Obtain the expression of electric field by a plane of infinite size and with uniform charge distribution.

Explanation

Let a be the uniform surface charge density of an infinite plane sheet.

Take the x-axis normal to the given plane. By symmetry, the electric field will not depend on y and z coordinates and its direction at every point must be parallel to the x-direction.
We can take the Gaussian surface to be a rectangular parallel piped of cross sectional area A, as shown.
Only the two faces 1 and 2 will contribute to the flux, electric field lines are parallel to the other faces and they do not contribute to the total flux.
The unit vector normal to surface 1 is in -x-direction while the unit vector normal to surface 2 is in the + x-direction.
Therefore, flux `vecE. DeltavecS` through both the surfaces are equal and add up.
Therefore the net flux through the Gaussian surface is 2 EA.
The charge enclosed by the closed surface is `sigmaA`.
Therefore by Gauss.s law,
`2EA = (sigmaA)/epsilon_(0)`
`therefore E = sigma/(2epsilon_(0))`
`therefore vecE =sigma/(2epsilon_(0)).hatn`...........(1)
where `hatn` is a unit vector normal to the plan and going away from it.
E is directed away from the plate if `sigma` is positive and toward the plate if `sigma` is negative.
For a finite large planar sheet equation (1) i approximately true in the middle regions of th planar sheet, away from the ends.