Question

An electric dipole of dipole moment \( \vec{p} \) is kept in a uniform electric field \( \vec{E} \). The amount of work done to rotate it from the position of stable equilibrium to that of unstable equilibrium will be:

• A. \( 2pE \)
• B. \( -2pE \)
• C. \( pE \)
• D. Zero

Hint:

The work done on a dipole in a uniform electric field depends on the change in potential energy, \( \Delta U = U_{\text{final}} - U_{\text{initial}} \).

Answer 1

The Correct Option is A

Work Done to Rotate the Dipole in an Electric Field

1: Work Done Formula
The work done \( W \) to rotate an electric dipole from an initial angle \( \theta_1 \) to a final angle \( \theta_2 \) in a uniform electric field is given by: \[ W = -\int_{\theta_1}^{\theta_2} \vec{\tau} \cdot d\vec{\theta} \] where:
- \( \vec{\tau} = pE \sin\theta \) is the torque on the dipole,
- \( p \) is the dipole moment,
- \( E \) is the electric field.
2: Work Done to Rotate from Stable to Unstable Equilibrium
For stable equilibrium: \( \theta_1 = 0^\circ \) (where \( \cos \theta = 1 \)),
For unstable equilibrium: \( \theta_2 = 180^\circ \) (where \( \cos \theta = -1 \)). Thus, the work done to rotate the dipole is: \[ W = -\int_{0}^{180^\circ} pE \cos\theta \, d\theta = 2pE \] Thus, the correct answer is: \[ \boxed{(A) \, 2pE} \]