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\begin{center}
\textbf{Vector Fields}

\textit{\textbf{Conservative Fields}}
\end{center}

\textbf{Question}

If the potential $\phi$ is given by $\phi(\un{r}) =
\frac{1}{|\un{r}-\un{r}_0|^2}$, then find the corresponding
three-dimensional vector field.


\textbf{Answer}

\begin{eqnarray*}
\phi(\un{r}) & = & \frac{1}{|\un{r}-\un{r}_0|^2}\\
\frac{\pa \phi}{\pa x} & = & - \frac{2}{| \un{r} - \un{r}_0|^3}
\frac{(\un{r} - \un{r}_0) \bullet \frac{\pa \un{r}}{\pa x}}{| \un{r} -
\un{r}_0 |}\\
& = & -\frac{2(x-x_0)}{| \un{r} - \un{r}_0 |^4}
\end{eqnarray*}
Similar formulae hold for other first partials of $\phi$.

\begin{eqnarray*}
\Rightarrow \un{F} & = & \nabla \phi\\
& =& -\frac{2}{|\un{r} - \un{r}_0|^4} [ (x-x_0)\un{i} + (y-y_0)\un{j} +
(z-z_0)\un{k}]\\
& = & -2 \frac{\un{r} - \un{r}_0}{| \un{r} - \un{r}_0 |^4}.
\end{eqnarray*}
This is the vector field with scalar potential $\phi$.

\end{document}