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On a similar note: when using Gauss' Law, do you even begin with Coulomb's law, or does one take it as given that flux is the surface integral of the Electric field in the direction of the normal to the surface at a point?

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While it's healthy to know these derivations, you should keep in mind that Gauss's law is more general than Coulomb's law. Coulomb's law is only true if the charges are stationary, there are no changing magnetic fields, etc. But Gauss's law is true under all circumstances. So Gauss's law is more than just a consequence of Coulomb's law. –  Steve B Nov 23 '12 at 13:31
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Let us for simplicity consider $n$ point charges $q_1$, $\ldots$, $q_n$, at positions $\vec{r}_1$, $\ldots$, $\vec{r}_n$, in the electrostatic limit, with vacuum permittivity $\epsilon_0$.

Now let us sketch one possible strategy to prove Gauss' law from Coulomb's law:

  1. Deduce from Coulomb's law that the electric field at position $\vec{r}$ is $$\tag{1} \vec{E}(\vec{r})~=~ \sum_{i=1}^n\frac{q_i }{4\pi\epsilon_0}\frac{\vec{r}-\vec{r}_i}{|\vec{r}-\vec{r}_i|^3} . $$

  2. Deduce the charge density $$\tag{2} \rho(\vec{r})~=~\sum_{i=1}^n q_i\delta^3(\vec{r}-\vec{r}_i). $$

  3. Recall the following mathematical identity $$\tag{3}\vec{\nabla}\cdot \frac{\vec{r}}{|\vec{r}|^3}~=~4\pi\delta^3(\vec{r}) .$$ (This Phys.SE answer may be useful in proving eq.(3), which may also be written as $\nabla^2\frac{1}{|\vec{r}|}=-4\pi\delta^3(\vec{r})$).

  4. Use eqs. (1)-(3) to prove Gauss' law in differential form $$\tag{4} \vec{\nabla}\cdot \vec{E}~=~\frac{\rho}{\epsilon_0} .$$

  5. Deduce Gauss' law in integral form via the divergence theorem.

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Is 3 Poisson's equation, a generalisation of it or a subdivision of it? And thanks for the answer- not too unfathomable. –  Alyosha Sep 26 '12 at 19:17
I updated the answer. Eq.(3) is a mathematical identity, while Poisson's equation has physical content. –  Qmechanic Sep 26 '12 at 19:35
Now I realize that the proof is also given near the bottom of this Wikipedia page. –  Qmechanic Sep 30 '12 at 16:16
I intuit equation (2), but why cube $\delta(\mathbf{r}-\mathbf{r_i})$? –  Alyosha Mar 12 '13 at 13:17
@Alyosha: It's standard notation for the 3-dimensional delta function $\delta^3(\vec{r})~:=~\delta(x)\delta(y)\delta(z)$. –  Qmechanic Mar 12 '13 at 14:27
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