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

Consider an infinite plane which carries the uniform charge per unit area σ. Let the plane coincide 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(a)A = σA/ε₀, where E(a) = -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 ε₀. It follows that E = σ/2ε₀.