Q. An electric dipole of length 1 cm is placed with the axis making an angle of 30° to an electric field of strength$10^{4}NC^{- 1}$. If it experiences a torque of $10\sqrt{2}$Nm, the potential energy of the dipole is
A) 0.245J
B) $2.45 \times 10^{- 4}J$
C) 0.0245 J
D) $24.5 \times 10^{- 4}J$
Explanation
(d) Torque,$\ \ \ \tau = pE\sin\theta$
$or\ \ \ \ \ \ \ \ \ \ \ \ \ p = \frac{\tau}{E\sin\theta}$
Potential energy, $U = pE\ cos\ \theta$
$= \frac{\tau}{E\sin\theta}.E\cos\theta$
$= \frac{\tau}{E\sin\theta} = \frac{10\sqrt{2}}{10^{4}\tan{30{^\circ}}}$
$= 24.5 \times 10^{- 4}J$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.