Verified answer

Consider one plate that has a surface charge density s. Draw a small cylinder whose z-axis is perpendicular to the surface of the plate (the end cap of the cylinder lies in a plane parallel to the surface of the plate). Now make one end cap of the cylinder be just above the surface of the plate in the region between the two plates, and the other cap is inside the plate. This makes the E field parallel to the axis of the cylinder and constant across the area of the end cap that is above the plate. So we can use Gauss' Law :

Integral_surface(E * dA) = q/e0 --> Now q = s*A_cap = s*pi*r^2 where r = radius of cylinder

integral_surface(E * dA) = s*pi*r^2/e0

Now for the integral - dA = 2*pi*r*dr k_hat where k_hat is a unit vector in the z direction (pointing along cylinder axis)

E = |E| k_hat where |E| = magnitude of E vector = constant

THus E * dA = 2*pi*|E|*rdr and

integral_surface(E * dA) = 2*pi*|E|*integral(r dr) = pi*|E|*r^2

So pi*|E|*r^2 = s*pi*r^2/e0 --> |E| = s/e0