If the area of the plates of a parallel plate capacitor is halved and the separation between the plates tripled,while the charge on the capacitor remains constant,then by what factor does the energy stored in the capacitor change?
A)increase by a factor of 2
B)decrease by a factor of 2/3
C)increase by a factor of 6
D)increase by a factor of 3/2
E)decrease by a factor of 1/6

Question

Quizplus · Accepted Answer

The energy stored in a parallel plate capacitor is given by the formula $U = \frac{1}{2} \frac{Q^2}{C}$, where $U$ is the energy, $Q$ is the charge, and $C$ is the capacitance. The capacitance of a parallel plate capacitor is given by $C = \frac{\varepsilon_0 A}{d}$, where $A$ is the area of the plates, $d$ is the separation between the plates, and $\varepsilon_0$ is the permittivity of free space. If the area of the plates is halved and the separation is tripled, the new capacitance $C'$ becomes $C' = \frac{\varepsilon_0 (A/2)}{3d} = \frac{1}{6}C$. Substituting the new capacitance into the energy formula gives $U' = \frac{1}{2} \frac{Q^2}{C'} = \frac{1}{2} \frac{Q^2}{\frac{1}{6}C} = 3 \times \frac{1}{2} \frac{Q^2}{C} = 3U$. Therefore, the energy stored in the capacitor increases by a factor of 6.

If the Area of the Plates of a Parallel Plate