 
term which vanishes at infinity.  The second term becomes
\beq
W=\frac12\iniv\rhx\Phx;
\eeq
Thus for a {\bf{linear}} dielectric, the original formula is valid.

\subsection{Force on a Dielectric}

	From the above, it is clear that $W$, as written, is a function
of the free charges, their positions, and of the positions of the dielectrics
through $\ep(\x)$ (which may vary from point to point).  Thus we may
write
\beq
W(\x,\rho)=
\frac{1}{8\pi}\iniv \frac{|\Dx|^2}{\ep(\x)} \;\;\; {\rm{linear\; only}}
\eeq
From this, it is clear that if the free charges (which produce $\D$)
are fixed, and we move one
of the dielectrics, then the energy is {\em{reduced}} if the change
makes $\ep$ increase in the region where $\Dx$ is finite.  In particular,
{\em{ the energy is reduced by having a dielectric move from a region
of low field to one of high field.  Thus the force on such a linear 
dielectric must always be such as to draw it into a region of greater 
fields}}.
 
\centerline{\psfig{figure=fig27.ps,height=2.25in,width=6.25in}}


	If the free charges are held fixed, then since $W$ depends
on the positions and magnitude of the charges and dielectrics, it
follows that the force on a dielectric is 
\beq
F_\et=-\lep\pde W\et\rip_Q
\eeq
where $F_\et$ is the $\et$-component of the force on the dielectric.
This is because the most stable state of the system is that with the
minimum $W$.

	In this calculation, it is important that the energy was a 
natural function of the charges and positions of the charges and dielectrics.
Then we could evaluate the total differential
\[
dW=\lep\pde{W}{\et}\rip_Q d\et + \lep\pde{W}{Q}\rip_{\et} dQ
\]
to obtain the force.  This is analogous to the situation in elementary
thermodynamics where the energy $U$ is a natural function of
the volume and temperature $U(V,T)$.  If we wanted to obtain a potential
which was a function of the entropy $S$ and V (suppose for example $S$
is change in such a way as to keep $T$ fixed, i.e. a system
in a heat bath), then we made a Legendre
transformation $F=U-TS$, and the most stable state of the system is
that with the minimum $F$ (For an elegant discussion of elementary
thermodynamics see {\em{Thermodynamics}} by Enrico Fermi, about 77pp.)

	Thus, since $W$ is a natural function of the positions and charges,
it is not appropriate for the case where the potentials are held fixed.
We need the potential which is a natural function of the positions and
potentials.  As in the paragraph above, the way to remedy this is a Legendre
transformation to a new function $W'$, defined by
\beq
W'=W-\frac{1}{4\pi}\iniv \Ex\cdot\Dx
\eeq
This is a general expression (not just for the linear case)
where $\E=-\grad\Phx$ is a natural function of the potentials, and
$\D$ is a function of the charges (since $\div \D=4\pi\rho$).  
A differential change in $W'$ is given by
\beq
\de W'=\de W-\frac{1}{4\pi}\iniv \Ex\cdot\de\Dx-
\frac{1}{4\pi}\iniv \Dx\cdot\de\Ex.
\eeq
Then, since
\[
\de W=\frac{1}{4\pi}\iniv \Ex\cdot\de\Dx
\]
\beq
\de W'=-\frac{1}{4\pi}\iniv \Dx\cdot\de\Ex
\eeq
Then since $\Ex$ is a natural function of the potential, $W'$ is a
natural function of potentials and positions, as desired.

	Thus the force on a dielectric in the presence of fixed
potentials (i.e. conductors connected to a battery) is
\beq
F_\et=-\lep\pde {W'}{\et}\rip_\Ph
\eeq
In the linear case, we can evaluate this in terms of $W$, since
\beq
W'=W-\frac{1}{4\pi}\iniv \Ex\cdot\Dx=-\frac{1}{8\pi}\iniv \Ex\cdot\Dx=-W
\eeq
Thus in {\bf{the linear case only}},
\beq
F_\et=+\lep\pde W\et\rip_\Ph
\eeq

\subsection{Forces on a Dielectric Revisited}
	
These force formulae may also be derived in a more pedestrian manner.

We can derive one for the
change in a system's energy when a piece of dielectric is moved
from one place to another under conditions of constant macroscopic charge
density. Consider that initially the macroscopic fields, charge density,
potential, and polarization are $\E_0$, $W_0$, $\D_0$, $\rh_0$, $\Ph_0$, and
$\P_0$. Let the final ones have subscript 1 instead of 0. Then
\beq
W_1=\frac1{8\pi}\iniv\E_1(\x)\cdot\D_1(\x)\hbox{ and }
W_0=\frac1{8\pi}\iniv\E_0(\x)\cdot\D_0(\x),
\eeq
so
\beq
\De W\equiv W_1-W_0=\frac1{8\pi}\iniv[\E_1\cdot\D_1-\E_0\cdot\D_0].
\eeq
By adding and subtracting identical terms, we can turn this expression into
\beq
\De W=\frac1{8\pi}\iniv[\E_0+\E_1]\cdot[\D_1-\D_0]+\frac1{8\pi}\iniv
[\E_1\cdot\D_0-\E_0\cdot\D_1].
\eeq
By doing an integration by parts (in the usual way), one can show that the
first term is zero if $\rh_1=\rh_0$, so we have
\beq
\De W=\frac1{8\pi}\iniv[\E_1\cdot\D_0-\E_0\cdot\D_1].
\eeq

As an example of the use of this formula, imagine that a dielectric having
$\ep=\ep_1$ is moved in from infinity to occupy some domain V where
formerly there was empty space. Everywhere else there is vacuum. Then
$\D_1=\ep_1\E_1$ in V, and $\D_1=\E_1$ elsewhere. Also, $\E_0=\D_0$
everywhere. Our formula for the change in energy gives
\beq
\De W=-\frac1{8\pi}\int_Vd^3x(\ep_1-1)\E_1\cdot\E_0=-\frac12\iniv
\P_1\cdot\E_0.
\eeq
This is the energy of the dielectric object placed in an external field
$\E_0$. The factor of 1/2 distinguishes it from the energy of a permanent
dipole placed in an external field which we derived earlier. It has to do
with the fact that in the present case the field has to do work to polarize
the dielectric in the first place.

We may also devise a formula for the force on a piece of dielectric. In the
case that the (macroscopic) charge is fixed, no work is done moving any charge
and so we have a conservative system in the sense that the change in the
field energy must be equal to the work that an external agent does on the
dielectric when the latter is moved. This force (recall our earlier
arguments of this kind) is equal and opposite to the electric field force
on the dielectric so we wind up concluding that
\beq
F_\et=-\lep\pde W\et\rip_Q
\eeq
where the ``$Q$'' means that the derivative with respect to displacement in
the $\et$ direction is taken at constant sources (constant $\rhx$). The
result is the force in the direction of $\et$, and this is the usual
expression for a conservative system.

A more difficult case is one in which
there is an external source of energy. A very common case of this kind
which frequently arises in electrostatics involves a set of conducting
objects or surfaces on which the macroscopic charge $\rh$ resides, and
keeping these surfaces at fixed potentials when the dielectric is moved.
The latter is easily achieved by connecting the conductors to fixed voltage
sources (batteries). To see what happens, imagine making a small
displacement of the dielectric in two steps. First, move it by $d\et$ while
maintaining $\rhx$ fixed. Then restore $\Phx$ to its original value at
those points where there is non-zero macroscopic charge density by
adjusting this charge density as necessary. We can calculate the change in
field energy during either of these steps by applying the general formula
for a linear system
\beq
W=\frac12\iniv\rhx\Phx,
\eeq
which gives, for small changes in $\rh$ and $\Ph$,
\beq
\de W=\frac12\iniv[\de\rhx\Phx+\rhx\de\Phx].
\eeq
In the first step described above, there is no change in $\rh$, so
\beq
\de W_1=\frac12\iniv\rhx\de\Ph_1(\x).
\eeq
This is the same as $\de W_Q$ since charges are fixed in this step:
\beq
\de W_1=\de W_Q.
\eeq
In step 2, $\rhx$ is adjusted so that $\Phx$ returns to its initial value
everywhere where the charge density does not vanish. In this step
\beq
\de W_2=\frac12\iniv\rhx\de\Ph_2(\x)+\frac12\iniv\de\rhx\Phx.
\eeq
However, at points where $\rhx$ in non-zero, $\de\Ph_2(\x)=-\de\Ph_1(\x)$
because in the second step we restore the potential to its original value
at these points. Hence we can rewrite $\de W_2$ as
\beq
\de W_2=-\frac12\iniv\rhx\de\Ph_1(\x)+\frac12\iniv\de\rhx\Phx.
\eeq
There is a second way to see what $\de W_2$ is; in this step we make an
infinitesimal change in the charge density, $\de\rhx$, and the change in
energy accompanying this adjustment is, to first order in infinitesimals,
\beq
\de W_2=\iniv\Phx\de\rhx.
\eeq
By comparing the two equations we have for $\de W_2$, we learn that
\beq
\frac12\iniv\Phx\de\rhx=-\frac12\iniv\rhx\de\Ph_1(\x).
\eeq
Using this relation in \eq{113}, we find that
\beq
\de W_2=-\iniv\rhx\de\Ph_1(\x)
\eeq
and this is the same as $-2\de W_1$. Consequently we can say that the total
change in energy, which we shall call $\de W_V$ (the ``$V$'' signifies
constant potentials at points where $\de\rhx$ is non-zero), is
\beq
\de W_V=\de W_1+\de W_2=-\de W_1=-\de W_Q.
\eeq
Consequently,
\beq
F_\et=-\lep\pde W\et\rip_Q=+\lep\pde W\et\rip_V.
\eeq
In other words, if we can calculate the energy as a function of $\et$, the
position of the dielectric, at constant potentials where $\rhx\ne0$,
we can find the force on the dielectric by taking the {\bf positive}
derivative of this energy with respect to the dielectric's position.

\section{Example: Dielectrophoresis}

A spherical dielectric particle of radius $a$ and $\ep=\ep_1$ is placed
in a dielectric fluid ($\ep_2\ne\ep_1$) contained within an annulus
with conducting walls.

% origin /home/wanderer/jarrell/text/courses/jackson/chap4/dielectro/
\centerline{\psfig{figure=diefig1.ps,height=3.5in,width=8.5in}}

\noindent The annulus is maintained at a relative potential $V_0$.  
Assuming that
$a\ll $ any other dimension in the problem, and that the densities
of the particle and fluid are the same:
\begin{itemize}
\item Show that the {\bf{net}} force on the particle is
\[
\F_{net}=
-\frac{1}{3}a^3AV_0^2r^{-3}\leb\ln\lep R_{out}/R_{in}\rip\rib^{-2}\rhat\,,
\]
where 
\[
A=\frac{\ep_2(4\ep_2+5)-\ep_1(\ep_2-1)}{(2\ep_2+\ep_1)^2}(\ep_1-\ep_2)\,.
\]
\item Discuss how $\F_{net}$ depends upon (1) $\ep_2$ relative to $\ep_1$
and (2) $\ep_1$ for fixed $\ep_2$.
\item The drift velocity $\v$ of the particle is given by
\[
\F_{net}=6\pi\et a \v\,,
\]
(1) Discuss the dependence of $\v$ on $a$ and $\ep_1$ for fixed $\ep_2$,
and (2) Suggest possible uses for this setup (dielectrophoresis).
\end{itemize}

\bigskip

{\bf{Solution}}.  To solve this we must first find the field between
the cylinders in the absence of the particle.  To accomplish this,
we use Gauss' law, the constitutive relations, and the fact that
$\D$ and $\E$ are purely radial.
\[
\div\D=4\pi\rho\;\; ;\;\; \D=\ep_2\E
\]

% origin /home/wanderer/jarrell/text/courses/jackson/chap4/dielectro/
\centerline{\psfig{figure=diefig2.ps,height=2.6in,width=8.5in}}

\[
l2\pi r\ep_2 E_r=4\pi\lambda l\;\; ; \;\; E_r=\frac{2\lambda}{\ep_2 r}
\]
\[
V_0=\int_{R_{in}}^{R_{out}} dr \pde{\Ph}{r}=
-\frac{2\lambda}{\ep_2}\ln\lep R_{out}/R_{in}\rip
\]
When we solve for $\lambda(V_0)$, we find that 
\[
\E_{cyl}=-\frac{V_0}{r\ln\lep R_{out}/R_{in}\rip} \rhat
\]

	Now we must solve for the field within the volume of the particle.
Since $a\ll R$, we will assume that $\E_{cyl}$ is essentially uniform over 
the diameter of the particle.  This problem then becomes very similar to
one we solved in class, that of a dielectric sphere in a uniform external
field $\E_0$.

% origin /home/wanderer/jarrell/text/courses/jackson/chap4/dielectro/
\centerline{\psfig{figure=diefig3.ps,height=1.5in,width=8.5in}}

\noindent Recall that for this problem, 
\[
\E_{sphere}=\frac{3}{\ep+2}\E_0
\]
where $\E_{sphere}$ is the electric field within the sphere.  However,
since the dielectric constant only enters through the boundary condition
in a relative way
\[
D_{1n}=D_{2n} \; \Rightarrow \; \ep_1E_{1n}=\ep_2E_{2n} \; \Rightarrow \; 
\frac{\ep_1}{\ep_2}E_{1n}=E_{2n},
\]
it must be that
\[
\E_{sphere}=\frac{3\ep_2}{2\ep_2+\ep_1} \E_{cyl}.
\]
The polarization of the sphere is then
\[
\P=\frac{\ep_1-1}{4\pi}\E_{sphere}=
\frac{1}{4\pi}\frac{3\ep_2(\ep_1-1)}{2\ep_2+\ep_1}\E_{cyl},
\]
which corresponds to a dipole moment of the sphere
\[
\p=\frac{4\pi}{3} a^3\P=a^3\frac{\ep_2(\ep_1-1)}{2\ep_2+\ep_1}\E_{cyl}
\]
Thus the electrostatic force on the sphere is given by
\[
\F_{elec}=\lep\p\cdot\grad\rip\E_{sphere}=
a^3\frac{3\ep_2^2(\ep_1-1)}{(2\ep_2+\ep_1)^2}\lep\E_{cyl}\cdot\grad\rip\E_{cyl}
\]
\[
\F_{elec}=\lep\p\cdot\grad\rip\E_{sphere}=
\frac{a^3}{2}
\frac{3\ep_2^2(\ep_1-1)}{(2\ep_2+\ep_1)^2}
\grad\left|\E_{cyl}\right|^2
\]

	In addition to this force, there is an electric Archimedes force.
This force is due to the fact that the particle displaces some of the 
dielectric fluid.  We can calculate it by applying Archimedes principle
just like we do for gravitational forces.  The electric Archimedes force
is minus the force that the sphere of displaced fluid experiences.
\[
\F_{elec-Arch}=-\F_{elec}(\ep_1\to\ep_2)
\]
As indicated, we can get this force by replacing $\ep_1$ by $\ep_2$
in the force equation above.
\[
\F_{elec-Arch}=-\frac{a^3}{6}\lep\ep_2-1\rip\grad\left|\E_{cyl}\right|^2
\]
Thus the net electric force on the sphere is
\[
\F_{net}=\frac{a^3}{2}
\leb\frac{3\ep_2^2(\ep_1-1)}{(2\ep_2+\ep_1)^2} -
\frac{\ep_2-1}{3}\rib 
\grad\left|\E_{cyl}\right|^2.
\]
Now evaluating the gradient, we get
\[
\F_{net}=-\frac{a^3 V_0^3}{r^3\lep\ln(R_{out}/R_{in})\rip^2}
\leb\frac{3\ep_2^2(\ep_1-1)}{(2\ep_2+\ep_1)^2} -
\frac{\ep_2-1}{3}\rib ,
\]
or, after a bit of algebra,
\[
\F_{net}=-\frac{a^3 A V_0^3}{3 r^3\lep\ln(R_{out}/R_{in})\rip^2}
\]
where
\[
A=\frac{\ep_2(4\ep_2+5)-\ep_1(\ep_2-1)}{(2\ep_2+\ep_1)^2}(\ep_1-\ep_2)
\]

	{\bf{Interpretation}}.  This setup has practical applications for 
sorting bits of dielectric particles with different $\ep_1$.  Thus
we should consider $A$ for fixed $\ep_2$ and different $\ep_1$ of
the different particles. If $A>0$, then the force is inward,
toward stronger fields, and if $A<0$, then the force is outward, toward
weaker fields.  The zeroes of $A$ occur when
\[
\ep_2=\ep_1\;\;\; ; \;\;\; \ep_1=\frac{\ep_2(4\ep_2+5)}{\ep_2-1},
\]
and the maximum for fixed $\ep_2$
\[
\pde{A}{\ep_1}=0 \Rightarrow \ep_1=2(\ep_2+1).
\]

	Now lets consider a numerical example.  Let $\ep_2=2.0$ and
vary $\ep_1$.

% origin /home/wanderer/jarrell/text/courses/jackson/chap4/dielectro/
\centerline{\psfig{figure=diefig4.ps,height=3.5in,width=6.6in}}

\noindent As shown in the figure, $A$ is not a monotonic function of $\ep_1$
In fact there are three regions
\begin{center}
\begin{tabular}{||l|l||}\hline\hline
$\ep_1<\ep_2$	& force is outward \\ \hline
$\ep_2<\ep_1<\ep_2(4\ep_2+5)/(\ep_2-1)$ 	& force is inward\\ \hline
$\ep_1>\ep_2(4\ep_2+5)/(\ep_2-1)$ 	& force is outward\\ \hline \hline
\end{tabular}
\end{center}

\noindent Thus, mildly and highly polar particles move out, while particles 
of intermediate polarizability more in.  At first it seems that 
this makes no sense; however,
consider the following explanation:

	Since the potential within the annulus is held fixed by an external
source, the appropriate thermodynamic potential for this system is
\[
W'(r,\Ph)=-\frac{1}{8\pi}\inv \ep(\x)\left|\Ex\right|^2
\]
The most stable state of the system is the one which minimizes this potential.
Thus the most stable state of the system is obtained by having the largest
field where $\ep(\x)$ is largest.  Reconsider the three regions.

\begin{tabbing}
XXXX\=XXXXXXXXXXXXXXXXXXX\=XXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXX \kill
(1) \>$\ep_1<\ep_2$ 	\> The particle has a lower $\ep$ than that of\\
	\>		\> the fluid, thus it is expelled to regions of\\
	\>		\> low filed.  Thus, $\ep(\x)$ is maximized where\\
	\>		\> $\E$ is large.\\
	\> \\
(2) \>$\ep_2<\ep_1<\ep_2(4\ep_2+5)/(\ep_2-1)$
			\> Now the particle has a higher $\ep$ than the fluid\\
	\>		\> thus it moves to regions of large field\\
	\> \\
(3) \>$\ep_1>\ep_2(4\ep_2+5)/(\ep_2-1)$ 
			\> The field inside the particle goes to zero as $\ep\to\infty$\\
	\>		\> (the sphere approaches a metal), and the potential\\
	\>		\> is minimized by having the sphere go to regions of\\
	\>		\> low field.\\
\end{tabbing}

	Now consider the motion of the particle.  The drift velocity of
the particle is given by
\[
\F_{net}=6\pi\eta\v.
\]
If we use the following reasonable parameters
\begin{center}
\begin{tabular}{||l|l|l||}\hline
parameter	&  value	& cgs value \\ \hline \hline
$a$		& 0.1 mm	& 0.01 cm   \\ \hline
$\ep_1$		& 4.6		& 4.6	    \\ \hline
$\ep_2$		& 2.3		& 2.3       \\ \hline
$\et$	& $6.5 \times 10^{-4}$ (MKSA) & $6.5\times 10^{-3}$ cgs \\ \hline
$R_{in}$		& 0.5 mm	& 0.05 cm     \\ \hline
$R_{out}$		& 1 cm		& 1 cm	      \\ \hline
$V_0$		& $4.0\times 10^{3}$ volts & 13.34 statvolts \\ \hline
$r$		& 0.5 cm	& 0.5 cm	\\ \hline \hline
\end{tabular}
\end{center}
The value of $\ep_2$ corresponds to that of organic solvents, in
this case Benzene.  Solving with these parameters,
we find that $v=0.031\;\;{\rm{cm./sec}}$. Slow.

	We could use this setup to separate living cells from dead ones
due to the very different water content (since $\ep_{water}\approx 81$).



\appendix

\section{Multipole Expansion: with Spherical Harmonics}

In this appendix, we will discuss the multipole expansion as it is done in
Jackson.

Consider the potential $\Phx$ produced by some localized charged
distribution $\rh(\x)$,
\beq
\Phx=\inivp\rhxp\xxpi.
\eeq
Substitute the spherical harmonic expansion for $\xxpi$ to have
\beq
\Phx=\sum_{l,m}\frac{4\pi}{2l+1}\leb\inivp\rhxp\ylmsp\frac{r_<^l}{r_>^{l+1}}
\rib\ylm.
\eeq
 
\centerline{\psfig{figure=fig28.ps,height=1.5in,width=6.25in}}

\noindent Now, if the origin of coordinates is chosen to be 
around the center of the
charge distribution, and if the field point $\x$ is such that $r$ is larger
than the distance of any source point (where $\rhxp\ne0$) $\xp$ from the
origin, then it is true that for all $\xp$ of importance in the integral,
$r>r'$ and so $r_<=r'$ and $r_>=r$. Thus,
\beq
\Phx=\sum_{l,m}\frac{4\pi}{2l+1}\leb\inivp r'^l\ylmsp\rhxp\rib\frac\ylm{r^{l+
1}}.
\eeq
This result may be written in the form
\beq
\Phx=\sum_{l=0}^\infty\sum_{m=-l}^l\frac{4\pi}{2l+1}q_{lm}\frac\ylm{r^{l+
1}}\equiv\sum_{l,m}\Ph_{lm}(\x),
\eeq
where
\beq
q_{lm}\equiv\iniv r^l\ylms\rhx
\eeq
is known as a {\em multipole moment} of the charge distribution. These
moments, which satisfy the identity
\beq
q_{l,m}=(-1)^mq^*_{l,-m}
\eeq
by virtue of the same property of the spherical harmonics,
completely determine the field outside of the domain where the charge
is located. Note, however, that they do not contain enough information to
tell us what the actual charge distribution is. The moments of greatest
interest are the ones with small values of $l$. We can understand this
statement from the fact that the moment $q_{lm}$ is
proportional to, as seen from \eq{5}, $a^l$, where $a$ is the size of
the charge distribution. Hence the potential produced by this moment is
proportional to $(q'/r)(a/r)^l$ where $q'$ is a characteristic charge in
the distribution (The actual total charge may vanish). This contribution to
the potential becomes very small for large $l$ given that
$r$ is significantly larger than $a$.

The components of the electric field associated with the $l,m$ multipole
are
\beqa
E_r=-\pde{\Ph_{lm}}{r}=\frac{4\pi(l+1)}{2l+1}q_{lm}\frac{\ylm}{r^{l+2}}
\nonumber\\
E_\th=-\frac1r\pde{\Ph_{lm}}{\th}=-\frac{4\pi}{2l+1}q_{lm}\frac1{r^{l+2}}
\pde{\ylm}{\th}\nonumber\\
E_\ph=-\frac1{r\sin\th}\pde{\Ph_{lm}}{\ph}=-\frac{4\pi}{2l+1}q_{lm}\frac
{im}{r^{l+2}\sin\th}\ylm.
\eeqa

The leading moments are
\beq
q_{00}=\iniv\rhx\frac1{\sqrt{4\pi}}=\frac q{\sqrt{4\pi}}
\eeq
where $q$ is precisely the total charge of the system. This term is the
{\em monopole moment} of the charge distribution; it is fundamentally just the
total charge. Similarly,
\beq
q_{10}=\sqrt{\frac3{4\pi}}\iniv\rhx r\cos\th=\sqrt{\frac3{4\pi}}\iniv\rhx z
\eeq
and
\beq
q_{11}=-q^*_{1,-1}=-\sqrt{\frac3{8\pi}}\iniv\rhx r\sin\th e^{-i\ph}
=-\sqrt{\frac3{8\pi}}\iniv\rhx(x-iy).
\eeq
From these equations we can see that the information contained in the
coefficients $q_{1m}$ is the same as what is contained in the components of
the electric dipole moment $\edm$ of the charge distribution,
\beq
\edm\equiv\iniv\rhx\x.
\eeq
The explicit connection is
\beqa
\edm=\lec\-\frac12\sqrt{\frac{8\pi}3}(q_{11}-q_{1,-1})\xh
-\frac{1}{2}\sqrt{\frac{8\pi}3}i(q_{11}+q_{1,-1})\yh+\sqrt{\frac{4\pi}3}q_{10}
\zh\ric
\eeqa

The $l=2$ moments, called {\em electric quadrupole moments}, are easily
shown to be
\beqa
q_{22}=\frac14\sqrt{\frac{15}{2\pi}}\iniv\rhxp(x-iy)^2\nonumber \\
q_{21}=-\sqrt{\frac{15}{8\pi}}\iniv\rhx(x-iy)z\nonumber \\
q_{20}=\frac12\sqrt{\frac5{4\pi}}\iniv\rhx(3z^2-r^2).
\eeqa
These multipole moments are traditionally written in terms of the components
of the {\em traceless quadrupole moment tensor}, defined by
\beq
Q_{ij}\equiv\iniv\rhx(3x_ix_j-r^2\de_{ij});
\eeq
the subscripts $i$ and $j$ stand for Cartesian components $x$, $y$, and
$z$, or 1,2,3. With a little algebra, one can show that
\beqa
q_{22}=\frac1{12}\sqrt{\frac{15}{2\pi}}(Q_{11}-2iQ_{12}-Q_{22}) \nonumber \\
q_{21}=-\frac13\sqrt{\frac{15}{8\pi}}(Q_{13}-iQ_{23}) \nonumber \\
q_{20}=\frac12\sqrt{\frac5{4\pi}}Q_{33}
\eeqa
It seems a little strange to be replacing at most five independent numbers
(contained in the moments $q_{2m}$) by nine numbers $Q_{ij}$; however, the
quadrupole moment tensor is symmetric, $Q_{ij}=Q_{ji}$, reducing the
number of possible independent components to six, and it also has, as its
name suggests and as may be shown easily from the definition, zero trace so
that $Q_{33}=-Q_{11}-Q_{22}$ and only two of the diagonal components are
independent. Thus the tensor can have at most five independent components
also.

 
\edo

