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\markright{{\sf Lecture 4 (Ch 2 part 1) Method of images}}
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\begin{document}

\chapter{Ch. 2, part 1}

\bigskip\ \bigskip

\textsc{Lecture 4 - Method of images}

\ \newpage\ 

Recall that a general solution of the boundary value problem can be obtained
if we know the Green function $G$ and that we can write $G(\mathbf{x,x}%
^{\prime })=1/\left| \mathbf{x-x}^{\prime }\right| +F(\mathbf{x,x}^{\prime
}) $, where $F(\mathbf{x,x}^{\prime })$ can be regarded as a potential due
to charges external to the volume of interest, that is, in practical
applications, to fictitious charges inside conductors. In a two-dimensional
problem (nothing depends on $z$), one can similarly write $G(\mathbf{x,x}%
^{\prime })=-2\ln \left| \mathbf{x-x}^{\prime }\right| +F(\mathbf{x,x}%
^{\prime })$.

In the cases a sphere (in 3-d) or a cylinder (in 2-d) one can attribute $F$
to a single ``image''. In the limit of infinite radius, the sphere reduces
to a plane, the simplest and most useful case.

\section{The conducting plane}

Placing a charge $q$ at the point $(0,0,z_o)$ and the ``image charge'' $-q$
at the mirror image point $(0,0,-z_o)$ assures that the potential 
\begin{equation}
\frac q{\sqrt{x^2+y^2+(z-z_o)^2}}-\frac q{\sqrt{x^2+y^2+(z+z_o)^2}}
\label{I1}
\end{equation}
vanishes on the $xy$ plane $(z=0)$. To be definite, we take $z_o>0$. The
actual potential of a charge $q$ in front of a conducting plane is given by
eq. (\ref{I1}) for $z>0,$ vanishes for $z<0$. Green's function is obtained
for $q=1$.

The force acting on $q$ is simply 
\begin{equation}
F_z=-\frac{q^2}{4z_o^2}  \label{F1}
\end{equation}
as if $q$ and $-q$ were two real charges at a distance $2z_o.$ The work
needed to bring $q$ from infinity to a distance $z_o$ from the plane is then 
\begin{equation}
W=-\frac{q^2}{4z_o}  \label{W1}
\end{equation}
If $q$ and $-q$ were two real charges at a distance $2z_o,$ the mutual
energy would be $-q^2/2z_o$, twice as large as $W$. The difference arises
because the image charge is not present to begin with.

The electric field component $E_z$ on the $xy$ plane (i.e. at the surface of
the conductor) is obtained by taking 
\[
-\frac \partial {\partial z}\left[ \frac q{\sqrt{x^2+y^2+(z-z_o)^2}}-\frac q{%
\sqrt{x^2+y^2+(z+z_o)^2}}\right] 
\]
and setting $z=0.$ It is 
\[
E_z=-\frac{2qz_o}{\left( x^2+y^2+z_o^2\right) ^{3/2}} 
\]
The parallel field components $E_x$ and $E_y$ vanish, as they should.

The ``induced'' surface charge density is given by $4\pi \sigma =E_z$. The
total induced surface charge is $-q$, as expected, since: 
\[
\frac 1{4\pi }\iint \frac{2z_o}{\left( x^2+y^2+z_o^2\right) ^{3/2}}%
\,dx\,dy=\int_0^\infty \frac{z_o}{\left( \rho ^2+z_o^2\right) ^{3/2}}\rho
\,d\rho =1 
\]

The field at the surface due to $q$ is $\frac 12E_z$. Thus the force exerted
by $q$ on the surface charge density is given by the integral of $\frac
12\sigma E_z$: 
\[
F_z=\frac 1{4\pi }\iint \frac{2q^2z_o^2}{\left( x^2+y^2+z_o^2\right) ^3}%
\,dx\,dy=\int_0^\infty \frac{q^2z_o^2}{\left( \rho ^2+z_o^2\right) ^3}\rho
\,d\rho 
\]
The integration gives $F_z=$ $q^2/4z_o^2,$ confirming the result (\ref{F1})
of the simpler argument.

It may also be desirable to confirm that the work (\ref{W1}) correctly
represents the energy balance in the field. Consider then the quantities:

\begin{itemize}
\item  $W_q(\infty )$ : the self energy of the charge $q$ when it is at
infinity (this is infinite, but never mind: it will cancel out)

\item  $W_q(z_o)$ : the self energy of the charge $q$ in its final position
(also infinite, but will cancel)

\item  $W_\sigma $ : the self-energy of the surface charge

\item  $W_{q\sigma }$ : the interaction energy between the charge $q$ in its
final position and the surface charge
\end{itemize}

The energy of the final configuration is $W_q(z_o)+W_\sigma +W_{q\sigma }$,
that of the initial configuration is $W_q(\infty )$. One can easily show
that the energy balance is given by $W_{q\sigma }$, because $%
W_q(z_o)+W_\sigma =W_q(\infty )$. But $W_{q\sigma }$ is one half the
interaction energy of the charges $q$ and $-q$ at distance $2z_o,$ because
it is obtained by integrating the corresponding energy density over half
space. Thus $W_{q\sigma }=-\frac 12(q^2/2z_o),$ in agreement with \ref{W1}.

------------------------------------------

To show that $W_q(z_o)+W_\sigma =W_q(\infty )$, note that all three
quantities are obtained by integrating the same energy density, over all
space for $W_q(\infty ),$ over the half-space $z>0$ for $W_q(z_o),$ and
(effectively) over the remaining half space for $W_\sigma $.

------------------------------------------

\section{The conducting sphere}

In this case, a point charge and its image are not equal in magnitude. We
put the charge $q$ along the $z$ axis at distance $s$ from the center of the
sphere. The image charge is then located on the $z$ axis at 
\[
s^{\prime }=\frac{a^2}s 
\]
where $a$ is the radius of the sphere, and has magnitude and sign given by 
\[
-q^{\prime }=q\frac as=q\frac{s^{\prime }}a 
\]
Strictly, the image method applies to the grounded sphere, and the result is
extended to other cases by superposition of solutions.

\subsection{Grounded sphere}

It can be shown directly that the potential 
\begin{equation}
\frac q{\sqrt{x^2+y^2+(z-s)^2}}+\frac{q^{\prime }}{\sqrt{x^2+y^2+(z-s^{%
\prime })^2}}  \label{I2}
\end{equation}
vanishes on the sphere $x^2+y^2+z^2=a^2$. To be definite, we assume that $%
s>a $, but eq. (\ref{I2}) is valid for all $s$. The solution is also
symmetric in $q,s$ and $q^{\prime },s^{\prime }$, so the ``interior''
solution $(s<a)$ can be obtained from the ``exterior'' solution by
interchanging $q,s$ with $q^{\prime },s^{\prime }.$ Green's function is
obtained for $q=1$.

The force acting on $q$ is simply 
\begin{equation}
F_z=\frac{qq^{\prime }}{\left( s-s^{\prime }\right) ^2}=-\frac{q^2\dfrac as}{%
\left( s-\dfrac{a^2}s\right) ^2}=-\frac{q^2as}{\left( s^2-a^2\right) ^2}
\label{F2}
\end{equation}
as if $q$ and $q^{\prime }$ were two real charges at a distance $s-s^{\prime
}.$ This falls off as $1/s^3$ for large $s,$ and for $s$ near $a$ reduces to 
\[
F_z=-\frac{q^2}{4\left( s-a\right) ^2} 
\]
in agreement with (\ref{F1}) for the plane. The work needed to bring $q$
from infinity to a distance $s$ from the center of the sphere is then 
\begin{equation}
W=\dint_\infty ^s\frac{q^2az_o}{\left( r^2-a^2\right) ^2}\,dr=-\frac 12\frac{%
aq^2}{s^2-a^2}  \label{W2}
\end{equation}
If $q$ and $q^{\prime }$ were two real charges at a distance $s-s^{\prime },$
the mutual energy would be $qq^{\prime }/\left( s-s^{\prime }\right) $,
twice as large as $W$. The difference arises because the image charge is not
present to begin with.

The electric field component $E_z$ on the sphere is obtained by taking 
\[
-\frac \partial {\partial z}\left[ \frac q{\sqrt{x^2+y^2+(z-s)^2}}+\frac{%
q^{\prime }}{\sqrt{x^2+y^2+(z-s^{\prime })^2}}\right] 
\]
which is

\[
\frac{q(z-s)}{\left( \sqrt{x^2+y^2+(z-s)^2}\right) ^3}+\frac{q^{\prime
}(z-s^{\prime })}{\left( \sqrt{x^2+y^2+(z-s^{\prime })^2}\right) ^3} 
\]
and setting $x^2+y^2+z^2=a^2.$ Keeping in mind that on the sphere 
\[
\frac q{\sqrt{x^2+y^2+(z-s)^2\ }}=-\frac{q^{\prime }}{\sqrt{%
x^2+y^2+(z-s^{\prime })^2}} 
\]
one gets 
\[
E_z=\frac{q(z-s)-q^{\prime }(q/q^{\prime })^3(z-s^{\prime })}{\left(
a^2+s^2-2sz\right) ^{3/2}}= 
\]
\[
q\,\frac{z\left[ 1-(q/q^{\prime })^2\right] -s\left[ 1-(q/q^{\prime
})^2(s^{\prime }/s)\right] }{\left( a^2+s^2-2sz\right) ^{3/2}} 
\]
Using 
\[
\left( \frac q{q^{\prime }}\right) ^2=\left( \frac sa\right) ^2=\left( \frac
a{s^{^{\prime }}}\right) ^2=\frac s{s^{\prime }} 
\]
one finds finally 
\[
E_z=\frac{qz\left( 1-(s/a)^2\right) \ }{\left( a^2+s^2-2sz\right) ^{3/2}} 
\]
Similarly: 
\[
E_x=\frac{qx\left( 1-(s/a)^2\right) \ }{\left( a^2+s^2-2sz\right) ^{3/2}} 
\]
\[
E_y=\frac{qy\left( 1-(s/a)^2\right) \ }{\left( a^2+s^2-2sz\right) ^{3/2}} 
\]
We see that the vector $\mathbf{E}$ on the sphere is radially directed, as
expected.

The ``induced'' surface charge density is then, for $s>a$, 
\begin{equation}
\sigma =\frac{qa}{4\pi }\frac{1\ }{\left( a^2+s^2-2sa\cos \gamma \right)
^{3/2}}\,\frac{a^2-s^2}{a^2}  \label{C2}
\end{equation}
where $\gamma $ is the angle from the direction of the charge (see Figs.
2.3, 2.4). The total induced surface charge is $q^{\prime }=-qa/s$, as
expected, since: 
\[
\frac a{4\pi }\int_0^\pi \,\frac{2\pi a^2\sin \gamma \,d\gamma \ }{\left(
a^2+s^2-2sa\cos \gamma \right) ^{3/2}}=\frac{a^2}{2s}\left( \frac
1{s-a}-\frac 1{s+a}\right) =\frac as\frac{a^2}{s^2-a^2}
\]
For $s<a$ the sign of $\sigma $ in (\ref{C2}) must be changed, because the
normal is inwards. The total induced charge is $-q$, because in this case 
\[
\frac a{4\pi }\int_0^\pi \,\frac{2\pi a^2\sin \gamma \,d\gamma \ }{\left(
a^2+s^2-2sa\cos \gamma \right) ^{3/2}}=\dfrac{a^2}{2s}\left( \dfrac
1{a-s}-\dfrac 1{s+a}\right) =\dfrac{a^2}{a^2-s^2}
\]

The field at the surface due to $q$ has $z$ component 
\[
\frac{q(z-s)}{\left( a^2+s^2-2sz\right) ^{3/2}} 
\]
Thus the force exerted by $q$ on the surface charge density is given by 
\[
F_z=\frac{q^2}2\left[ 1-(s/a)^2\right] \int_0^\pi \,\frac{a\ (a\cos \gamma
-s)}{\left( a^2+s^2-2sa\cos \gamma \right) ^3}\,a^2\sin \gamma \,d\gamma 
\]
$\,$ The integration gives $F_z=$ $q^2s/\left( s^2-a^2\right) ^2$,
confirming the result (\ref{F2}) of the simpler argument.

It may also be desirable to confirm that the work (\ref{W2}) correctly
represents the energy balance in the field, but we will skip it.

\subsection{Insulated sphere}

We can change the charge on the sphere, and its potential, by adding a
fictitious charge at the origin. The potential 
\begin{equation}
\frac q{\sqrt{x^2+(y-s)^2+z^2}}+\frac{q^{\prime }}{\sqrt{x^2+(y-s^{\prime
})^2+z^2}}+\frac{Q-q^{\prime }}{\sqrt{x^2+y^2+z^2}}  \label{I3}
\end{equation}
describes a charge $q$ at a distance $s$ from the center of a sphere of
radius $a$ carrying a charge $Q$. If the sphere is insulated, $Q$ stays
constant as $q$ moves about.

The force acting on $q$ is simply 
\[
F_z=\frac{qq^{\prime }}{\left( s-s^{\prime }\right) ^2}+\frac{q(Q-q^{\prime
})}{s^2} 
\]
Using the previous result (\ref{F2}), this becomes 
\begin{equation}
F_z=-\frac{q^2as}{\left( s^2-a^2\right) ^2}+\frac{qQ}{s^2}-\frac{q^2a}{s^3}
\label{F3}
\end{equation}
$F_z$ is always attractive for an uncharged sphere, and is even attractive
at small $s-a$ when $Q$ has the same sign as $q$. See Fig. 2.5. This
explains why it is not so difficult to add charge to an already charged
metal sphere, or a similar body. And why the charge stays on the body.
However, the image force has nothing to do with the work function of metals.
Jackson is wrong here.

\subsection{Sphere at a fixed potential $V$}

Just replace $Q-q^{\prime }$ with $Va$ in eq. (\ref{I3}) and the following.
The force is now 
\begin{equation}
F_z=-\frac{q^2ar}{\left( s^2-a^2\right) ^2}+\frac{qVa}{s^2}  \label{F4}
\end{equation}
It is still attractive for small $s-a$, even when $qV$ is positive.

\subsection{Sphere in a uniform field}

The sphere could be either grounded or uncharged and insulated, since an
applied uniform field does not induce a net charge. The trick is to generate
the uniform field $E_z$ by putting a charge $Q$ at $z=-R$ and a charge $-Q$
at $z=R$, then letting $Q$ and $R$ go to infinity (Fig. 2.6). The potential
of these charges 
\begin{equation}
\frac Q{\sqrt{x^2+y^2+(z+R)^2}}-\frac Q{\sqrt{x^2+y^2+(z-R)^2}}  \label{I1b}
\end{equation}
reduces to 
\begin{equation}
\frac Q{R+z}-\frac Q{R-z}=-\frac{2Qz}{R^2}=-E_0z  \label{Inf}
\end{equation}
where $E_0=2Q/R^2$ is the applied field. The image of $Q$ is a charge $-Qa/R$
at $z=-a^2/R$\/; the image of $-Q$ is a charge $Qa/R$ at $z=a^2/R.$ In the
limit of large $R/a$\thinspace these two charges are a dipole of dipole
moment 
\[
D=\frac{Qa}R\,\frac{2a^2}R=a^3E_0 
\]
and generate the dipolar potential 
\[
D\frac z{r^3}=a^3E_0\frac z{r^3}=a^3E_0\frac{\cos \theta }{r^2} 
\]
The total potential for $r>a$ is then 
\[
\Phi =-E_0\left( r-\frac{a^3}{r^2}\right) \cos \theta 
\]
The induced surface charge density is 
\[
\sigma =-\frac 1{4\pi }\left. \dfrac{\partial \Phi }{\partial r}\right|
_{r=a}=\frac 3{4\pi }E_0\cos \theta 
\]
It is worth remembering that a $\cos \theta $ \thinspace charge distribution
on a spherical shell gives a uniform field inside the shell, and a dipole
fileld outside. It is also notable that the polarizability of a conducting
sphere of radius $a$ is simply 
\[
\frac D{E_0}=a^3 
\]

\subsection{Green's function}

We simply rewrite the point-charge solution, 
\begin{equation}
\frac q{\sqrt{x^2+y^2+(z-s)^2}}-\frac{q^{\prime }}{\sqrt{x^2+y^2+(z-s^{%
\prime })^2}}  \label{I2}
\end{equation}
putting $q=1$ and placing the charge at the generic point $\mathbf{x}%
^{\prime }=(x^{\prime },y^{\prime },z^{\prime })$ instead of $(0,0,s)$. The
image charge then goes at $(a/\left| \mathbf{x}^{\prime }\right| )^2$ $%
(x^{\prime },y^{\prime },z^{\prime })$ and we obtain 
\[
G(\mathbf{x,x}^{\prime })=\frac 1{\left| \mathbf{x-x}^{\prime }\right| }-%
\frac{a/\left| \mathbf{x}^{\prime }\right| }{\left| \mathbf{x-}a^2\mathbf{x}%
^{\prime }/\left| \mathbf{x}^{\prime }\right| ^2\right| } 
\]
which can be rewritten 
\[
G(\mathbf{x,x}^{\prime })=\frac 1{\left| \mathbf{x-x}^{\prime }\right|
}-\frac a{\sqrt{\left| \mathbf{x}\right| ^2\left| \mathbf{x}^{\prime
}\right| ^2+a^4-2a^2\mathbf{x}\cdot \mathbf{x}^{\prime }}} 
\]
For boundary-value problems, we need $\partial G/\partial n^{\prime }$ on
the surface of the sphere, $\left| \mathbf{x}^{\prime }\right| =a$. This is
given by Jackson, which then goes on to use the result to compute the field
of a sphere with hemispheres at different potentials, an uninteresting
problem that we will skip. However, it may be useful as guidance for problem
2.3.

\section{The conducting cylinder}

The treatment is similar to that for the sphere and is left by Jackson as
problem 2.7. Problem 2.11 actually gives part of the solution, and problem
2.4 is closely related. The following points are worth noting:

\begin{itemize}
\item  The image charges are $q$ and $-q$ (unlike the case of the sphere)

\item  You cannot make the potential vanish both on the cylinder and at $%
\rho =\infty $. With the image charge alone, the potential on the cylinder
is not zero (unlike the case of the sphere). You can add a charge to the
cylinder (as in the case of the sphere), but then the potential will not go
to zero at $\rho =\infty $.

\item  All the equipotentials are cylinders. Thus you can solve the problem
2.4 of two oppositely charged cylinders, external to each other or inside
each other, in general not coaxial. One result of the problem is the
capacitance $C$ that was quoted in lecture 3: it looks different, but 
\[
\cosh ^{-1}x=\ln \left( x+\sqrt{x^2-1}\right) 
\]
\end{itemize}

\end{document}