Electric Field inside a nonconducting sheet

[forestmine]

Homework Statement

A huge (essentially infinite) horizontal nonconducting sheet 10.0 cm thick has charge uniformly spread over both faces. The upper face carries +95.0 nC/m2 while the lower face carries -25.0 nC/ m2. What is the magnitude of the electric field at a point within the sheet 2.00 cm below the upper face? (\varepsilon_0 = 8.85 × 10-12 C2/N · m2)

Homework Equations

electric flux = ∫E*da = q(enc)/ε_0

The Attempt at a Solution

I figured this problem should be relatively easy, but the fact that the charge densities are different on each side is throwing me off...

First, I thought to find the E field at some particular point, I ought to use a Gaussian surface, and my thought was to use a cylinder, sticking in as far as .02m below the upper face. Doing so, gives me E=σA/ε_0*A, and the A's cancel, of course.

What I'm really confused about is how to take into account the fact that it's only .02m into the sheet. And for that matter, because the charge density is different on each side, I'm not sure how to go about finding the charge density at that particular point of .02m inside the sheet.

If anyone could help clear some of this up, I'd really appreciate it.

Thanks!

 

[Simon Bridge]

Principle of superposition.
You can do it OK as a single sheet of charge right?

 

[forestmine]

Thanks for the reply.

That's right, just confused about the varying charge density. So if I understand you correctly, I should find the field of the one side, the other side, and then again for 2cm in? How do I determine a correct charge density for the point that's 2 cm in, however?

 

[Simon Bridge]

Your description says the charge is spread uniformly over the faces only - not penetrating into the sheet. So what you have is two parallel 2D sheets of charge.

so put the +Q sheet at z=+5cm and the -Q sheet at z=-5cm and work out the field at z=+3cm.

 

[asteriatic]

Your approach using a Gaussian surface and the equation for electric flux is correct. To take into account the distance of .02m into the sheet, you can use the fact that the electric field is constant within a nonconducting sheet. This means that the electric field at the point .02m below the upper face will be the same as the electric field at the surface of the sheet, which is given by E = σ/ε_0, where σ is the surface charge density. To find the surface charge density at that particular point, you can use the given charge densities on each face and the fact that the charge must be conserved. This means that the total charge on the upper face must be equal to the total charge on the lower face. From there, you can solve for the surface charge density at the point of interest and use it to calculate the electric field using E = σ/ε_0 as mentioned before. I hope this helps clarify the problem for you.