Electric Field Due To An Infinite Plane Sheet Of Charge
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 p1near the sheet, distant r in front of the sheet. To evaluate the field at p1 we choose another point p2 on the other side of sheet such that p1and p2are 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 p1and p2.suppose the flat ends of p1and P2have 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
Φ1=E.dS=EdS cos 00=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 900=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
Comparing equations (1) and (2),we get
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
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.