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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?

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  • $\begingroup$ I believe that what you a looking for is not for integrable charge densities, but generalized densities in the sense of distribution theory, and if there is an equivalent for gauss law in that scenario. Try looking for books on distribution theory. $\endgroup$
    – Hydro Guy
    Feb 13, 2015 at 12:49
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    $\begingroup$ @HydroGuy Yes, I guess the problem it is more general and not restricted to electromagnetism. I don't have much knowledge on distribution theory, there are no ways to prove it using analysis? $\endgroup$
    – NNec
    Feb 13, 2015 at 13:14
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    $\begingroup$ @0celo7 The divergence should be zero everywhere except at the points where $ \mathbf{r} = \mathbf{r'} $ where it should be proportional to a Dirac delta function. But I am not sure of it, especially because I have little knowledge of distribution theory and I would like to know other ways to prove it related to mathematical analysis. $\endgroup$
    – NNec
    Feb 13, 2015 at 13:22
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    $\begingroup$ @0celo7 Thanks for your hints, but it still turns out the use of Dirac delta functions: $\endgroup$
    – NNec
    Feb 13, 2015 at 15:00
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    $\begingroup$ I might be misunderstanding here. You dont't want to use delta functions? If so, then Hydro's comment makes sense: you'll have to look at distribution/measure theory. $\endgroup$
    – Ryan Unger
    Feb 13, 2015 at 15:01

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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.

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    $\begingroup$ It would be interesting to have a proof without Dirac's delta functions. I think the original proof (that I cannot find) is older than the theory of delta sequences and thus shouldn't use Dirac functions. $\endgroup$
    – NNec
    Feb 13, 2015 at 17:10

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