Question

electric field at a point near an infinite thin sheet of charge conductor is ??

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

900 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