A parallel plate capacitor with plates of area A and plate separation d is charged so that the potential difference between its plates is V. If the capacitor is then isolated and its plate separation is decreased to d/2, what happens to the potential difference between the plates?

Question

Quizplus · Accepted Answer

When the plate separation is decreased to d/2, the capacitance doubles because capacitance C is inversely proportional to the distance between the plates (C = εA/d). Since the charge Q on the capacitor remains constant (as it is isolated), the potential difference V across the plates is given by V = Q/C. Thus, if the capacitance doubles, the potential difference is halved, resulting in a final potential difference of 0.5V.

A Parallel Plate Capacitor with Plates of Area a and Plate