located at $(x_{0},y_{0})$


\begin{align}\phi = & \limit_{a \rigtharrow 0} \frac{m}{2\pi}\left[\ln\sqrt{(x-x...
...c{\mu}{2\pi}\frac{(x-x_{0})\cos\alpha+(y-y_{0})\sin\alpha}{r} \notag
\end{align}
where $r = \sqrt{(x-x_{0})^{2}+(y-y_{0})^{2}}$.


Timothy S Choe
2003-03-15