Divergence Theorem, mathematical approach to Gauss's Law?

Question

Let $D$ be a compact region in $\mathbb{R}^3$ with a smooth boundary $S$. Assume $0 \in \text{Int}(D)$. If an electric charge of magnitude $q$ is placed at $0$, the resulting force field is $q\vec{r}/r^3$, where $\vec{r}(x)$ is the vector to a point $x$ from $0$ and $r(x)$ is its magnitude. Show that the amount of charge $q$ can be determined from the force on the boundary by proving Gauss's law:$$\int_S \vec{F} \cdot \vec{n}\,dA = 4\pi q.$$

I am familiar with the approach in basic textbooks, but I would be interested in seeing a derivation/proof using the language of differential topology.

Answer 1

First, we will compute $\text{div}\,F$. The partial derivatives are given by$${{\partial F}\over{\partial x_i}} = q {\partial\over{\partial x_i}}\left({{x_i}\over{r^3}}\right) = q\left({1\over{r^3}} - {{3r^2 {{x_i}\over{r}} x_i}\over{r^6}}\right) = 0.$$Thus, $\text{div}\,F = 0$ away from the origin. Consider now a ball $B$ of radius $r$ centered at the origin contained entirely in the interior of $D$. Then$$\int_B F \cdot \mathbf{N}\,dA = q \int_b {r\over{r^3}}\,dA = {q\over{r^2}} \int_B dA = {{4\pi r^2 q}\over{r^2}} = 4\pi q.$$Finally, consider the manifold $M$ consisting of the space between $B$ and $D$ with those as boundary. Then, by the Divergence Theorem,$$\int_B F \cdot \mathbf{N}\,dA - \int_D F \cdot \mathbf{N}\,dA = \int_M \text{div}\,F\,dV = 0 \implies \int_D F \cdot \mathbf{N}\,dA = 4\pi q.$$