Given:

Thickness of sheet=d

Volume charge density=ρ

Let the surface area of sheet be A

Consider a cuboidal Gaussian surface of width x from the central plane such that the central plane becomes one face of the cuboidal surface.

Volume of this cuboidal surface is given by

Charge enclosed by this gaussian surface = (volume charge density)× (volume of gaussian surface containing charge)

..(i)

We know that,

By Gauss’s law, flux of net electric field (E⃗ ) through a closed surface S equals the net charge enclosed by the surface (q_in) divided by ϵ₀

……..(i)

By symmetry arguments we can say that electric field is normal to the plane and has same magnitude at all points which are at same distance from the central plane.

Here flux only pass through the face which is parallel to the central plane which is at distance x from the central plane . for remaining faces of the gaussian surfaces the angle between the electric field vector and area vector =90.

Therefore no flux pass through these faces .

Therefore total flux passing through the gaussian surface is given by

…….(ii)

Using gauss’s law(i) and eqn.(ii) we can write

…..(iii)

Therefore , the electric field at a point P inside the plate at a distance x from the central plane is given by ρx/ϵ₀

Eqn(iii) is valid only inside the sheet i.e. when x<d/2

(as x is measured from the central plane )

Now for outside the sheet (i.e. x>d/2)

Consider again a similar Gaussian surface of width x where x is measured from the central plane

Charge enclosed by this gaussian surface =(volume charge density × volume of gaussian surface containing charge)

…(iv)

Using gauss’s law(i) and eqn(ii) and (iv)

Which is independent of x

Therefore electric field at a point P outside the sheet at a distance x is constant and given by ρd/2ϵ₀

Graph of electric field for 0<x<d is given by