Question

A parallel plate capacitor has plate area $A$, plate separation $d$, and magnitude of charge, $Q$, on each plate. The capacitor is isolated from any voltage source. The plates of this capacitor are then separated to a new distance, $1.5d$.
In terms of the given variables and fundamental constants, how much work was required by an outside force to separate the plates?

• A. $\frac{Q^2}{4C}$
• B. $\frac{Q^2}{2C}$
• C. $\frac{3Q^2}{4C}$
• D. $\frac{Q^2}{C}$
• E. There is not enough information to determine the work required.

Answer 1

Answer: (A)

$\frac{Q^2}{4C}$

Initial capacitance $C_1=\frac{A{ϵ}_0}{d}$

Final capacitance $C_2=\frac{A{ϵ}_0}{1.5d}=\frac{2A{ϵ}_0}{3d}$

Since the capacitor was isolated, the charge stored remains the same.

Thus work done=$\frac{Q^2}{2C_2}-\frac{Q^2}{2C_1}$

$=\frac{Q^2}{4C}$