JEE Physics 2025 Question with Solution

Given: A parallel plate capacitor has area $A$, plate separation $d$, electric field $E$, and permittivity $\epsilon_0$.

Find: The potential energy stored in the capacitor.

For a capacitor, the stored energy is

$$U = \frac{1}{2}CV^2$$

For a parallel plate capacitor,

$$C = \frac{\epsilon_0 A}{d}$$

Also, the electric field and potential difference are related by

$$E = \frac{V}{d}$$

so

$$V = Ed$$

Substituting into the energy formula,

$$U = \frac{1}{2}C(Ed)^2$$

Now substitute $C = \frac{\epsilon_0 A}{d}$:

$$U = \frac{1}{2}\left(\frac{\epsilon_0 A}{d}\right)(Ed)^2$$

Simplifying,

$$U = \frac{1}{2} \epsilon_0 E^2 A d$$

Therefore, the potential energy stored in the capacitor is $\frac{1}{2} \epsilon_0 E^2 A d$, so the correct option is A.

Given: $C = \frac{\epsilon_0 A}{d}$ and $V = Ed$ for a parallel plate capacitor.

Find: Stored energy $U$.

Use

$$U = \frac{1}{2}CV^2$$

Substitute both relations directly:

$$U = \frac{1}{2}\left(\frac{\epsilon_0 A}{d}\right)(Ed)^2$$

This reduces to

$$U = \frac{1}{2} \epsilon_0 E^2 A d$$

The shortcut works because the question already gives $E$, so expressing $V$ as $Ed$ immediately converts the standard energy formula into the required form. Hence, the correct option is A.