% BB Dec 03 2016.
%\documentclass[aps,twocolumn]{revtex4}
%\documentclass[aps,preprint]{revtex4}
\documentclass[aps,12pt]{revtex4}
\usepackage{epsfig}
\newcommand{\be}{\begin{equation}}
\newcommand{\ee}{\end{equation}}
\newcommand{\ba}{\begin{array}}
\newcommand{\ea}{\end{array}}
\newcommand{\tq}{\tilde{q}}
                                                                                
\begin{document}

\centerline{\bf Electrodynamics A (PHY 5346) Fall 2016 
Solutions Test on Homework December~1.} \bigskip

\begin{enumerate}

\item Time in Minkowski space (from HW~5, set~2).
\smallskip

We use natural units, $c=1$, and give all results in units of seconds.

\begin{enumerate}
\item It follows from $x^1=4\,t/5=t-15$ that the coordinates of 
      B$_0$ are $(t,x)=(75,60)$. 
\item The proper time of B is then $\tau=\sqrt{t^2-x^2}=45$.
\item At position A$_2$ the time on the clock of A is $75+60=135$.
\end{enumerate}
\bigskip

\item Potential in a rectangular box (HW 26, set~8). \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


\item Dielectric sphere (HW 41, set~12).  \smallskip

\noindent (a) We expand the potential into spherical harmonics. Due 
to the axial symmetry we have only $m=0$ contributions, which are
Legendre polynomials.
\begin{eqnarray} \nonumber
  {\rm Inside:}~~~~\Phi_{\rm in} &=& \sum_{l=0}^{\infty} A_l\,r^l\,
  {P}_l(\cos\theta)\,, \\ \nonumber
  {\rm Outside:}~~~~\Phi_{\rm out} &=& \sum_{l=0}^{\infty} \left[
  \,B_l\,r^l + C_l\,r^{-(l+1)}\,\right]\,{P}_l(\cos\theta)\,.
\end{eqnarray}
For $r\to\infty$: $\Phi\to -E_0\,z = -E_0\,r\,\cos\theta$. This
implies that the only non-vanishing coefficient $B_l$ is
$$ B_1\ =\ -E_0\ .$$
BCs at $r=R$:
\begin{eqnarray} \nonumber
  {\rm Tangential:}~~~~\frac{1}{R}\,\left. \frac{\partial\Phi_{\rm in}}
  {\partial\theta}\right|_{r=R} &=& \frac{1}{R}\,\left. 
  \frac{\partial\Phi_{\rm out}}{\partial\theta}\right|_{r=R}\,, 
  \\ \nonumber
  {\rm Normal:}~~~~\epsilon_1\,\left. \frac{\partial\Phi_{\rm in}}
  {\partial r}\right|_{r=R} &=& \epsilon_2\,\left. 
  \frac{\partial\Phi_{\rm out}}{\partial r}\right|_{r=R}\,. 
\end{eqnarray}

\noindent (b) 
Matching for the tangential BC $\partial{P}_l/\partial\theta$ 
term by term (they are independent functions of $\theta$), we find:
\begin{eqnarray} \nonumber
  A_1\,R\,=\,B_1\,R + \frac{C_1}{R^2}\ \Rightarrow\
  A_1 &=& -E_0 + \frac{C_1}{R^3}\\ \nonumber
{\rm and\ for}~~l\ge 2:~~ A_l &=& \frac{C_l}{R^{(2l+1)}}\ .
\end{eqnarray}
Similarly we match for the normal BC $P_l$ term by term and find:
\begin{eqnarray} \nonumber
  \epsilon_1\,A_1 &=& -\epsilon_2\,E_0-2\,\epsilon_2\,\frac{C_1}{R^3}
  \\ \nonumber {\rm and\ for}~~l\ge 2:~~ \epsilon_2\,l\,A_l &=& 
                  - (l+1)\,\epsilon_2\,\frac{C_l}{R^{(2l+1)}}\ .
\end{eqnarray}
Putting the $l\ge 2$ equations together we get
$$ \frac{C_l}{R^{(2l+1)}} = - \frac{(l+1)\,\epsilon_2}{l\,\epsilon_1}\,
   \frac{C_l}{R^{(2l+1)}}\ \Rightarrow\ C_l=0,~~l\ge 2\
   \Rightarrow\ A_l=0,~~l\ge 2\ .  $$
Let us define
$$ \epsilon\ =\ \frac{\epsilon_1}{\epsilon_2}\ .$$
With this notation we have
$$  \epsilon\,A_1\,=\,-E_0-2\,\frac{C_1}{R^3}\ .$$
Combining this with our other equation for $A_1$ gives
$$ A_1\,=\, -\left(\frac{3}{\epsilon+2}\right)\,E_0\,,\qquad
   C_1\,=\, \left(\frac{\epsilon-1}{\epsilon+2}\right)\,R^3\,E_0\,.$$
Therefore,
\begin{eqnarray} \nonumber
  \Phi_{\rm in} &=& -\left(\frac{3}{\epsilon+2}\right)\,E_0\,r\,
  \cos(\theta) \,=\, -\left(\frac{3}{\epsilon+2}\right)\,E_0\,z\,,
  \\ \nonumber \Phi_{\rm out} &=& -E_0\,z + 
  \left(\frac{\epsilon-1}{\epsilon+2}\right)\,E_0\,\frac{R^3}{r^2}\,
  \cos(\theta)\,.
\end{eqnarray}
Note that the last term is the potential of a dipole. The electric
fields are then
\begin{eqnarray} \nonumber
  \vec{E}_{\rm in} &=& -\nabla\,\Phi_{\rm in} \,=\,
  \left(\frac{3}{\epsilon+2}\right)\,\vec{E}_0\,, \\ \nonumber
  \vec{E}_{\rm out} &=& -\nabla\,\Phi_{\rm out} \,=\, \vec{E}_0 - 
  \left(\frac{\epsilon-1}{\epsilon+2}\right)\,R^3\,\left(
  \frac{r^2\,\vec{E}_0-3\vec{r}\,(\vec{r}\cdot\vec{E}_0)}{r^5}
  \right)\,.
\end{eqnarray}

\noindent (c) The surface charge density is
\begin{eqnarray} \nonumber
  \sigma_{\rm pol} &=& \frac{1}{4\pi}
  \left( E^r_{\rm out}-E^r_{\rm in} \right) \,=\, \frac{1}{4\pi}\,
  \hat{r}\cdot\left(\vec{E}_{\rm out}-\vec{E}_{\rm in}\right) 
  \\ \nonumber &=& \frac{1}{4\pi}\,E_0\,\cos(\theta)\,
  \left[\left(1-\frac{3}{\epsilon+2}\right) - \left(
  \frac{\epsilon-1}{\epsilon+2}\right)\,\left(1-3\right)\right]
  \\ \nonumber &=& \frac{3}{4\pi}\,\left(
  \frac{\epsilon-1}{\epsilon+2}\right)\,E_0\,\cos(\theta)\ .
\end{eqnarray}


\end{enumerate} 
\end{document}
