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

{\bf Set 10:} \bigskip

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

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\item Exercise E.39: Paul Trap.
\smallskip

In the quasi-static approximation the field is the electrostatic field 
with the given boundary conditions at the time in question. Due to the 
cylindrical symmetry we have $\Phi=\Phi(\rho,z;t)$. To get the boundary 
conditions on the end electrodes right, we set
$$ \Phi\ =\ \left(z^2-\frac{1}{2}\,\rho^2-d^2\right)\,F(\rho,z,t)\,. $$
The boundary condition on the ring electrode implies 
$\rho^2/2=z^2+d^2/2$ and on this boundary
$$ \Phi\ =\ V_0\,\sin(\omega t)\ =\ -\frac{3}{2}\,d^2\,F(\rho,z,t)\,.$$
Therefore, we try
$$ F(t)\ =\ - \frac{2}{3}\,\frac{V_0}{d^2}\,\sin(\omega t)\,,$$
as solution. For it the potential becomes
$$ \Phi\ =\ - \frac{2}{3}\,\frac{V_0}{d^2}\,
   \left(z^2-\frac{1}{2}\,\rho^2-d^2\right)\,\sin(\omega t)\,, $$
which is now easily seen to be the solution:
$$ \nabla^2\Phi = - \frac{2}{3}\frac{V_0}{d^2}\,\sin(\omega t)
   \,\nabla^2 \left(z^2-\frac{1}{2}\,\rho^2-d^2\right)\ =\ 0\,,$$
which is unique due to the Dirichlet BCs. The electric field is given 
by
$$ \vec{E}\ = - \nabla\Phi\ =\ -\left(\hat{\rho}\,
                \frac{\partial\Phi}{\partial\rho} 
              + \hat{z}\,\frac{\partial\Phi}{\partial z}\right)\
   =\ \frac{2}{3}\,\frac{V_0}{d^2}\,\left(2\,z\,\hat{z}
              -\rho\,\hat{\rho}\right)\,\sin(\omega t)\,. $$
\bigskip


\item Exercise: Potential of two point charges.
\smallskip 

See: Method of images for a sphere, p.50 to~52 lecture notes.
\bigskip


\item Exercise E.49: Potential inside a sphere with Dirichlet BC.
\smallskip 

\noindent Given in the problem we have (substitute $A_{lm}$)
$$ \Phi =\sum_{l=0}^{\infty } \left( \frac{r}{R} \right)^{l} 
\int d\Omega^{\prime}\, \sum_{m=-l}^l 
\Phi (\theta ^{\prime },\phi^{\prime },R)\,  
\overline{Y}_l^m (\theta ^{\prime },\phi ^{\prime })\,
Y_l^m (\theta ,\phi )\ . $$
Using the addition formula for spherical harmonics this becomes
$$ \Phi = \frac{1}{4\pi} \sum_{l=0}^{\infty } (2l+1)\,\left(  
   \frac{r}{R}\right)^{l}\int d\Omega^{\prime}\,P_{l}(\cos\gamma)\ .$$
Starting with the generating function
$$ \frac{1}{\sqrt{1-2xt+t^{2}}}=\sum_{l=0}^{\infty } P_{l}(x)t^l$$
we take the derivative on both sides and multiply by $2t$ to get
$$ \frac{2xt-2t^{2}}{(1-2xt+t^2)^{3/2}} = 
  \sum_{l=0}^{\infty }2lP_{l}(x)t^{l}\ . $$
Adding the last two equations gives
\begin{eqnarray} \nonumber
  &~& \frac{2xt-2t^{2}}{(1-2xt+t^2)^{3/2}}+\frac{1}{\sqrt{1-2xt+t^{2}}}
  \\ \nonumber &=& \frac{1-t^{2}}{(1-2xt+t^2)^{3/2}}\
  =\ \sum_{l=0}^{\infty } (2l+1) P_{l}(x)\, t^{l} \,.
\end{eqnarray}
We define $t=r/R$ and $x=\cos\gamma$ to get
$$ \frac{1-(r/R)^2}{(1-2r\cos \gamma -(r/R)^2)^{3/2}}
   = \sum_{l=0}^{\infty } (2l+1)\,P_{l}(\cos \gamma )\,
   \left(\frac{r}{R}\right)^{l}\ . $$
Finally we substitute back into our second equation to arrive at
$$ \Phi=\frac{R\,(R^{2}-r^{2})}{4\pi }\int d\Omega'\,\frac{\Phi
   (\theta',\phi',R)} {(R^{2}-2Rr\cos\gamma+r^2)^{3/2}}\ .$$
\bigskip

\end{enumerate}
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