Question

using gauss's theorem derive an expression for the electric field at a point near an infinitely large charged sheet of surface density σ​

Answer 1

Answer:

Consider an infinite plane which carries the uniform charge per unit area σ

 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(a)A=ϵ0σA,

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=2ϵ0σ.