Question

the electric field at a point near an infinite thin sheet of charged counductor

Answer 1

Φ1=E.dS=EdS cos 0°=EdS

At the points on the curved surface,the field vector E and area vector dS make an angle of

90° with each other.

So, φ2=E.dS=EdS cos 90°=0

Therefore,cylindrical surface does not contribute to the flux.

Hence, the total flux through the closed surface is

Φ=φ1+φ1+ φ2 (there are two end caps)

Or φ=EdS+EdS+0=2EdS (1)

Now according to Gauss’s law for electrostatics

Φ=q/ε0 (2)

Comparing equations (1) and (2),we get

2EdS=q/ε0

Or E=q/2ε0dS (3)

The area of sheet enclosed in the Gaussian cylinder is also dS. Therefore,the charge contained in the cylinder,q=σdS (σ=q/dS)

Substituting this value of q in equation (3),we get

E=σdS/2ε0dS

Or E=σ/2ε0