Question

Use Gauss's theorem to find the electric field due to a uniformly charged infinitely large plane thin sheet with surface charge density $\sigma$.

Answer 1

Consider an infinite plane which carries the uniform charge per unit area $\sigma$
Let the plane coincides with the $y-z$ plane.
Let us draw a cylindrical gaussian surface, whose axis is normal to the plane, and which is cut in half by the plane.
Let the cylinder run from $x=-a$ to $x=+a$, and let its cross-sectional area be A. According to Gauss' law,
$$2E\left(a\right)A=\frac{\sigma A}{{ϵ}_0},$$
where $E\left(a\right)=-E(-a)$ is the electric field strength at $x=+a$. Here, the left-hand side represents the electric flux out of the surface.
The only contributions to this flux come from the flat surfaces at the two ends of the cylinder. The right-hand side represents the charge enclosed by the cylindrical surface, divided by ${ϵ}_0$. It follows that
$$E=\frac{\sigma }{2{ϵ}_0}.$$