Potential due to two infinite parallel planes

[Lisa...]

Consider two infinite parallel planes of charge, one in the yz plane and the other at distance x = a. (a) Find
the potential everywhere in space when V = 0 at x = 0 if the planes carry equal positive charge densities + σ. (b)
Repeat the problem with charge densities equal and opposite, and the charge in the yz plane positive.

I haven't had any problems with part a. As for part b, E=0 if x > a (the field lines of both planes cancel) , so I figured the potential needs to be 0 too, but the answer should be V= $$\frac{- \sigma a}{\epsilon_0}.$$ Why is that?

EDIT: I just figured it out by myself :)

 

[Nate810]

The potential difference between the planes is V = \frac{- \sigma a}{\epsilon_0}, so the potential at x=0 must be - \frac{ \sigma a}{\epsilon_0}.

 

[kyle8921]

The potential is not affected by the field outside the two planes, so the potential at x > a is just the potential due to the charge on the yz plane, which is V = \frac{\sigma a}{\epsilon_0}. However, at x < a, the potential is affected by both planes and is given by V = \frac{\sigma a}{\epsilon_0} + \frac{\sigma a}{\epsilon_0} = \frac{2 \sigma a}{\epsilon_0}, which simplifies to V = \frac{- \sigma a}{\epsilon_0}. This is because the charge on the yz plane creates a potential that is opposite in sign to the potential created by the charge on the other plane, resulting in a cancellation at x < a. Therefore, the potential at x < a is equal to the potential at x > a, but with a negative sign. This is why the final answer is V = \frac{- \sigma a}{\epsilon_0}.