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 $d$. (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: (A), (D)

The correct options are
A $\frac{Q^2}{4\pi {ϵ}_{0}d}$
D $\frac{Q^{2}d}{{ϵ}_{0}A}$

Energy stored in capacitor before the charge=$\frac{Q^2}{2C}$

where $C=\frac{A{ϵ}_0}{d}$

After the plates are moved to distance $3d$ after isolation, the charge on the plates remains same with new capacitance being,

$C^{\prime }=\frac{A{ϵ}_0}{3d}=\frac{C}{3}$

Energy stored in capacitor in this configuration $=\frac{Q^2}{2C^{\prime }}=\frac{3Q^2}{2C}$

Thus energy required to separate the plates $=\frac{3Q^2}{2C}-\frac{Q^2}{2C}$$=\frac{Q^2}{C}=\frac{Q^{2}d}{{ϵ}_{0}A}$