A pair of parallel plates,forming a capacitor,are charged.The plates are pulled apart to triple the original separation,the charges on the plates remaining the same.What is the ratio of the final energy stored to the original energy stored?
A) 9
B) 3
C) 1
D) 1/3
E) 1/9

Question

A pair of parallel plates,forming a capacitor,are charged.The plates are pulled apart to triple the original separation,the charges on the plates remaining the same.What is the ratio of the final energy stored to the original energy stored?
A) 9
B) 3
C) 1
D) 1/3
E) ​1/9

Quizplus · Accepted Answer

The energy stored in a capacitor is given by $U=\frac{1}{2}CV^2$, where C is the capacitance and V is the potential difference across the plates. 
Since the charge remains constant and the separation is tripled, the capacitance will decrease by a factor of three. However, the potential difference will also increase by a factor of three (since the charges are the same and the electric field is inversely proportional to the distance between the plates).
Thus, the final energy stored will be:
$U'=\frac{1}{2}(C/3)(3V)^2=\frac{1}{2}CV^2	imes 3 = 3U$
The ratio of final energy to original energy is then:
$U'/U=3/1=3$

A Pair of Parallel Plates,forming a Capacitor,are Charged