Most of the books about electromagnetism prove Gauss' law for a point charge in vacuum:
$$ \Phi = \int_{\Sigma} \mathbf{E} \centerdot d \mathbf{S} = q/\epsilon_0 $$
and then simply state that for a continuous charge distribution the charge is
$$ q= \int_{V'} \rho (\mathbf{r'}) dV' $$ and thus the application of the divergence theorem gives the differential form of Gauss' law: $$ {\rm div} (\mathbf{E}) = \rho/\epsilon_0 .$$
But it is always true that given any (integrable) charge density $ \rho(\mathbf{r'}) $ distributed over an arbitrary volume $V'$, such that the produced electric field is:
$$
\mathbf{E}(\mathbf{r})=1/{4\pi \epsilon_0}\int_{V'} \rho(\mathbf{r'})\frac{(\mathbf{r}-\mathbf{r'})}{\mid \mathbf{r} – \mathbf{r'}\mid^3}dV',
$$
that $$ {\rm div} (\mathbf{E})=\frac{\rho}{\epsilon_0} ~?$$
How could this be proved rigorously?
Best Answer
For proving that the Coulomb E satisfies the Gauss law in differential form, simply take the divergence of Coulombian E. You will need the Dirac delta and some of its properties.
For deriving the divergence of E field, do just the same as above...
You can also derive the Gauss law in integral form from the differential form by applying the divergence theorem (= Gauss theorem --- I guess you know, why :)).
The integral form of the Gauss law from the Coulomb E can be directly derived by writing the flux integral for the Coulomb E, using the linear independency of the two integrals (primed vs. non-primed), and then recognizing a common vector integral expression.