Question

how do we find the electric field using gauss' theore​

Answer 1

Explanation:

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=

ϵ

0

σA

,

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 ϵ

0

. It follows that

E=

2ϵ

0

σ

.