# Electrostatics and Gauss law [duplicate]

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?

• You can prove this with a fairly difficult intigration – Sourabh Mar 3 '19 at 14:20
• From what assumptions do you want to prove this claim? – ACuriousMind Mar 3 '19 at 14:24
• $\int E\cdot dS$ is the definition of flux. How can you prove it is zero without using its definition? – garyp Mar 3 '19 at 16:37
• But how can you even talk about flux at all without referencing its definition? I can talk about an automobile because I have a definition of "automobile" implicit in my memory. How can I talk about a "Xxfghold" without defining what a "Xxfghold" is? How can I prove flux is zero without knowing what flux is? – garyp Mar 3 '19 at 16:41

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 $$\oint \vec{\mathbf{E}}\cdot\mathrm{d}\vec{\mathbf{S}} = \frac{Q}{\varepsilon_0}$$