Question

You have a parallel plate capacitor with charge q on its plates. A battery, connected to the capacitor, keeps constant the potential difference across the plates. You decide to pull apart the two plates to twice their original distance. Determine the new amount of charge on the plates.

	A,	2q
	B,	q/2
	C,	4q
D,		q

Answer 1

As we know that Q= CV

C = epsilon A/ d

if V= constant then

q = V (epsilon A/ d)

in second case distance between plate = 2d

so C' = epsilon A/ 2d

so new charge = Q = V (epsilon A/ 2d)

Q= q/2

answer B