# Understanding Gauss' Law on Charged Sheets and Oppositely Charged Plates

I'm grappling with understanding Gauss' Law as applied to charged sheets and oppositely charged plates.

From what I've gathered, when using a Gaussian pillbox encompassing both sides of an infinite sheet, we can derive the electric field to be $$2EA=\frac{\sigma A}{\epsilon_0}$$​, which simplifies to $$E=\frac{\sigma}{2\epsilon_0}$$.

Now, if I consider two oppositely charged plates, I understand that the resultant field outside both plates would be zero. However, I'm uncertain about the field within the plates.

Could it be conceptualized by taking a Gaussian surface only encompassing half the height of each plate and, consequently, accounting for only half the charge on each plate? Would this approach result in $$AE=\frac{(\sigma/2)A}{\epsilon_0}$$​, and thus $$E=\frac{\sigma}{2\epsilon_0}$$​ for each plate? Consequently, would the electric field inside both plates then become $$E=\frac{\sigma}{\epsilon_0}$$?

And what about if my plates are perfectly conducting? Then the charge at the top of a plate and bottom could be different right? This would also affect the resulting outcome if a charge is induced on them?

I'm seeking clarification on this interpretation of Gauss' Law in this scenario. Any insights or corrections would be greatly appreciated.