 Solution

The integral Gauss law in vacuum is ${\oint }_{\partial V}\mathbf{E}\cdot d\mathbf{S}=Q_{\mathrm{enc}}/{ϵ}_0$. A single isolated point charge has spherical symmetry, so its electric field is radial and constant in magnitude on a centred sphere. The flux is $4\pi r^2{E}_r$, giving Coulomb's lawIn electrostatic equilibrium, a conductor is an equipotential: a tangential electric field would move its mobile charges. The electric field inside the conducting material is zero. A thin pillbox straddling its surface therefore gives $E_{n}A=\sigma A/{ϵ}_0$, henceHere $\sigma$ is the charge density on the particular face and $\hat{\mathbf{n}}$ points into vacuum. For an isolated thin plate with total charge divided equally between two faces, each face instead has half that total areal density. The local boundary result must not be confused with the $\sigma /(2{ϵ}_0)$ field of an isolated nonconducting infinite charged sheet.