Derivation of Gauß' Law

Question

I am reading a book about special relativity and the author derives Gauss law the following way: $$\int_{S_R}E\dot{}d\textbf{S}=\frac{q}{4\pi R^2\epsilon_0}\int_{S_R}dS=\frac{q}{\epsilon_0}$$ (where $S_R$ is the surface of a sphere with radius $R$) from the fact that $$\textbf{E}=\frac{q\textbf{r}}{4\pi \epsilon_0||\textbf{r}||^3} .$$ I don't understand how he arrives at the last term, could somebody explain this to me?

Answer 1

From the form stated, the magnitude of the field $\mathbf{E}$ is constant and the direction is radial, i.e. it is a vector normal to the sphere.

Recall that $d\mathbf{S}$ is a vector whose magnitud is the area of the surface and points outwards. Therefor $\mathbf{E}\cdot d\mathbf{S}=EdS$ (where the scalar product in $\mathbb{R}^3$ is performed as usual, taking in account that $\mathbf{E}$ and $d\mathbf{S}$ are paralel). Consider now a sphere whith origin in the center. It is clear that, since $E$ depends solely in $r$, must be constant along the surface, so it can be factored out the integral.

Since $\int_{S_R}d\mathbf{S}$ is the area of the sphere, the integral is $4\pi R^2$. Puting all together you recover $\frac{q}{\epsilon_0}$.

Answer 2

Nevermind, I figured it out (that was a really dumb question): $\int_{S_R}{d\textbf{S}\dot{}\textbf{E}}$ is the same as $\int_{S_R}dS \ \textbf{E}\dot{}\textbf{n}$ (where $\textbf{n}$ is the unit vector pointing away from the sphere). Because $\textbf{n}=\frac{\textbf{r}}{||\textbf{r}||}$ the dot product becomes $\frac{q}{4\pi R^4\epsilon_0}\textbf{r}\dot{}{\textbf r}$ and $\textbf {r} \dot{} \textbf {r}=R^2$. The Integral $\int_{S_R}dS$ is equal to the surface area of a sphere, so the integral becomes $\frac{q}{\epsilon_0}$.