# Gauss's Law

1. Jul 17, 2006

### 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?