Let us today discuss another application of gauss law of electrostatics that is Electric Field Due To An Infinite Plane Sheet Of Charge:-

Consider a portion of a thin, non-conducting, infinite plane sheet of charge with constant surface charge density σ. Suppose we want to find the intensity of electric field E at a point p_{1}near the sheet, distant r in front of the sheet. To evaluate the field at p_{1 }we choose another point p_{2} on the other side of sheet such that p_{1}and p_{2}are equidistant from the infinite sheet of charge(try to make the figure yourself). Now we draw a small closed Gaussian cylinder with its circular ends parallel to the sheet and passes through the points p_{1}and p_{2}.suppose the flat ends of p_{1}and P_{2}have equal area dS.The cylinder together with flat ends from a closed surface such that the gauss’s law can be applied.

By symmetry,the magnitude of electric field E at all the points of infinite plane sheet of charge on either sides end caps is same and along the outward drawn normal,for positively charged sheet.

Therefore, the electric flux through each cap is

Φ_{1}=E.dS=EdS cos 0^{0}=EdS

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

90^{0} with each other.

So, φ_{2}=E.dS=EdS cos 90^{0}=0

Therefore,cylindrical surface does not contribute to the flux.

Hence, the total flux through the closed surface is

Φ=φ_{1}+φ_{1}+ φ_{2} (there are two end caps)

Or φ=EdS+EdS+0=2EdS (1)

Now according to Gauss’s law for electrostatics

Φ=q/ε_{0} (2)

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

2EdS=q/ε_{0}

Or E=q/2ε_{0}dS (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/2ε_{0}dS

Or E=σ/2ε_{0}

This is the relation for electric filed due to an infinite plane sheet of charge.

Thus, the field is uniform and does not depend on the distance from the plane sheet of charge.

nicely explained 😊