Solving Electric Fields with Gauss's Law: Doubts and Calculations

[Saketh]

Hello, everyone. I have a problem that I solved using Gauss's Law. However, I am unconfident in my answers, as I have very little experience with Gauss's Law.

The surfaces of two large (i.e. infinite) parallel conducting plates have charge densities as follows: $$\sigma_1$$ on the top of the top plate; $$-\sigma_2$$ on the bottom of the top plate; $$\sigma_2$$ on the top of the bottom plate; $$-\sigma_1$$ on the bottom of the bottom plate; $$\sigma_1 > \sigma_2$$. Use Gauss's law and symmetry to calculate the electric field below, between, and above the plates.​

Here's how I did it:
I made a gaussian surface - a cylinder that cut through both of the plates with radius $$r$$ - in order to find the electric field above and below the plates. Then, I used Gauss's law:
$$\oint \vec{E}\cdot \,d\vec{A} = E\oint \,dA = \frac{q_{inside}}{\epsilon_0}$$.
This is the part that I doubt myself:
$$E(\pi r^2 + \pi r^2) = \frac{\pi r^2(\sigma_1 - \sigma_2 + \sigma_2 - \sigma_1)}{\epsilon_0} = 0$$
So I concluded that below and above the plates the electric field was zero. But this did not make sense to me. Are my calculations correct? Did I put the gaussian surface in the correct place to find the electric field above and below the plates?

For between the plates, I made two different gaussian surfaces - one for each plate - that had one face in one plate and the other face in the space between the plates, such that the inside charge included both $$\sigma_1$$ and $$\sigma_2$$. But is this the correct gaussian surface, or am I supposed to make surfaces such that the $$\sigma_1$$'s are left out? This is very confusing!

I continued, and eventually calculated that the electric field in between the plates is $$-\frac{2\sigma_2 - 2\sigma_1}{\epsilon_0}$$, which means that's it's going downward. Is this correct?

Thank you for your help!

 

[maverick280857]

First of all, draw a diagram and sketch the direction of the fields. Realize that in the section between the plates, the field must be equivalent to that of a parallel plate capacitor. Is the field zero there?

Secondly, what is the net charge enclosed by the Gaussian Surface that you have chosen? Do you know that on a conductor (in electrostatic equilibrium) charge can reside only on the surface and not in the bulk? Can you give an argument using Gauss's Law to justify this given that the field in the bulk of a conductor has to be zero in electrostatics?