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Using Gauss’s law, prove that the electric field at a point due to a uniformly charged infinite plane sheet is independent of the distance from it.

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Let electric charge be uniformly distributed over the surface of a thin, non-conducting infinite sheet. Let the surface charge density (i.e., charge per unit surface area) be σ. We need to calculate the electric field strength at any point distant r from the sheet of charge.

To calculate the electric field strength near the sheet, we now consider a cylindrical Gaussian surface bounded by two plane faces A and B lying on the opposite sides and parallel to the charged sheet and the cylindrical surface perpendicular to the sheet (fig). By symmetry the electric field strength at every point on the flat surface is the same and its direction is normal outwards at the points on the two plane surfaces and parallel to the curved surface.

Total electric flux

\(\therefore\) Total electric flux = 2Ea

As σ is charge per unit area of sheet and a is the intersecting area, the charge enclosed by

Gaussian surface = σa

According to Gauss’s theorem,

Total electric flux = \(\cfrac{1}{e_0}\times\) (total charge enclosed by the surface)

Thus electric field strength due to an infinite flat sheet of charge is independent of the distance of the point.

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