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

{\bf Set 9:} \bigskip

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% \end{figure} \centerline{Potential as function of $z$.}

\begin{enumerate}

\setcounter{enumi}{26}

\item Exercise E.41: Potential over a plane.
\smallskip

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

The potential on the $z$ axis is then
$$\Phi(\vec{r}) = -
  \int_{-\infty}^{+\infty}dx'\int_{-\infty}^{+\infty}dy'\,
  \Phi(x',y',0)\,\left. \frac{\partial G_D}{\partial z'}\right|_{z'=0}$$
where the potential $\Phi(x',y',0)$ in the $x$-$y$-plane is given. In
our case the integration is only over the upper half-plane as $\Phi=0$
on the lower half-plane. On the $z$-axis $\rho=0$ holds and
\begin{eqnarray} \nonumber
  \Phi(z)&=&\int_{-\infty}^{+\infty}dx'\int_0^{\infty}dy'\,\
  \frac{z\,\Phi_1}{2\pi\,\left[(\vec{\rho}^{\,'})^2+z^2\right]^{3/2}}
  \\ \nonumber &=& 
  \frac{z\,\Phi_1}{2\pi}\int_0^{\pi}d\phi'\int_0^{\infty}\rho'\,
  d\rho'\,\frac{1}{(\rho'^2+z^2)^{3/2}}\ =\ \frac{\Phi_1}{2}\,. 
\end{eqnarray}
\bigskip


\item Exercise E.44: Angled plates.  \smallskip

\noindent (a) The potential satisfies $\nabla^2\Phi=0$. Choosing 
cylindrical coordinates and recognizing that by symmetry reasons
$\Phi$ does not depend on $\rho$ and $z$, we have
$$\nabla^2\Phi\ =\ \frac{1}{\rho^2}\,\frac{d^2\Phi}{d\phi^2}\ =\ 0\,.$$

\noindent (b) The boundary conditions are $\Phi(0)=0$ and $\Phi(\beta)=
\Phi_0$ and the solution is
$$ \Phi(\phi)\ =\ a\,\phi + b\ =\ \Phi_0\,\frac{\phi}{\beta}\,.$$

\noindent (c) The electric field is
$$ \vec{E}\ =\ -\nabla\Phi\ =\ -\frac{\hat{\phi}}{\rho}\, \frac{d\Phi}
   {d\phi}\ =\ -\frac{\Phi_0}{\beta\,\rho}\,\hat{\phi}\,.$$

\noindent (d) Charge densities and total charge follow from 
               Gauss' law. On each plate with opposite signs:
$$ \int_S d\vec{a}\cdot\vec{E}\ =\ 4\pi\,Q\,,\qquad
   - \hat{\phi}\cdot\vec{E}\ =\ 4\pi\,\sigma\,.              $$
Therefore,
$$  \frac{\Phi_0}{\beta\,\rho}\,\hat{\phi}=4\pi\,\sigma\ \Rightarrow\
    \sigma = \frac{\Phi_0}{4\pi\,\beta\,\rho}  $$
and the total charge on each plate is
$$ Q\ =\ \int_{\rho_1}^{\rho_2} d\rho \int_0^h dz\,
   \frac{\Phi_0}{4\pi\,\beta\,\rho}\ =\
   \frac{\Phi_0\,h}{4\pi\,\beta}\,\ln(\rho_2/\rho_1)\, .  $$

\noindent (d) The capacitance is
$$ C\ =\ \frac{Q}{\Phi_0}\ =\ 
   \frac{h}{4\pi\,\beta}\,\ln(\rho_2/\rho_1)\,.$$

\noindent (e) The energy is
$$ U\ =\ \frac{1}{2}\,C\,\Phi_0^2\ =\ \frac{h\,\Phi_0^2}{8\pi\,\beta}\,
   \ln(\rho_2/\rho_1)\,.$$
From this the torque follows to be
$$ \vec{\tau}\ =\ \frac{dU}{d\beta}\,\hat{z}\ =\ -
   \frac{h\,\Phi_0^2}{8\pi\,\beta^2}\,\ln(\rho_2/\rho_1)\,\hat{z}\,.$$
\bigskip


\item Exercise: Expansion in cylindrical coordinates. \smallskip
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% numbers of our solution.

As there is no azimuthal dependence in the BC, there will be
not $\phi$ dependence in the result and the expansion reads
$$ \Phi(\rho,z)\ =\ \sum_{m=0}^{\infty} \sum_{n=1}^{\infty} 
   A_{mn}\,J_m(k_{mn}\rho)\,\sinh(k_{mn}z)\, $$
where due to the BC the only surviving coefficient is $A_{01}$. So,
$$ \Phi(\rho,z)\ =\ A_{01}\,J_0(k_{01}\rho)\,\sinh(k_{01}z)\,.$$
Now,
$$ 1\ =\ \rho_0\ =\ \frac{x_{01}}{k_{01}}\ \Rightarrow\ k_{01}\ 
      =\ x_{01}\ \approx 2.405 $$
and the BC at $z=1$ is matched with $A_{01}=1/\sinh(x_{01})$.

\begin{enumerate}

\item The potential inside the cylinder is 
$$ \Phi(\rho,z)\ =\ J_0(x_{01}\rho)\,
   \frac{\sinh(x_{01}z)}{\sinh(x_{01})}\,.$$

\item For $\rho=0$, $z=1/2$: $J_0(x_{01}\rho)=J_0(0)=1$, $\sinh
(x_{01}/2)=1.514$.  Further, $\sinh(x_{01})=5.494$ and we find
$$ \Phi\left(0,\frac{1}{2}\right)\ =\ J_0(0)\,\frac{1.514}{5.495}\ 
   =\ 0.2756\,.$$

\item For $\rho=1/2$ we have $J_0(x_{01}/2=0.670)$ and find
$$ \Phi\left(\frac{1}{2},\frac{1}{2}\right)\ =\ 0.670\,
   \frac{1.514}{5.495}\ =\ 0.1846,.$$

\end{enumerate}
\bigskip\bigskip

\item Exercise: Calculate $L_-$. See (C.17) to (C.23) lecture notes.

\end{enumerate}
\end{document}
