Calculating Electric Field for Infinite Plane Slab with Uniform Charge Density

[Midas_Touch]

Problem: An infinite plane slab, of thickness 2d carries a uniform charge density rho. Find the electric field as a function of y, where y=0 at the center. Plot E versus y calling E positive when it point in the +y direction and negative when it points in the -y direction.

Okay, so I worked out the electric field, which gave me
E = 2d*rho/epsilon_0

Does this look right?

 

[Tide]

No. The field is not constant for -d < y < d.

 

[Midas_Touch]

No. The field is not constant for -d < y < d.

I used the integral of E.da = Qenc/epsilon_0

E.4d^2 = rho*8d^3/epsilon

What am I doing wrong?

 

[Tide]

Set up a "pill box" (cylinder) whose center is at y = 0 and whose flat faces are parallel to the plane y = 0 and equidistant from it. By symmetry, the electric field is perpendicular to the flat surfaces (but perpendicular to the curved surface) so the only contribution to the flux is from the flat faces.

That flux is related to the total charge enclosed by the surface (Gauss' Law) and depends on the height of the pill box. Clearly, when the height of the box is 0 then the charge enclosed is also zero so the flux and electric field at y = 0 is 0. If you make the box taller, then the amount of charge it contains increases. You need to figure out how much charge is contained and it will be proportional to 2xE(y) x Area of flat surfaces.