Suppose the two plates of a parallel-plate capacitor have fixed charges $Q$ and $-Q$. If we double the gap between the plates, the:
-(a) potential difference between the plates,
A) quadruples.
B) doubles.
C) remains the same.
D) is reduced by a factor of two.
E) is reduced by a factor of four.
F) some other result (specify).

Question

Quizplus · Accepted Answer

The potential difference (V) between the plates of a capacitor is directly proportional to the distance (d) between the plates when the charge (Q) on the plates and the permittivity of the medium between the plates are constant. Mathematically, V = Qd/(εA), where ε is the permittivity of the medium and A is the area of the plates. Doubling the distance between the plates doubles the potential difference.

Suppose the Two Plates of a Parallel-Plate Capacitor Have Fixed QQQ

Suppose the Two Plates of a Parallel-Plate Capacitor Have Fixed $Q$