How is Gauss' Law (integral form) arrived at from Coulomb's Law, and how is the differential form arrived at from that?

Question

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?

Answer 1

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.