Defining the problem
I (believe) am getting a wrong result when I try to prove Gauss's Law for a charge $q$ inside a closed surface $\partial \Omega$. The (mathematical) problem goes like this:
Given a closed surface $\partial \Omega \subset \mathbb{R}^3$, the constants $k_e, q \in \mathbb{R}$ and the vector field $\mathbf{E} = k_e q \mathbf{r} / |\mathbf{r}|^3$, calculate the total (electric) flux of the vector field $\mathbf{E}$ through the surface $\partial \Omega$: $$ \Phi_e = \iint_{\partial \Omega} \mathbf{E}\cdot d\mathbf{A} $$ Where $d\mathbf{A}$ is just the differential area on the closed surface $\partial \Omega$. (I couldn't render the closed surface integral $\unicode{x222F}$ using Mathjax, sorry).
Trying to solve it
Defining $\Omega$ as all the points inside the closed surface and using the divergence theorem, we can write the integral as: $$ \Phi_e = \iiint_{\Omega} \nabla \cdot \mathbf{E}\,dV $$
Using the Einstein's summation convention and cartesian coordinates to represent $\mathbf{r}$ as $\mathbf{r} = x_i \mathbf{e}_i $, the divergence $\nabla\cdot\mathbf{E}$ is expressed as :
$$ \nabla\cdot \mathbf{E} = \frac{\partial}{\partial x_i} \frac{k_e q x_i}{|\mathbf{r}|^3} = k_e q \left( x_i \frac{\partial}{\partial x_i} \frac{1}{|\mathbf{r}|^3} + \frac{1}{|\mathbf{r}|^3}\frac{\partial}{\partial x_i} x_i \right) = k_e q \left( x_i \frac{\partial}{\partial x_i} \frac{1}{|\mathbf{r}|^3} + \frac{3}{|\mathbf{r}|^3} \right) $$ For $\partial |\mathbf{r}|^{-3} / \partial x_i$ we have: $$ \frac{\partial}{\partial x_i} \frac{1}{\left(\mathbf{r\cdot r}\right)^{3/2}} = -\frac{3}{2}(\mathbf{r\cdot r})^{-5/2} (2\mathbf{r})\cdot \frac{\partial \mathbf{r}}{\partial x_i} = -\frac{3\mathbf{r}}{|\mathbf{r}|^5} \cdot \frac{\partial x_j \mathbf{e}_j}{\partial x_i} = -\frac{3x_i}{|\mathbf{r}|^5} $$ Substituting back into the expression for $\nabla \cdot \mathbf{E}$, we have: $$ \nabla\cdot\mathbf{E} = k_e q \left( -\frac{3x_i x_i}{|\mathbf{r}|^5} + \frac{3}{|\mathbf{r}|^3} \right) = k_e q \left( -\frac{3|\mathbf{r}|^2}{|\mathbf{r}|^5} + \frac{3}{|\mathbf{r}|^3} \right) = k_e q \left( -\frac{3}{|\mathbf{r}|^3} + \frac{3}{|\mathbf{r}|^3} \right) = 0 $$
Substituting on the integral for $\Phi_e$ will just give $0$, too.
Wrong result:
Here is where I know there is a mistake: Although with the given electric field $\mathbf{E}$ the charge $q$ is placed in the origin, I haven't said anything about the closed surface $\partial \Omega$: I have not said if the closed surface contains or not the origin, this is, I haven't said if the charge is inside the closed surface or not.
Gaus's Law states that if the charge is inside $\partial \Omega$, then $\Phi_e = 4\pi k_e q$ (If the origin is inside $\partial \Omega$, for the given $\mathbf{E}$)
If the charge is outside, then $\Phi_e = 0$. (If the origin is outside $\partial \Omega$ for the given $\mathbf{E}$)
So,
- Since I am getting $0$ as an answer, in which moment I am assuming that the charge is outside?
- How can I get the two results, $0$ and $4\pi k_e q$?