Two point charges $-4 \mu \mathrm{c}$ and $4 \mu \mathrm{c}$, constituting an electric dipole, are placed at $(-9,0,0) \mathrm{cm}$ and $(9,0,0) \mathrm{cm}$ in a uniform electric field of strength $10^4 \mathrm{NC}^{-1}$. The work done on the dipole in rotating it from the equilibrium through $180^{\circ}$ is :

<p>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:</p> <p>$W = -pE(\cos \theta_2 - \cos \theta_1)$</p> <p>where:</p> <p><p>$ p = q \cdot d $ is the dipole moment,</p></p> <p><p>$ E $ is the electric field strength,</p></p> <p><p>$ \theta_1 $ and $ \theta_2 $ are the initial and final angles between the dipole moment and the electric field.</p></p> <p>Given:</p> <p><p>Charge $ q = 4 \, \mu \mathrm{C} = 4 \times 10^{-6} \, \mathrm{C} $</p></p> <p><p>Distance $ d = 18 \, \mathrm{cm} = 0.18 \, \mathrm{m} $</p></p> <p><p>Electric field $ E = 10^4 \, \mathrm{N/C} $</p></p> <p><p>Initial angle $\theta_1 = 0^\circ$ (aligned with the field, equilibrium position)</p></p> <p><p>Final angle $\theta_2 = 180^\circ$</p></p> <p>First, calculate the dipole moment $ p $:</p> <p>$$ p = q \cdot d = 4 \times 10^{-6} \, \mathrm{C} \cdot 0.18 \, \mathrm{m} = 7.2 \times 10^{-7} \, \mathrm{Cm} $$</p> <p>Calculate the work done $ W $:</p> <p>$W = -pE(\cos 180^\circ - \cos 0^\circ)$</p> <p>$W = -7.2 \times 10^{-7} \, \mathrm{Cm} \cdot 10^4 \, \mathrm{N/C} \cdot (-1 - 1)$</p> <p>$W = 7.2 \times 10^{-7} \, \mathrm{Cm} \cdot 10^4 \, \mathrm{N/C} \cdot 2$</p> <p>$W = 1.44 \times 10^{-2} \, \mathrm{J} = 14.4 \, \mathrm{mJ}$</p> <p>Therefore, the work done on the dipole in rotating it from the equilibrium position through $180^\circ$ is <strong>14.4 mJ</strong> (Option D).</p>