 Solution

Gauss's law statesThe surface is closed and oriented outward, and $Q_{\mathrm{enc}}$ includes all enclosed charge. The integral law follows from the local form through the divergence theorem. It is universally valid in classical electromagnetism, but calculating the field directly from the flux requires enough symmetry.

For spherical charge symmetry, $E(r)=Q(<r)/(4\pi {ϵ}_0{r}^2)$. A uniform infinite line gives $E=\lambda /(2\pi {ϵ}_{0}r)$ from a coaxial cylinder. An infinite sheet gives $E=\sigma /(2{ϵ}_0)$ on either side from a pillbox, with the direction set by charge sign. The normal field jump is $\sigma /{ϵ}_0$; the electrostatic field inside a conductor is zero. Zero enclosed charge implies zero net flux, not generally zero field. ^[ref]