JEE Main 2025 — Electrostatics Question with Solution

$$u=\frac{1}{2}{\epsilon }_0{E}^2$$

The energy density of an electric field in free space is given by the expression above. For a parallel plate capacitor with plate area $$A$$ and separation $$d$$, the volume between the plates is

$$V=Ad.$$

Multiplying the energy density by the volume gives the total energy stored:

$$U=u\times V=\frac{1}{2}{\epsilon }_0{E}^2\times Ad=\frac{1}{2}{\epsilon }_0{E}^{2}Ad.$$

Thus, the potential energy stored in the capacitor is

$$\frac{1}{2}{\epsilon }_0{E}^{2}Ad.$$

This corresponds to the option:

$$\text{Option D: }\frac{1}{2}{\epsilon }_0{E}^{2}Ad.$$