Capacitor #1 has parallel plates of length L and width W,plate separation d,and is filled with dielectric with constant κ.Capacitor #2 has parallel plates of length 2L and width 2W,plate separation 2d,and is filled with dielectric with constant 2κ.If a voltage V is applied to capacitor #1 and a voltage 2V is applied to capacitor #2,which capacitor stores more energy and by what factor?
A)Capacitor #2 by a factor of 8
B)Capacitor #2 by a factor of 16
C)Capacitor #1 by a factor of 4
D)Both capacitors store the same energy.

Question

Quizplus · Accepted Answer

The energy stored in a capacitor is given by $ U = \frac{1}{2} C V^2 $. The capacitance $ C $ for a parallel plate capacitor is $ C = \frac{\kappa \epsilon_0 A}{d} $, where $ A $ is the area of the plates. For capacitor #1, $ C_1 = \frac{\kappa \epsilon_0 LW}{d} $ and for capacitor #2, $ C_2 = \frac{2\kappa \epsilon_0 (2L)(2W)}{2d} = \frac{8\kappa \epsilon_0 LW}{2d} = \frac{4\kappa \epsilon_0 LW}{d} $. Thus, $ C_2 = 4C_1 $. The energy for capacitor #1 is $ U_1 = \frac{1}{2} C_1 V^2 $ and for capacitor #2 is $ U_2 = \frac{1}{2} C_2 (2V)^2 = \frac{1}{2} \times 4C_1 \times 4V^2 = 8 \times \frac{1}{2} C_1 V^2 = 8U_1 $. Therefore, capacitor #2 stores 8 times more energy.

Capacitor #1 Has Parallel Plates of Length L and Width