Question

A parallel-plate capacitor with an air gap has magnitude of charge $Q$ applied on each plate. The plates have area $A$ and are separated by a distance dd. (The distance $d$ is very small compared to the side lengths of the plates.) The capacitor is then isolated from the voltage source.
How much work will be required by an outside force to separate the plates to a distance $3d$?

• A. $\frac{Q^2}{4\pi {ϵ}_{0}d}$
• B. $\frac{Q^2}{2\pi {ϵ}_{0}d}$
• C. $\frac{Q^{2}d}{2{ϵ}_{0}A}$
• D. $\frac{Q^{2}d}{{ϵ}_{0}A}$
• E. $\frac{3Q^{2}d}{{ϵ}_{0}A}$

Answer 1

Answer: (D)

$\frac{Q^{2}d}{{ϵ}_{0}A}$

Potential energy stored in the capacitor charged to $Q$ at a separation $d$ of plates=$U_1=\frac{Q^2}{2C}$

where $C=\frac{A{ϵ}_0}{d}$

Hence $U_1=\frac{Q^{2}d}{2A{ϵ}_0}$

When the capacitor is isolated, the charge on the capacitor remains the same.

$C^{\prime }=\frac{A{ϵ}_0}{3d}$

Thus $U_2=\frac{3Q^{2}d}{2A{ϵ}_0}$

Hence work required by outside force=$U_2-U_1=\frac{Q^{2}d}{{ϵ}_{0}A}$

Correct answer is option D.