Electrostatics and Gauss law

Question

How can we prove that surface integral of the electric fieldfor a point charge that is outside a gaussian surface, $$\int\mathbf E\cdot\,\mathrm d\mathbf S,$$ without actually using the concept of flux?

Answer 1

Let us start from Coulomb's law: $$\vec{\mathbf{E}}(\vec{\mathbf{r}}) = \frac{1}{4\pi\varepsilon_0}\frac{q}{r^2}\hat{\mathbf{r}}$$ where we take $\vec{\mathbf{r}}$ to be our coordinate, and $\varepsilon_0$ to be the vacuum permittivity.

Of course, you should be able to see how we can express this same law in terms of the charge density, instead of the charge itself, through integrating over space: $$\vec{\mathbf{E}}(\vec{\mathbf{r}}) = \frac{1}{4\pi\varepsilon_0}\int\frac{\rho(\vec{\mathbf{r}}_0)}{||\vec{\mathbf{r}}_0-\vec{\mathbf{r}}||^2}\hat{\mathbf{r}}~\mathrm{d}^3\vec{\mathbf{r}}$$

where $\rho$ is the aforementioned charge density. Here, we can take as a theorem the divergence of an inverse-square vector field to be $$\vec{\nabla}\cdot\frac{1}{r^2}\vec{\mathbf{r}} = 4\pi\delta(\vec{\mathbf{r}})$$

Using this, we now can take the divergence of our charge density-defined electric field equation: $$\vec{\nabla}\cdot\vec{\mathbf{E}}(\vec{\mathbf{r}}) = \frac{1}{\varepsilon_0}\int\rho(\vec{\mathbf{r}})\delta(\vec{\mathbf{r}}_0-\vec{\mathbf{r}})~\mathrm{d}^3\vec{\mathbf{r}}$$

which is obviously $$\vec{\nabla}\cdot\vec{\mathbf{E}}(\vec{\mathbf{r}}) = \frac{\rho(\vec{\mathbf{r}})}{\varepsilon_0}$$

By the divergence theorem, we can see that this is equivalent to $$\oint \vec{\mathbf{E}}\cdot\mathrm{d}\vec{\mathbf{S}} = \frac{Q}{\varepsilon_0}$$

which is your desired result.