Question

The electric field at a point near an infinite thin sheet of charged conductor is

Answer 1

$\bold{ans : E = \frac{\sigma}{2\epsilon_0}}$

explanation:-
first understand what is difference between sheet and plate in electrostatic ?
sheet => both side charge 
plate => one side charge 

here , mentioned sheet , then surely charge will appear in both side of conductor .

now, Use the Gauss concept for finding electric field .
Let Q present in charged sheet conductor .
because , I clearify that both side charge presents in sheet . 
so, each side charge share = Q/2 

now, Gauss formula, 
Φ = Qnet/2ε₀
here , Qnet at one side = Q/2 
also we know, Φ { electric flux near the surface of sheet } = E{electric field near the surface} × A { cross section Area of gaussian surface is chosen by us } 

so, EA = Q/2ε₀
E = Q/2Aε₀= {Q/A}/2ε

we know, Q/A = surface charge density = σ
E = σ/2ε₀

hence, electric field at a point near an infinite sheet of charged conductor is σ/2ε₀

Answer 2

Electric field at a point near an infinite thin sheet of charged conductor is σ/2ε₀  

To find the electric field at a point near an infinite thin sheet of charged conductor;

In order to find the electric field at a point P1 on an infinite thin sheet of positive charge another point P2 is taken on the other side of the sheet, so that P1 and P2 are equidistant from the sheet. Then a cylindrical Gaussian surface is constructed through the plane which extends equally on two sides of the plane. It is seen that the electric field is uniform and does not depend on the distance from the charge sheet.

To find the electric field at a point P which is near but outside the conductor: The Gaussian surface is constructed and it is seen that the electric field near a plane charged conductor is twice the electric field. This is in accordance with Coulomb's theorem which states that the electric field at any point very close to the surface of a charged conductor is equal to charge density of the surface divided by free space permittivity.