An electric dipole of dipole moment \(2\times10^{-8}\ \mathrm{C\,m}\) makes an angle of \(60^\circ\) with a uniform electric field of \(3\times10^{4}\ \mathrm{N\,C^{-1}}\). The work needed to rotate it slowly until it points opposite to the field is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(9\times10^{-4}\ \mathrm{J}\)
The work needed to turn a dipole slowly from angle \( \theta_0 \) to \( \theta_1 \) is \( W=pE(\cos\theta_0-\cos\theta_1) \). Here \( \theta_0=60^\circ \) and \( \theta_1=180^\circ \).
\( \begin{aligned} W &= pE(\cos60^\circ-\cos180^\circ) \\ &= 2\times10^{-8}\times3\times10^{4}\times(0.5+1) \\ &= 9\times10^{-4}\ \mathrm{J} \end{aligned} \)
NCERT (2026-27):
Potential energy of a dipole in an external field – “… Suppose an external torque τext is applied in such a manner that it just neutralises this torque and rotates it in the plane of paper from angle θ0 to angle θ1 at an infinitesimal angular speed and without angular acceleration. The amount of work done by the external torque will be given by”
\( W=pE(\cos\theta_0-\cos\theta_1) \)\( \qquad (2.31) \)