Can someone please verify my proof of Gauss's Law? I directly copied it from my assignment, and it is an extra credit assignment, so I just want someone to tell me if it is completely correct or whether I made an assumption or there is something missing in my proof.
We prove Gauss's Law for a single point charge, then use the superposition principle of the electric field to generalize it for multiple charges. We start from Coulomb's Law for the Electric Field: $$\mathbf{E}=\frac{1}{4\pi\varepsilon_0}\frac{Q}{r^2}\hat{\mathbf{r}}$$ where $+Q$ is a point charge. Negative charges change the direction of the entire field, so this proof will work for negative charges as well. Define the origin to be the location of this charge. Mathematically, we can write Coulomb's Law as $$\mathbf{E}=\frac{1}{4\pi\varepsilon_0}\frac{Q}{\lVert{\mathbf{r}\rVert}^3}\mathbf{r}$$ where $\mathbf{r}$ is the outward radial vector relative to the charge. \\ Consider an arbitrary closed surface $S$ that completely surrounds the charge. Consider an arbitrary tiny differential area element vector $d\mathbf{S}$ on $S$ that is normal to $S$ at that point. Then the differential electric flux $d\Phi_E$ is given by $$d\Phi_E=\mathbf{E}\cdot d\mathbf{S}$$ so if we take the closed surface integral around the surface, we get $$\Phi_E=\iint_S\mathbf{E}\cdot d\mathbf{S}$$ or $$\Phi_E=\frac{Q}{4\pi\varepsilon_0}\iint_S\frac{\mathbf{r}}{\lVert{\mathbf{r}\rVert}^3}\cdot d\mathbf{S}$$ We now show that the integral $$\iint_S\frac{\mathbf{r}}{\lVert{\mathbf{r}\rVert}^3}\cdot d\mathbf{S}=4\pi$$ We begin with the Divergence Theorem, which states that $$\iiint_V \nabla\cdot\mathbf{E}\;dV=\iint_{\partial V} \mathbf{F}\cdot d\mathbf{S}$$ where $V$ is a simply-connected closed region and $\partial V$ is the boundary surface of $V$. For simplicity, let $$\mathbf{e}=\frac{\mathbf{r}}{\lVert{\mathbf{r}\rVert}^3}$$ It can be shown that $\nabla\cdot\mathbf{e}=0$ by writing $$\mathbf{e}=\frac{\mathbf{r}}{\lVert{\mathbf{r}\rVert}^3}=\frac{x\mathbf{i}+y\mathbf{j}+z\mathbf{k}}{(x^2+y^2+z^2)^{3/2}}$$ and directly using the definition of divergence. We cannot, however, use the divergence theorem directly on the region enclosed by $S$ on $\mathbf{e}$ since $\mathbf{e}$ is not defined at the origin. \\ We now consider a new region $E$ that is such that the outer boundary surface of $E$ is $S$ and the inner boundary of the surface is a sphere centered at the origin with radius $a$ such that the sphere is entirely contained within $S$. Call the sphere $S_0$. In this way, the origin is excluded from $E$ which allows us to use the divergence theorem. Now, $$\iiint_E \nabla\cdot\mathbf{e}\: dV = \iint_{\partial E} \mathbf{e}\cdot d\mathbf{S}=\iint_{\partial E} \mathbf{e}\cdot\mathbf{n}\: dS$$ where $\mathbf{n}$ is the outward unit normal vector. Now, $$\iint_{\partial E}\mathbf{e}\cdot\mathbf{n}\: dS=\iint_S \mathbf{e}\cdot\mathbf{n}_1\: dS+\iint_{S_0}\mathbf{e}\cdot(-\mathbf{n}_2)\: dS$$ Where $\mathbf{n}_1$ represents the outward normal of $S$ and $\mathbf{n}_2$ represents the outward normal of $S_0$. Note that since the outward normal of $\partial E$ is actually the inward normal of $S_0$, we must add a negative sign to account for this. As a result, we have $$\iiint_E \nabla\cdot\mathbf{e}\: dV=\iint_S \mathbf{e}\cdot d\mathbf{S}-\iint_{S_0}\mathbf{e}\cdot d\mathbf{S}$$ However, we said earlier that $\nabla\cdot\mathbf{e}=0$, so we get $$\iint_S\mathbf{e}\cdot d\mathbf{S}=\iint_{S_0}\mathbf{e}\cdot d\mathbf{S}$$ This makes it so that the flux of the electric field through any closed surface is equal to the flux of the electric field through a sphere, which easy to find. We have $$\iint_{S_0}\mathbf{e}\cdot d\mathbf{S}=\iint_{S_0}\mathbf{e}\cdot\mathbf{n} \: dS=\iint_{S_0}\frac{\mathbf{r}}{\lVert{\mathbf{r}\rVert}^3}\cdot\frac{\mathbf{r}}{\lVert{\mathbf{r}\rVert}}\: dS=\iint_{S_0}\frac{1}{a^2} dS $$ where the last part is found by using the unit normal vector of a sphere and the fact that $\mathbf{r}\cdot\mathbf{r}=\lVert{\mathbf{r}\rVert}^2$. This gets us $$\iint_{S_0}\frac{1}{a^2}dS=\frac{1}{a^2}4\pi a^2=4\pi$$ We just showed that the flux through the sphere is $4\pi$ which means that the flux through the arbitrary surface $S$ is also $4\pi$. Going back to the initial equation, we get $$\Phi_E=\frac{Q}{4\pi\varepsilon_0}\iint_S \frac{\mathbf{r}}{\lVert{\mathbf{r}\rVert}^3}\cdot d\mathbf{S}=\frac{4\pi Q}{4\pi\varepsilon_0}=\frac{Q}{\varepsilon_0}$$ This shows that Gauss's Law holds for one charge. \\ Now, if there are multiple charges $q_1, q_2, \ldots, q_n$ in the surface, then the superposition principle states that the total electric field $\mathbf{E}$ is just the vector sum of the individual electric fields, or $$\mathbf{E}=\mathbf{E}_1+\mathbf{E}_2+\ldots+\mathbf{E}_n$$ Thus, $$\Phi_E=\iint_S\mathbf{E}\cdot d\mathbf{S}=\iint_S (\mathbf{E}_1+\mathbf{E}_2+\ldots+\mathbf{E}_n)\cdot d\mathbf{S}=\iint_S\mathbf{E}_1\cdot d\mathbf{S}+\iint_S\mathbf{E}_2 \cdot d\mathbf{S}+\ldots+\iint_S\mathbf{E}_n\cdot d\mathbf{S}$$ This equals $$\frac{q_1}{\varepsilon_0}+\frac{q_2}{\varepsilon_0}+\ldots+\frac{q_n}{\varepsilon_0}=\frac{q_1+q_2+\ldots+q_n}{\varepsilon_0}=\frac{q_{enc}}{\varepsilon_0}$$ Using this addition, we can see that Gauss's Law works for continuous charge distributions as well since a continuous charge distribution is simply made up of infinitely many point charges whose electric fields add due to superposition. \\ Finally, say that the charge is outside the closed surface. This means that the closed surface does not enclose the origin. Since $\nabla\cdot\mathbf{E}=0$, the Divergence Theorem gives $$\iint_S \mathbf{E}\cdot d\mathbf{S}=\iiint_E \nabla\cdot\mathbf{E}\: dV=0$$ where $E$ is the region with $S$ as its boundary. Now because $\mathbf{E}$ is defined everywhere inside the region (as the only place $\mathbf{E}$ isn't defined is the origin, which we have already said is outside the region), we see that $\Phi_E=0$. This easily extends to multiple charges outside the surface, since the divergence operator has a distributive property if each electric field is defined in the region, which it is. That is, $$\nabla\cdot(\mathbf{E}_1+\mathbf{E}_2+\ldots+\mathbf{E}_n)=\nabla\cdot\mathbf{E}_1+\nabla\cdot\mathbf{E}_2+\ldots+\nabla\cdot\mathbf{E}_n$$ which equals zero since each term equals zero. \\ We have thus proved Gauss's Law for the electric field.