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\begin{center}
{Mathematical Physics --- PHZ 3113}\\\vspace {2mm} 
{\bf Divergence,} (January 25, 2013) \\
\end{center}
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\begin{enumerate}

\item Solution
\begin{eqnarray}
  \nabla\cdot\vec{r}\ =\ 
  \sum_{i=1}^3 \frac{\partial x_i}{\partial x_i}\ =\ 3\,.
\end{eqnarray} 

\item Solution
\begin{eqnarray} \nonumber
  \nabla\cdot\,\vec{r}\,f(r) &=& f(r)\,\nabla\cdot\vec{r} +
  \vec{r}\cdot\nabla f\\ =\ 3\,f(r)+\vec{r}\cdot\hat{r}\,
  \frac{df}{r} &=& 3\,f(r)+r\,\frac{df}{r}\,. \label{vecrfr}
\end{eqnarray} 

\item Example: Calculate (\ref{vecrfr}) for $f(r)=r^{n-1}$. Solution:
\begin{eqnarray} \nonumber
  \nabla\cdot\,\vec{r}\,r^{n-1} &=& 3\,r^{n-1}+r\,\frac{dr^{n-1}}{d\,r}
  \\ =\ 3\,r^{n-1}+(n-1)\,r^{n-1} &=& (n+2)\,r^{n-1}\,.
\end{eqnarray} 

\item Solution for $\vec{r}\ne\vec{0}$
\begin{eqnarray}
  \nabla\cdot\vec{E} &=& \nabla\cdot\frac{q\,\vec{r}}{4\pi\,\epsilon_0
  \,r^3}\\ \nonumber &=& (-2+2)\,\frac{q}{4\pi\,\epsilon_0\,r^3}\ =\ 0\,.
\end{eqnarray} 

\end{enumerate}

\end{document}

