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\markright{{\sf Lecture 3 (Ch 1 part 3) Capacitance}}
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\begin{document}

\chapter{Ch.1, part 3}

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

\vspace{1.0in}

\textsc{Lecture 3: Capacitance and capacitors}\ 

\newpage\ 

\section{A single conductor}

If an isolated conductor is raised to the potential $V$, the potential
everywhere is $\Phi (\mathbf{x)}=V\phi (\mathbf{x})$, where $\phi (\mathbf{x)%
}$ $=1$ on the conductor and vanishes at infinity. The charge on the
conductor, as given by Gauss's theorem, is 
\[
4\pi Q=-V\int_{\mathcal{S}}\frac{\partial \phi }{\partial n}\,da 
\]
where $\mathcal{S}$ is (conveniently) the surface of the conductor. We see
that $Q$ is proportional to $V$. By definition, the capacitance is 
\begin{equation}
C=\frac QV=-\frac 1{4\pi }\int_{\mathcal{S}}\frac{\partial \phi }{\partial n}%
\,da  \label{C}
\end{equation}

The energy of the charged conductor is 
\[
W=\frac 12QV=\frac 12\ CV^2=\frac 12\frac{Q^2}C 
\]
Another expression for $W$ is the usual 
\[
W=\frac 1{8\pi }\int_{\mathcal{V}}\left( V\text{\thinspace }\mathbf{\nabla }%
\phi \right) ^2d^3x 
\]
where $\mathcal{V}$ is the volume external to the conductor. Comparing these
two expressions for $W$, we get another formula for $C$%
\[
C=\frac 1{4\pi }\int_{\mathcal{V}}(\mathbf{\nabla }\phi )^2d^3x 
\]
(The earlier formula is recovered from this with an integration by parts).
We see that $C$ is always positive. This follows also from the fact that $%
\phi (\mathbf{x)}$ has its maximum, $\phi =1$, on the surface $\mathcal{S}$,
so that $\partial \phi /\partial n$ is negative on $\mathcal{S}$.

\noindent ------------------------------

In esu, the unit of $C$ the $\mathrm{cm}$. In the MKSA system, it is the $%
\mathrm{farad}$:

\[
\mathrm{farad}=\frac{\mathrm{coul}}{\mathrm{volt}}=\frac{3\cdot 10^9\mathrm{%
statcoul}}{\left( 1/300\right) \,\mathrm{statvolt}}=9\cdot 10^{11}\mathrm{cm}
\]

For a sphere of radius $a$, the (esu) capacitance is $C=a$.

The farad is a huge capacitance. In common use is the picofarad: 
\[
\mathrm{pF}=\mu \mu \mathrm{F}=0.9\,\mathrm{cm} 
\]

\noindent ---------------------------------

For a simple body, $C$ is of the order of the size of the body. For a sphere
of radius $a,$ we have the basic result $C=a$ (in gaussian units, of course;
in SI units $C=4\pi \epsilon _0a$). If body A can be inscribed in body B, it
can be shown that $C_A<C_B$ (Jackson's problem 1.20). Thus the radii of the
inscribed and circumscribed sphere give simple bounds on $C$. For a convex
body without sharp edges, corners, or points, $C$ is almost the same as the
capacitance of a sphere having the same surface area $S,$ or $C<\sqrt{S/4\pi 
}$.

\smallskip\ 

\textbf{\ Examples of capacitance for simple bodies:}

\begin{itemize}
\item  cube, side $a$ (numerically from Van Bladen) and $2/\pi =.6366198$%
\[
0.665\,a<C<0.6678\,a
\]

\item  prolate ellipsoid, axes $a,b,b$, focal distance $c=\sqrt{a^2-b^2}$: 
\[
C=\frac{2c}{\ln \dfrac{a+c}{a-c}}=\dfrac c{\ln \dfrac{a+c}b}
\]

\begin{itemize}
\item  needle ($a>>b$): 
\[
C=\dfrac a{\ln \dfrac{2a}b}
\]
\end{itemize}

\item  oblate ellipsoid, axes $a,a,b$, focal distance $c=\sqrt{a^2-b^2}$: 
\[
C=\frac c{\arctan \dfrac cb}
\]

\begin{itemize}
\item  disk ($a>>b$): 
\[
C=\frac{2a}\pi 
\]
\end{itemize}

\item  thin toroidal ring, ring radius $a,$ cross-section radius $b:C=\dfrac{%
\pi a}{\ln \dfrac{8a}b}$

\item  spherical bowl, radius $a,$ subtended angle $2\theta :$%
\[
C=\frac a\pi \left( \theta +\sin \theta \right) 
\]
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Note that the results for the ellipsoids and for the bowl reduce to $C=a$ in
the appropriate limit ( $b=a$ for the ellipsoids, $\theta =\pi $ for the
bowl). Also, for small $\theta $ the bowl reduces to a disk of radius $%
a\theta $ and its capacitance is $C=2a\theta /\pi $ as expected.
\end{itemize}

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

\section{Many conductors}

For two bodies, the relation between charges and voltages is \label{C2} 
\begin{eqnarray}
Q_1 &=&C_{11}V_1+C_{12}V_2  \label{C2} \\
Q_2 &=&C_{21}V_1+C_{22}V_2  \nonumber
\end{eqnarray}
with $C_{21}=C_{12}.$ It can be shown that $C_{11}$ and $C_{22}$ are
positive, $C_{12}$ is negative, and $C_{11}\geq \left| C_{12}\right| $ as
well as $C_{22}\geq \left| C_{12}\right| $.

To obtain these results, start from the two basic configurations where $%
V_1=1,V_2=0$ and $V_1=0,V_2=1$. Call the corresponding potentials $\phi
_1(x) $ and $\phi _2(x)$. In the general configuration the potential is $%
\Phi (x)=V_1\phi _1(x)+V_2\phi _2(x)$ and Gauss's theorem gives 
\[
4\pi Q_1=-V_1\int_{\mathcal{S}_1}\frac{\partial \phi _1}{\partial n}%
\,da-V_2\int_{\mathcal{S}_1}\frac{\partial \phi _2}{\partial n}\,da 
\]
\[
4\pi Q_2=-V_1\int_{\mathcal{S}_2}\frac{\partial \phi _1}{\partial n}%
\,da-V_2\int_{\mathcal{S}_2}\frac{\partial \phi _2}{\partial n}\,da 
\]

This argument clearly can be generalized to any number of conductors.

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

\textsc{Homework Assignment}

Analogously to the one-conductor case, give explicit expressions for $C_{ij}$
and use energy arguments, or flux arguments, to show that $C_{ii}$ is
positive, $C_{ij}$ negative for $i\neq j$, and $\sum_jC_{ij}$ is
non-negative. Use Green's theorem to show that $C_{ij}=C_{ji}$.

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

\subsection{Capacitors}

Of practical interest is the configuration with zero net charge, $Q_1=Q$ and 
$Q_2=-Q$.

\begin{itemize}
\item  In an ideal \textit{capacitor }$C_{12}=-C_{11}$ (so that no flux
escapes): then eq. (\ref{C2}) gives $Q=C_{11}\Delta V$, where $\Delta
V=V_1-V_2$ (note that $C_{11}$ is not the same as the capacitance of the
isolated conductor 1).

\item  In the case of a symmetrical capacitor, $V_1=\Delta V/2$ and $%
V_2=-\Delta V/2$.

\item  Quite generally we can solve for $V_1$ and $V_2$ to obtain $\Delta
V=Q/C$, with the \textit{internal capacitance }now given by 
\begin{equation}
C=\frac{C_{11}C_{22}-C_{12}C_{21}}{C_{11}+C_{22}+C_{12}+C_{21}}  \label{Cm}
\end{equation}
\end{itemize}

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

\textsc{Homework Assignment}

Obtain eq. (\ref{Cm}) and its simpler forms that apply when:

\begin{itemize}
\item  The two conductors are equal

\item  Conductor 2 surrounds conductor 1
\end{itemize}

A spherical capacitor consists of a metal sphere of radius $a$ and a metal
shell of inner radius $b_1$, outer radius $b_2$. Find $C$ and all the $%
C_{ij} $ elements for this capacitor. When the capacitor is charged ($Q$ on
the inner sphere, $-Q$ on the outer spherical shell), what are the
potentials $V_1$ and $V_2$? How is the charge distributed on the outer shell
(how much is on the inner surface, how much on the outer surface?

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

Here is a list of $C$ for various capacitors.

\begin{itemize}
\item  Two closely spaced parallel plates of area $A$, separation $d$ ($%
A>>d^2$): 
\[
C=\frac A{4\pi d} 
\]

\item  Two concentric cylinders, radii $a$ and $b$ with $a>b$, length $%
L>>(a-b)$: 
\[
C=\frac L{2\ln \dfrac ab} 
\]

\item  Two parallel cylinders, both of radius $a$, separation $D>2a$, length 
$L>>D:$%
\[
C=\frac L{4\ln \left( \dfrac D{2a}+\sqrt{\left( \dfrac D{2a}\right) ^2-1}%
\right) } 
\]

\begin{itemize}
\item  for $D>>a$, this reduces to 
\[
C=\frac L{4\ln \dfrac Da} 
\]
\end{itemize}

\item  Any two parallel cylinders, radii $a,b$, separation $D,$ length $L:$ 
\[
C=\frac L{2\ln \left( x+\sqrt{x^2-1}\right) } 
\]
where 
\[
x=\left| \frac{D^2-a^2-b^2}{2ab}\right| 
\]
This formula applies when the cylinders are external to each other $(D>a+b)$
and $L>>D,$ as well as when they are inside each other $(D<\left| a-b\right|
)$ and $L>>\left| a-b\right| .$ The case of two intersecting cylinders is
different, since they must be at the same potential.

\item  Two concentric spheres, radii $a$ and $b$, with $a>b:$%
\[
C=\frac 1{\dfrac 1b-\dfrac 1a} 
\]
\end{itemize}

\subsection{Forces on charged conductors}

Forces can be obtained simply by differentiating at constant $Q_i$ the
energy formula 
\[
W=\frac 12\sum_iV_iQ_i=\frac 12\sum_{ij}\left( \Bbb{C\,}^{-1}\right)
_{ij}Q_iQ_j 
\]
where $\Bbb{C\,}^{-1}$ is the inverse of the capacitance matrix. The force
in the general direction of the parameter $\xi $ is 
\[
F_\xi =-\left( \frac{\partial W}{\partial \xi }\right) _Q=-\frac
12\sum_{ij}Q_iQ_j\frac \partial {\partial \xi }\left( \Bbb{C\,}^{-1}\right)
_{ij} 
\]

\begin{itemize}
\item  For a single conductor 
\[
W=\frac{Q^2}{2C} 
\]
and the force in the general direction of the parameter $\xi $ is 
\begin{equation}
F_\xi =-\left( \frac{\partial W}{\partial \xi }\right) _Q=-\frac{Q^2}2\frac{%
\partial (1/C)}{\partial \xi }=\frac{Q^2}{2C^2}\frac{\partial C}{\partial
\xi }  \label{FQ}
\end{equation}
For example, a charged sphere of radius $r$ experiences a radial force 
\[
F_r=-\frac{\partial W}{\partial r}=\frac{Q^2}{2r^2} 
\]
As expected, this force pushes outward, expanding the sphere.

\item  For a capacitor the force between the plates is also given by eq. (%
\ref{FQ}) where $C$ is now the internal capacitance. For example, the mutual
force between the plates of a thin capacitor ($C=A/4\pi x)$ is 
\[
F_x=-Q^2\frac \partial {\partial x}\left( \frac 12\frac{4\pi x}A\right) =-%
\frac{2\pi Q^2}A 
\]
As expected, this force is attractive and its magnitude per unit area is $%
2\pi \sigma ^2$, where $\sigma =Q/A$ is the surface charge density.
\end{itemize}

\noindent ---------------------------------

If we want to obtain the forces by differentiating at constant $V$, we must
use the formula (with the seemingly ``wrong'' sign) 
\begin{equation}
F_x=\left( \frac{\partial W}{\partial x}\right) _V=\frac \partial {\partial
x}\left( \frac 12CV^2\right) =\frac{V^2}2\frac{\partial C}{\partial x}
\label{F2}
\end{equation}

The reason for the ''wrong'' sign is that the virtual work at constant $V$
is $\delta W-V\delta Q$, where the extra term is the work (typically done by
batteries) to restore the charge. Since $\delta W=\frac 12V\delta Q$, we
find that $\delta W-V\delta Q=-\delta W$.

\end{document}
