# Proof of Gauss's law

1. Oct 9, 2009

Hi guys. My question is related to proving Gauss's law by using Columb's law. Let start with a charge density $$\rho(\vec{r})$$ in $$R^3$$ . by Columb's law we have :
$$E(\vec{r})=\int{\frac{\rho(\vec{r'})d^3r'(\vec{r'}-\vec{r})} {|\vec{r'}-\vec{r}|^3}}$$

suppose that domain of function $$\rho(\vec{r})$$ is finite in $$R^3$$. Clearly it can be showed that above integral exists for any $$\vec{r}$$ out of the domain. & it's divergence is zero out of the domain. So for any Gauss's surface out of domain we can use Divergence theoreom to prove that
$$\int{E(\vec{r}).\vec{dS}}=\int{\rho(\vec{r})d^3\vec{r}}=\frac{Q}{\epsilon 0}$$

But for a Gauss's surface that goes through domain of $$\rho(\vec{r})$$ we have two problems :
1. it is not clear that for all $$\vec{r}$$ the integral of electric field exist and converges.
2. Suppose that E converges for any $$\vec{r}$$. But it is not easy to bring devergence operator in the integral. Divergence theoreom doesn't work here.

What is your idea? Please don't use delta dirac function. I don't understand it. Because I haven't studied distributions in mathematics.
Similar question can be asked when we have surface charge density on the Gauss's surface.

Last edited: Oct 9, 2009
2. Oct 9, 2009