**Field due to a uniformly charged infinite plane sheet**

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

We can take the Gaussian
surface to be a rectangular parallelepiped of cross sectional area A, as shown.
(A cylindrical surface will also do.) As seen from the figure, only the two
faces 1 and 2 will contribute to the flux; electric field lines are parallel to
the other faces and they, therefore, 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 E.ΔS 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 σ A. Therefore by Gauss’s law,

where nˆ is a unit vector normal to the plane and going away from
it. E is directed away from the plate if σ is positive and toward the plate if
σ is negative.

Note that the above
application of the Gauss’ law has brought out an additional fact: E is
independent of x also

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