JEE Physics 2025 Question with Solution

Given: Dipole moment $p = 6 \times 10^{-6} \, \text{C m}$ and electric field $E = 10^6 \, \text{V/m}$. Initially, the dipole is parallel to the field, so $\theta_i = 0^\circ$. Finally, it is opposite to the field, so $\theta_f = 180^\circ$.

Find: The work required to rotate the dipole from parallel to opposite direction.

The potential energy of an electric dipole in a uniform electric field is

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

Initially,

$$U_i = -pE\cos 0^\circ = -pE$$

Finally,

$$U_f = -pE\cos 180^\circ = +pE$$

Therefore, the work done on the dipole is the increase in potential energy:

$$W = U_f - U_i = pE - (-pE) = 2pE$$

Substituting the given values,

$$W = 2 \times (6 \times 10^{-6}) \times (10^6)$$ $$W = 12 \, \text{J}$$

The first the solution also writes $2pE$, which gives $12 \, \text{J}$. Although the page shows inconsistent final numbers elsewhere, the working clearly supports $12 \, \text{J}$.

Therefore, the required work is $12 \, \text{J}$.

Given: $p = 6 \times 10^{-6} \, \text{C m}$, $E = 10^6 \, \text{V/m}$, initial angle $0^\circ$, final angle $180^\circ$.

Find: Work done in rotating the dipole through $180^\circ$ in the uniform field.

A dipole in an electric field has potential energy depending only on orientation. Hence the required work depends only on the change in potential energy, not on the path taken.

At $\theta = 0^\circ$,

$$U_1 = -pE\cos 0^\circ = -pE$$

At $\theta = 180^\circ$,

$$U_2 = -pE\cos 180^\circ = -pE(-1) = +pE$$

So,

$$\Delta U = U_2 - U_1 = pE - (-pE) = 2pE$$

Now substitute:

$$\Delta U = 2 \times 6 \times 10^{-6} \times 10^6$$ $$\Delta U = 12 \, \text{J}$$

Hence, the work that must be done on the dipole is $12 \, \text{J}$. The numerical values $6 \times 10^{-3}$ and $6$ shown on the page are inconsistent with the displayed formula and substitution.