An electric dipole with dipole moment \( 3\times10^{-9}\,\mathrm{C\,m} \) is aligned at \( 30^\circ \) with the direction of uniform electric field of magnitude \( 6\times10^{4}\,\mathrm{N\,C^{-1}} \). What will be the magnitude of the torque acting on the dipole?
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \( 9\times10^{-5}\,\mathrm{N\,m} \)
In a uniform field the forces on the two charges of a dipole are equal and opposite, so they form a couple. The torque on a dipole is \( \vec{\tau} = \vec{p}\times\vec{E} \), of magnitude
\( \displaystyle \tau = pE\sin\theta \)
\( \displaystyle \tau = 3\times10^{-9}\,\mathrm{C\,m} \times 6\times10^{4}\,\mathrm{N\,C^{-1}} \times \sin 30^\circ \)
\( \displaystyle \tau = 1.8\times10^{-4} \times 0.5\,\mathrm{N\,m} = 9\times10^{-5}\,\mathrm{N\,m} \)
Note: NTA picked this question from an NCERT (2026-27) exercise and changed the figures. Do not ignore NCERT while practising numericals.
NCERT (2026-27):
Exercises – …
