Section 15.1 Differential Form of Gauss' Law
Recall that Gauss' Law says that
\begin{gather*}
\Int_{\textrm{box}} \EE \cdot d\AA
= \frac{1}{\epsilon_0} \, Q_{\textrm{inside}} .
\end{gather*}
But the enclosed charge is just
\begin{gather*}
Q_{\textrm{inside}} = \Int_{\textrm{box}} \rho \> \dV
\end{gather*}
so we have
\begin{gather*}
\Int_{\textrm{box}} \EE \cdot d\AA
= \frac{1}{\epsilon_0} \Int_{\textrm{box}} \rho \> \dV .
\end{gather*}
Applying the Divergence theorem to the left-hand side of this equation tells us that
\begin{gather*}
\Int_{\textrm{inside}} \!\! \grad\cdot\EE \>\> \dV
= \frac{1}{\epsilon_0} \Int_{\textrm{box}} \rho \> \dV
\end{gather*}
for any closed box. This means that the integrands themselves must be equal, that is,
\begin{gather*}
\grad\cdot\EE = \frac{\rho}{\epsilon_0} .
\end{gather*}
This conclusion is the differential form of Gauss' Law, and is one of Maxwell's Equations. It states that the divergence of the electric field at any point is just a measure of the charge density there.