The expression for electric field intensity at a points outside uniformly charged thin plane sheet is (where, d is the distance of point from plane sheet)

Explanation

Electric field intensity at a point outside uniformly charged thin plane sheet :
Consider an infinite thin plane sheet of positive charge having a unform surface charge density `sigma` on both sides of the sheet. By symmetry, electric field perpendicular to plane sheet and directed outwards. Electric field intensity has same magnitude at a given distance on either sides of the sheet.

Let P be the point at which electric field intensity is to be found, E be electric field intensity at point P, S be Gaussian surface and PS is plane sheet.
To find electric field intensity at a point P due to uniformly charged infinite thin plane sheet, an imaginary cylinder is constructed around P with its axis perpendicular to plane sheet carrying charge with ends having cross-sectional area ds. The plane sheet passes through the middle of cylinder's length in such a way that the ends of cylinder are equidistant from the plnae sheet carrying charge.
Electric field intensity `(vec(E))` is perpendicular to the ends of cylinder, hence the electric flux througth each end is Eds. Since `vec(E)` is perpendicular to the plane sheet, it is parallel to the curved surface of Gaussian cylinder.
`:.` Electric flux passing througth curved surface =0
Total normal electric induction over Gaussian surface = `epsiE(2ds)`
Where, ds is surface area of end faces of cylinder.
Algebraic sum of charges enclosed by Gaussian cylinder `=sigmads`
As per Gauss' law,
`epsiE(2ds)=sigmads`
`:.E=(sigma)/(2epsi)`
This is the expression of electric field intensity at a point outside uniformly charged thin plane sheet.