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\begin{center}
\textbf{Phys 742 - EM I}

\textbf{In-class test 1 - 29 February 1996}
\end{center}

\smallskip\ 

\textbf{1.}

Two conducting spheres of radius $a$ carry opposite charges.

\textbf{(a) }Find the capacitance $C$ of this system when the spheres are
separated by a distance $b>>a$ (One can obtain an expansion in $a/b.$ The
first term of this expansion is enough for full credit on this problem, but
you can get the next term for extra credit if time permits). Give a number
(in cm and in $\mu F$) when $a=1$ \negthinspace cm and $b=\!10$ cm.

\textbf{(b) }What is the force between the spheres? Obtain the answer
directly and check that it is also given by a general formula involving%
\textbf{\ }$C$ with the $C$ of part (a)\textbf{\ }

\smallskip\ 

\textbf{2.}

Consider the problem of finding the potential in a charge-free region
surrounded by a surface $S$ when

\textbf{(a)} the potential is assigned on $S$

\textbf{(b) }the field is assigned on $S$

\textbf{(c)} both the potential and the field have assigned values on $S$

Discuss these three cases. Can the potential be assigned arbitrarily on the
boundary and is then the solution unique? Can the field components be
assigned arbitrarily on the boundary, or which components can be assigned
arbitrarily, and is then the solution unique? And what about case (c)? You
are not expected to prove your answers, but they must be precise answers.

\smallskip\ \ 

\textbf{3.}

\textbf{(a)} We know that the real part of the complex potential $i\arcsin
(z/a)$ represents the electrostatic potential of a charged strip. What is
the charge per unit length on the strip? Sketch the field lines and the
equipotentials.

\textbf{(b)} What electrostatic potential problem is solved by the real part
of $\arcsin (z/a)$? (If you can follow the branches of the complex $\arcsin $
you will think that the answer is obvious; in any case do the following).
Find the electric field all along the $x$ axis. From its behavior for $x>a$
argue that you have there a conducting plate. Similarly for $x<a.$ By
integrating $E_x$ from $-a$ to $a,$ find the potential difference between
the plates.

\smallskip\ 

\textbf{4.}

A grounded conducting sphere of radius $a$ is immersed in a uniform field $%
E_0.$

\textbf{(a) }What is charge density on the surface of the sphere?

\textbf{(b)} What is the induced dipole moment? \textbf{\ }What is the
induced quadrupole moment?

\smallskip\ 

\textbf{5.}

Sketch the field lines and equipotentials of a dipole.

\newpage 

\addtolength{\topmargin}{-0.6in}

\begin{center}
\textbf{Solution}
\end{center}

\textbf{1.}

\textbf{(a) }This is very similar to Jackson's 1.7. Let the spheres have
charges $\pm q.$ The potential on the sphere with charge $+q$ is
approximately $q/a-q/b.$ On the other sphere, it is $-q/a+q/b.$ Taking the
difference and dividing by $q,$%
\[
\frac 1C\simeq 2\left( \frac 1a-\frac 1b\right) \;\quad \text{or\quad }%
C\simeq \frac 12\,\frac{ab}{b-a}\simeq \frac 12\left( a+\frac{a^2}b\right) 
\]
For $a=1$ cm and $b=10$ cm, $C=(5/9)$ cm $=0.556$ cm\thinspace $=0.617$ pF.
(The other formula gives 0.55 cm). See also the \textbf{Comments.}

\textbf{(b) }The force is clearly $F=-q^2/b^2.$ This is consistent with 
\[
F=\left. -\frac \partial {\partial b}\left( \frac{q^2}{2C}\right) \right|
_{q=const}=-\frac{q^2}2\,\frac \partial {\partial b}\left( \frac 1C\right) 
\]

\smallskip\ 

\textbf{2.}

When the potential is assigned on $S,$ there is always a unique solution
(Dirichlet problem). When the \textsl{normal }component of the field is
assigned on $S$, there is always a solution that is unique up to an additive
constant (Neumann problem). In part (c) the problem is overspecified and
has\ no solution in general.

\smallskip\ 

\textbf{3.}

\textbf{(a)} If $F=i\arcsin (z/a)$, the complex field is 
\[
-\frac{dF}{dz}=\frac{-i/a}{\sqrt{1-z^2/a^2}}=\frac{-1}{\sqrt{z^2-a^2}} 
\]
The complex potential of a unit line charge is $-2\ln z$ and the complex
field is $2/z.$ Comparing the two fields for large $z,$ we see that the
charge per unit strip length is $-1/2.$

\textbf{(b) }Recall that $\dfrac{dF}{dz}=-E_x+iE_y.$ For $F=\arcsin (z/a)$, 
\[
\dfrac{dF}{dz}=\dfrac 1{\sqrt{a^2-z^2}} 
\]
It follows that $E_x$ vanishes for $\left| x\right| >a,$ meaning that the
halfplane for $x>a$ is at a constant potential, and the halfplane for $x<-a$
is too. The potential difference is 
\[
V_{right}-V_{left}=-\int_{-a/2}^{a/2}E_xdx=\int_{-a/2}^{a/2}\dfrac{dx}{\sqrt{%
a^2-x^2}}=\left. \arcsin \frac xa\right| _{-a}^a=\pi 
\]
One can get this directly by examining $\func{Re}F,$ which is just $\arcsin
(x/a)$ for $\left| x\right| <a$ and $y=0.$

\smallskip\ 

\textbf{4.}

The charge density is $(3/4\pi )E_0\cos \theta ,$ the dipole moment is $%
E_0a^3,$ the quadrupole moment is zero.

\newpage\ 

\begin{center}
\textbf{Comments}
\end{center}

\noindent \textbf{1. }

\textbf{(a)} \textit{(Alternative method) }Using $C_{21}=C_{12}$ (always
true) and $C_{22}=C_{11}$ (true in this case), 
\begin{equation}
C=\frac{C_{11}C_{22}-C_{12}^2}{C_{11}+C_{22}+2C_{12}}=\frac{C_{11}-C_{12}}2
\label{C}
\end{equation}
To find $C_{11}$ and $C_{12}$, we must put sphere 1 at potential 1 and
sphere 2 at potential 0.  To leading order, the charge on sphere 1 must be $%
q=a$ (so that $q/a=1$) and we must put on sphere 2 an image charge $%
q^{\prime }=-qa/b=-a^2/b.$ By definition, $C_{11}=q$ and $C_{12}=q^{\prime }.
$ Putting this back into eq. (\ref{C}), we get again the result
\[
C=\frac 12\left( a+\frac{a^2}b\right) 
\]
The disadvantage of this method is that it is limited to spheres, because it
uses explicitly the method of images. The other method is more general.
Suppose that the spheres are replaced by two bodies of capacitances $C_1$
and $C_2.$ ($C_1$ is the capacitance of body 1 all by itself and is only
approximately equal to $C_{11}.$ Similarly for $C_2.$) Then the potential on
body 1 is approximately $q/C_1-q/b$ and the potential on body 2 is
approximately $-q/C_2+q/b$.  Taking the difference and dividing by $q,$%
\[
\frac 1C\simeq \frac 1{C_1}+\frac 1{C_2}-\frac 2b
\]
\[
C\simeq \,\frac{C_1C_2}{C_1+C_2-2C_1C_2/b}\simeq \frac{C_1C_2}{C_1+C_2}%
\left( 1+\frac 2b\frac{C_1C_2}{C_1+C_2}\right) 
\]
Comparing this formula with eq. (\ref{C}), we see that, to leading order, $%
C_{12}=-C_1C_2/b.$

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