JEE Physics 2025 Question with Solution

Given: Charges are $-4 \, \mu C$ and $4 \, \mu C$. Their positions are $(-9,0,0) \, cm$ and $(9,0,0) \, cm$, so the separation is $18 \, cm = 0.18 \, m$. The uniform electric field is $E = 10^4 \, N/C$.

Find: The work done in rotating the dipole from equilibrium to $180^\circ$.

For a dipole in a uniform electric field, potential energy is

$$U = -pE\cos\theta$$

and the work done in rotating it from $\theta_1$ to $\theta_2$ is

$$W = U(\theta_2) - U(\theta_1)$$

First, calculate the dipole moment:

$$p = qd = 4 \times 10^{-6} \times 0.18 = 7.2 \times 10^{-7} \, C\,m$$

At equilibrium, $\theta_1 = 0^\circ$, so

$$U(0^\circ) = -pE\cos 0^\circ = -(7.2 \times 10^{-7})(10^4) = -7.2 \times 10^{-3} \, J$$

After rotation through $180^\circ$, $\theta_2 = 180^\circ$, so

$$U(180^\circ) = -pE\cos 180^\circ = +(7.2 \times 10^{-3}) \, J$$

Therefore,

$$W = U(180^\circ) - U(0^\circ) = 7.2 \times 10^{-3} - (-7.2 \times 10^{-3}) = 14.4 \times 10^{-3} \, J$$

So the work done is $14.4 \, mJ$. The correct option is A.

Given: $p = qd$, $E = 10^4 \, N/C$, and the dipole is rotated from $0^\circ$ to $180^\circ$.

Find: The work done.

Using

$$W = -pE\cos\theta_2 + pE\cos\theta_1 = pE(\cos\theta_1 - \cos\theta_2)$$

with $\theta_1 = 0^\circ$ and $\theta_2 = 180^\circ$,

$$W = pE(1 - (-1)) = 2pE$$

Now,

$$p = qd = 4 \times 10^{-6} \times 0.18 = 7.2 \times 10^{-7} \, C\,m$$

Hence,

$$W = 2 \times 7.2 \times 10^{-7} \times 10^4 = 14.4 \times 10^{-3} \, J = 14.4 \, mJ$$

This works because rotating from stable equilibrium to the opposite direction changes $\cos\theta$ from $1$ to $-1$, doubling the factor $pE$. Therefore, the correct option is A.