= Solution
In <electrostatics>, the relevant <Maxwell equations> reduce to $\nabla\cdot\mathbf E=\rho/\varepsilon_0$ and $\nabla\times\mathbf E=0$. For a volume $V$ with outward unit normal $\mathbf n$, integrate the first equation and apply the <divergence theorem>:
$$
\boxed{\oint_{\partial V}\mathbf E\cdot\mathbf n\,dS=\int_V\nabla\cdot\mathbf E\,dV=\frac1{\varepsilon_0}\int_V\rho\,dV=\frac{Q_V}{\varepsilon_0}.}
$$
This is \b[the integral form of Gauss's law], with $Q_V$ the enclosed <electric charge>. The zero <curl> is the accompanying condition allowing an <electrostatic potential> locally, and globally when the domain permits one.
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