The expression for torque '\( \vec{\tau} \)' experienced by an electric dipole of dipole moment '\( \vec{P} \)' in an external uniform electric field '\( \vec{E} \)' is given by :
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \( \vec{\tau} = \vec{P}\times\vec{E} \)
In a uniform field the forces \( +q\vec{E} \) and \( -q\vec{E} \) on the two charges of the dipole are equal and opposite, so the net force is zero, but they form a couple. The torque of the couple has magnitude
\( \displaystyle \tau = qE \times 2a\sin\theta = pE\sin\theta \)
and it acts to turn \( \vec{p} \) towards \( \vec{E} \). A vector of magnitude \( pE\sin\theta \) perpendicular to both \( \vec{p} \) and \( \vec{E} \) is their cross product:
\( \displaystyle \vec{\tau} = \vec{p}\times\vec{E} \)
Option (1) \( \vec{\tau} = \vec{P}\cdot\vec{E} \) — Incorrect. A dot product is a scalar; \( \vec{p}\cdot\vec{E} \) appears in the potential energy, \( U = -\vec{p}\cdot\vec{E} \), not in the torque.
Option (2) \( \vec{\tau} = \displaystyle\frac{\vec{P}}{\vec{E}} \) — Incorrect. Division by a vector is not defined, and the torque must grow with \( E \), not fall.
Option (3) \( \vec{\tau} = \displaystyle\frac{\vec{E}}{\vec{P}} \) — Incorrect. Division by a vector is not defined, and the torque must grow with \( p \).
Option (4) \( \vec{\tau} = \vec{P}\times\vec{E} \) — Correct. \( \vec{\tau} = \vec{p}\times\vec{E} \), magnitude \( pE\sin\theta \).
NCERT (2026-27):
Dipole in a Uniform External Field – “… The net force on the dipole is zero, since E is uniform. However, the charges are separated, so the forces act at different points, resulting in a torque on the dipole. … The magnitude of \( \mathbf{p} \times \mathbf{E} \) is also pE sinθ and its direction is normal to the paper, coming out of it. Thus,
\( \displaystyle \boldsymbol{\tau} = \mathbf{p} \times \mathbf{E} \qquad (1.22) \)”