Question

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

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
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Answer 2

infinite thin sheet of charged conductor
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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