( ii ) APPLICATION OF GAUSS'S LAW

( ii ) Electric Field due to a Uniformly Charged Infinite Plane Sheet or Sheet of Charge: -

Let s  = uniform surface charge density on an infinite plane sheet.

P and P’ are points at a perpendicular distance r on either sides of the charged plane. By symmetry, electric intensity at P and P’ will have equal magnitude and opposite direction. f the charge on the plane is positive/negative, the direction of the electric intensity will be away/towards the plane. Consider a closed cylinder with equal lengths on either side of The plane, from P to P’. As the electric intensity is perpendicular to the plane, the flux inked with the curved surface of the cylinder is zero. As the points P and P’ are equidistant  from the charged plane, the magnitude of electric intensity are the same.

Therefore  Ep = Ep’ = E and EpA + Ep’A = 2EA is the total flux coming out of the cross- sectional area, A, of the cylinder. The closed cylindrical surface encloses the charge

q = sA.If two uniformly charged infinite plates, having surface charge density s1 and s2, are kept parallel to each other, then the magnitudes and directions of electric intensity at points between and on either

sides of planes If s1 = - s 2 = s, then the electric intensity between the plates will be s / e0 and on either sides of the two plates will be zero.