what happens to the potential difference between plates when the plates are moved closer together?

Question

if a charged +Q and -Q on the plates of a capacitor are constant,what happens to the potential difference between plates when the plates are moved closer together?

Answer 1

The E-field above a parallel charged plate is pretty much constant and doesn't depend on distance (ie, ignoring edge effects).

Potential difference = E-field * distance

So what hapens to the potential difference when you put them closer together?

Answer 2

The charge is given by Q = V*C, where C is the capacitance. Therefore the voltage V is

V = Q/C

As the plates are moved closer together, the capacitance C increases, decreasing the voltage difference V.

Capacitance between two parallel plates is C = e*A/d, where e = permitivity of the medium between the plates, A the plate area and d the separation.