Question

A parallel-plate vacuum capacitor with plate area $A$ and separation $x$ has charges $+Q$ and $-Q$ on its plates. The capacitor is disconnected from the source of charge, so the charge on each plate remains fixed. What is the total energy stored in the capacitor?

• A. $\frac{Q^{2}x}{{ϵ}_{o}A}$
• B. $\frac{Q^{2}x}{2{ϵ}_{o}A}$
• C. $\frac{Q^{2}x}{4{ϵ}_{o}A}$
• D. $0$

Answer 1

The correct option is B $\frac{Q^{2}x}{2{ϵ}_{o}A}$

The capacitance of parallel plate capacitor with vacuum is $C=\frac{A{ϵ}_0}{x}$

As the charges on plates is constant so the energy stored in the capacitor is :

$U=\frac{Q^2}{2C}=\frac{Q^{2}x}{2A{ϵ}_0}$