Past exam of the physics 2 paper of the University of Oxford / 2013 / 2 / 2

For a point charge $q_j$, Coulomb's law gives ${\mathbf{E}}_j=q_j\hat{\mathbf{r}}/(4\pi {ϵ}_0{r}^2)$. The flux through an oriented surface element isHere $d\Omega$ is its signed solid angle seen from the charge. A closed surface subtends total oriented solid angle $4\pi$ for a point inside and zero for a point outside. Thus the flux of each inside charge is $q_j/{ϵ}_0$ and each outside charge has zero net flux. By superposition,This is Gauss's law; a surface passing through a point-charge singularity is excluded.