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I'm working on mathematically proving the Gauss' Law, and I'm having proving the approximations of the electric field integral converge uniformly. The steps are:

First I take discrete charges $q_i$ at locations $a_i\in \mathbb{R}^3$. The electric field is $$E(x)=\sum_{i=1}^k\frac{1}{4\pi\varepsilon}q_i\frac{\underline{x}-\underline{a}_i}{||\underline{x}-\underline{a}_i||^3}$$ where $\varepsilon$ is the vacuum permittivity. For an open set $U\subset\mathbb{R}^3$ if $\partial U$ is a smooth surface taking the surface integral of the second kind yields: $$\iint_{\partial U} E\cdot\hat{n}\,dS=\sum_{\{i;q_{i}\in U\}}\frac{q_i}{\varepsilon}$$

Next, we look at a charge density $\rho(x)$, continuous on the closure of an open set $U\subset\mathbb{R}^3$. The electric field now takes an integral form:$$E(x)=\frac{1}{4\pi\varepsilon}\int_{U}\frac{x-y}{||x-y||^{3}}\rho(y)dy$$ I use approximations for the field - Taking cubes of sidelength $s$ centered at $s\mathbb{Z}^3$ we conseider the cubes $\{C_i\}$ fully contained in $U$, and approximate the integrand by it's value at the centre point $\frac{x-a_{i}}{||x-a_{i}||^{3}}\rho(a_{i})$. I am able to show that we have pointwise convergence: $$E(x)=\lim_{s\to0}\sum_i\frac{1}{4\pi}\int_{C_i}\frac{x-a_i}{||x-a_i||^3}\rho(a_i)\,dy=\frac{1}{4\pi}\cdot\lim_{s\to0}\sum_i\frac{x-a_i}{||x-a_i||^3}\rho(a_i)\,Vol(C_i)$$ I want to use the first result, and push the surface integral through the electric field integral to get: $$\iint_{\partial U}E\cdot\hat{n}\,dS=\frac{1}{4\pi}\cdot\sum_i 4\pi\rho(a_i)\,Vol(C_i)=\lim_{s\to0}\sum_i\rho(a_i)\,Vol(C_i)=\frac1\epsilon\iiint_{U}\rho$$ $$\Rightarrow \iint_{\partial U}E\cdot\hat{n}\,dS=\frac{Q_{in}}\varepsilon$$ which is Gauss' Law.

The problem is I can't seem to be able to prove uniform convergence of these approximations. The integrand is not continuous and goes to $\infty$ when $x=y$, and I don't know how to work it to get uniform convergence. Any help will be greatly appriciated!

I'll just add that I know you can take the divergence of $E$ under the integral sign to get ${\rm div} E=\rho/\varepsilon$ and then use the divergence theorem, but the question specifically asks to do the steps above above and only then to conclude that ${\rm div} E=\rho/\varepsilon$.

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  • $\begingroup$ I'm not entirely sure if I know exactly what you have in mind, but my guess is that you won't be able to prove uniform convergence to $E(x)$ by approximating $\rho(x)$ with point charges. Instead, if $\rho(x)$ is Riemann integrable or continuous, you might consider using uniformly charged balls of finite radius, or cubes instead. Beyond that, the nature of convergence probably depends on the regularity of $\rho(x)$; the 'irregular' part (e.g. sets of point charges, or diverging/cuspal surface charges) may need to be treated exactly or asymptotically to overcome a.e. pointwise convergence. $\endgroup$
    – TLDR
    Aug 10 at 19:56
  • $\begingroup$ @TLDR We are under the assumption that $\rho$ is continuous on the closure of $U$, my bad for omitting that - I'll fix it. In any case, how do you think I could use uniformly charged balls or cubes to overcome this problem? Thanks. $\endgroup$
    – Harpalyke
    Aug 10 at 21:06

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