QJEEM25Phcc17: JEE Main 2025 Physics - Class 12 JEEM25Phcc17 Solution: Two point charges and , constituting an electric dipole, are…

The work done on an electric dipole in a uniform electric field when it is rotated from an angle $\theta_1$ to an angle $\theta_2$ is given by:

$$W=-pE(\cos {\theta }_2-\cos {\theta }_1)$$

where:

$p = q \cdot d$ is the dipole moment,

$E$ is the electric field strength,

$\theta_1$ and $\theta_2$ are the initial and final angles between the dipole moment and the electric field.

Given:

Charge $q = 4 \, \mu \mathrm{C} = 4 \times 10^{-6} \, \mathrm{C}$

Distance $d = 18 \, \mathrm{cm} = 0.18 \, \mathrm{m}$

Electric field $E = 10^4 \, \mathrm{N/C}$

Initial angle $\theta_1 = 0^\circ$ (aligned with the field, equilibrium position)

Final angle $\theta_2 = 180^\circ$

First, calculate the dipole moment $p$ :

$$p=q\cdot d=4\times {10}^{-6}\mathrm{C}\cdot 0.18\mathrm{m}=7.2\times {10}^{-7}\mathrm{Cm}$$

Calculate the work done $W$ :

$$W=-pE(\cos {180}^{\circ }-\cos 0^{\circ })$$

$$W=-7.2\times {10}^{-7}\mathrm{Cm}\cdot {10}^4\mathrm{N}/\mathrm{C}\cdot (-1-1)$$

$$W=7.2\times {10}^{-7}\mathrm{Cm}\cdot {10}^4\mathrm{N}/\mathrm{C}\cdot 2$$

$$W=1.44\times {10}^{-2}\mathrm{J}=14.4\mathrm{mJ}$$

Therefore, the work done on the dipole in rotating it from the equilibrium position through $180^\circ$ is 14.4 mJ (Option D).