Question

Energy per unit volume for a capacitor having area $A$ and separation $d$ kept at potential difference $V$ is given by

• A. $\frac{1}{2}\frac{{ϵ}_0{V}^2}{d^2}$
• B. $\frac{{ϵ}_0{V}^2}{d^2}$
• C. $\frac{1}{2}C{V}^2$
• D. $\frac{Q^2}{2C}$

Answer 1

The correct option is A $\frac{1}{2}\frac{{ϵ}_0{V}^2}{d^2}$

Total energy stored in a capacitor is given by

$E=\frac{1}{2}C{V}^2=\frac{1}{2}(\frac{{ϵ}_{0}A}{d})V^2$

Since volume between plate is $(A\times d)$

Hence, energy per unit volume will be

$\frac{E}{Ad}=\frac{\frac{1}{2}(\frac{{ϵ}_{0}A}{d})V^2}{Ad}=\frac{1}{2}\frac{{ϵ}_0{V}^2}{d^2}$

Hence, option $\left(a\right)$ is correct.

Key concept- Energy per unit volume is defined as energy density.