A pair of parallel plates,forming a capacitor,are charged.The plates are pulled apart to double 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)4
B)2
C)1
D)1/2

Question

Quizplus · Accepted Answer

The energy stored in a capacitor is given by $ U = \frac{1}{2} \frac{Q^2}{C} $. When the separation is doubled, the capacitance $ C $ is halved (since $ C \propto \frac{1}{d} $). Therefore, the energy becomes $ U' = \frac{1}{2} \frac{Q^2}{C/2} = 2U $. Thus, the ratio of the final energy to the original energy is 2.

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