%% Bernd Berg oct 21 2016.
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\begin{document}

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\centerline{\bf Electrodynamics A (PHY 5346) Fall 2016 
Solutions} \bigskip

{\bf Set 8:} \bigskip

% \begin{figure}[tb] % \vspace{5pc}
% \centerline{\hbox{ \psfig{figure=hsphere,width=12.5cm} }}
% \caption{for assignment~14.} \label{fig_lader} \vskip -10pt
% \end{figure} \centerline{Potential as function of $z$.}

\begin{enumerate}

\setcounter{enumi}{23}

\item Exercise (E.40): Method of images for a point charge in front of 
an infinite grounded plane. \smallskip

\noindent For the grounded plane we may choose the $(x-y)$ plane
and position the charge $q$ at $z=+a$ on the $z$-axis. By symmetry 
reasons the image charge, which ensures $\Phi=0$ on the plane, is 
$-q$ at $z=-a$.
\smallskip

\noindent (a) By Gauss law $Q_{\rm surface} = -q\,$.
\smallskip

\noindent (b) $F=|\vec{F}|=q^2/(2a)^2$ by Coulomb's law.
\smallskip

\noindent (c) Using cylindrical coordinates [$x=\rho\cos(\phi)$, 
$y=\rho\sin(\phi)$], the potential due to charge and image charge is  
$$ \Phi(\rho,z) = \frac{q}{\sqrt{\rho^2+(z-a)^2}} 
                - \frac{ q}{\sqrt{\rho^2+(z+a)^2}}\ . $$
The induced surface charge density is
\begin{eqnarray} \nonumber
  \sigma\,=\, -\left. \frac{1}{4\pi}\,\frac{\partial\Phi}{\partial z}
         \right|_{z\to 0^+} &=& -\frac{q}{4\pi}\,\lim_{z\to 0^+}\,
          \left\{ \frac{z+a}{[\rho^2+(z+a)^2]^{3/2}} -
          \frac{z-a}{[\rho^2+(z-a)^2]^{3/2}} \right\}\\ \nonumber
         &=& -\frac{q}{2\pi}\, \frac{a}{[\rho^2+a^2]^{3/2}}\, =\,
          -\frac{q}{2\pi}\, \frac{a}{r^3}
\end{eqnarray} 
where $r$ is the distance of the point on the plane from the position
of the charge $q$.

\noindent (d) The corresponding Dirichlet Green function is 
$$ G_D(\vec{r},\vec{r}^{\,'}) = -
   \frac{1}{4\pi\sqrt{(\vec{\rho}-\vec{\rho}^{\,'})^2+(z-z')^2}} +
   \frac{1}{4\pi\sqrt{(\vec{\rho}-\vec{\rho}^{\,'})^2+(z+z')^2}}\ . $$
with the derivative
$$ \left. \frac{\partial G_D}{\partial z'}\right|_{z'=0} =\ -\frac{z}
   {2\pi\,\left[(\vec{\rho}-\vec{\rho}^{\,'})^2+z^2\right]^{3/2}}\ .$$
\bigskip


\item Exercise (E.42): Potential inside a sphere from 
distinct BCs on the half-spheres.

\begin{figure}[th] % \vspace{5pc}
\centerline{\hbox{ \psfig{figure=figures/hsphere,width=12.5cm} }}
\caption{Potential asfunction of $z$.} \label{fig_hsphere}
% \vskip -10pt
\end{figure}

\noindent
A. We have to calculate
$$  \Phi(r) = R^2 \int_S d\Omega'\, \Phi_0 \left. 
    {\partial G_D\over \partial r'} \right|_{r'=R}\,,  $$
where the surface is the upper half-sphere and the derivative of the 
Green function is given by
$$ \left. {\partial G_D\over \partial r'} \right|_{r'=R} =
    {R^2-r^2\over 4\pi\,R\,\left(R^2+r^2-2Rr\,\cos\gamma\right)^{3/2}}$$
with $\cos\gamma=\cos (\theta)\,\cos (\theta')+\sin (\theta)\,
\sin(\theta')\,\cos(\phi-\phi')$. For $r=z$ on the $z$-axis we have
$\theta=0$ and, hence, $\cos\gamma=\cos\theta'$. The integral over
the upper half-sphere becomes
$$  \Phi(z) = \Phi_0\,\frac{R}{2}\,\left(R+z\right)\,
    \left(R-z\right)\,\int_0^1 {d\cos\theta'\over 
    \left(R^2+z^2-2R\,z\,\cos\theta'\right)^{3/2}}\,.$$
With $a=R^2+z^2$ and $b=2Rz$ the integral is
\begin{eqnarray} \nonumber 
   \int_0^1 {dx\over\left(a-b\,x\right)^{3/2} } &=& 
   \left. {2\over b\,\left(a-b\,x\right)^{1/2} }\right|_0^1\ 
   =\ {2\over b\,\sqrt{a-b}}-{2\over b\,\sqrt{a}} \\ \nonumber
 &=& {1\over Rz\,\sqrt{(R-z)^2}}-{1\over Rz\,\sqrt{R^2+z^2}}
\end{eqnarray}
and
\begin{eqnarray} \nonumber 
  \Phi(z) = \Phi_0\,\frac{1}{2z}\,(R+z)\,\left(1 -
  \frac{R-z}{\sqrt{R^2+z^2}} \right)\,.
\end{eqnarray}
Special values:
\begin{eqnarray} \nonumber 
  z\ =\ R\ &\Rightarrow&\ \Phi(R)\ =\ \Phi_0\,\frac{2R}{2R}
   \ =\ \Phi_0\,,\\ \nonumber
  z\ =\ -R\ &\Rightarrow&\ \Phi(-R)\ =\ 0\,,\\ \nonumber
  z\ \to\ 0\ &\Rightarrow&\ \Phi(0)\ =\ \Phi_0\,\frac{R}{2z}\,
     \left(1-\frac{R-z}{R}\right)\ \to\ \frac{1}{2}\,\Phi_0\,.
\end{eqnarray} \bigskip

\noindent B. The plot is shown in the figure.
\bigskip


\item Exercise (E.46): Potential in a rectangular box. \smallskip

\noindent By separation of variables, we get
$$ \Phi (x,y,z)=\sum_{nm} A_{nm}\sin \left( \frac{n\pi x}{a} \right)
\sin \left( \frac{m\pi y}{b} \right)\, \sinh \left( z\pi 
\sqrt{\frac{n^{2}}{a^{2}}+\frac{m^{2}}{b^{2}}} \right) $$
The BC condition given in the problem is
$$ \Phi (x,y,c)=\sin \left( \frac{\pi x}{a} \right)\, \sin \left( 
   \frac{2\pi y}{b} \right) + \sin \left( \frac{3\pi x }{a} \right)\,
   \sin \left( \frac{\pi y}{b} \right)\ . $$
From this we find that only $(n=1,m=2)$ and $(n=3,m=1)$ contribute to
the expansion
\begin{eqnarray} \nonumber
 \Phi (x,y,z) & = & A_{12}\, \sin \left( \frac{\pi x}{a}\right)\, 
 \sin \left( \frac{2\pi y}{b} \right)\, \sinh \left( z\pi 
 \sqrt{\frac{1}{a^{2}}+\frac{4}{b^{2}}} \right) \\ \nonumber
 & + & A_{31}\, \sin \left( \frac{3\pi x}{a} \right)\, 
 \sin \left( \frac{\pi y}{b} \right)\, \sinh \left( 
 z\pi\sqrt{\frac{9}{a^{2}}+\frac{1}{b^{2}}} \right)\ .
\end{eqnarray}
Comparing to 
$$ \Phi (x,y,c)=\sin \left(\frac{\pi x}{a}\right)\, 
  \sin \left(\frac{2\pi y}{b}\right) + \sin \left(\frac{3\pi x }{a}
  \right)\, \sin \left(\frac{\pi y}{b}\right) $$
we find
$$ A_{31}={1\over \sinh \left(c\pi\sqrt{9/a^2+1/b^2}\right)}~~{\rm 
and}~~ A_{12}={1\over \sinh \left(c\pi\sqrt{1/a^2+4/b^2}\right)}\ .$$
The same result is obtained by using the integral definitions
of the coefficients and performing the integrations.
\bigskip


\end{enumerate}
\end{document}
