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JEE Advanced Physics Electrostatics 2026 JEE Advanced 2026 (Paper 2)

JEE Advanced Physics Question (2026) — Solution

Question

Consider an electric dipole comprising two charges +q and -q each with mass m, separated by a fixed distance d and initially at rest with its dipole moment pointing along i . A uniform electric field E j is turned on at time t = 0 and it is turned off at t = t_f, when the dipole moment makes an angle _f with i . Neglecting any sources of energy loss, correct option(s) is/are:

Options

  1. A. The center of mass of the dipole is deflected towards j in the presence of the field.
  2. B. If the magnitude of the final angular velocity _f = 2qE md , then _f = 6 .
  3. C. If _f = /3, then the change in kinetic energy of the dipole is given by 2 3 \, qEd.
  4. D. For _f = /4, the dipole rotates around its center of mass with a constant angular velocity after t > t_f.

Answer

D. For _f = /4, the dipole rotates around its center of mass with a constant angular velocity after t > t_f.

Step-by-step solution

The net force on the dipole in a uniform electric field is zero, as the forces on the two charges are equal and opposite (qE j and -qE j ). Therefore, the center of mass remains at rest and is not deflected. Option (A) is incorrect. The moment of inertia of the dipole about its center of mass is I = m ( d 2 )^2 + m ( d 2 )^2 = md^2 2 . The torque on the dipole is = p E . The angle between the dipole moment p and the electric field E is 90^ - . The magnitude of the torque is = pE (90^ - ) = qdE . The work done by the electric field as the dipole rotates from = 0 to = _f is equal to the change in kinetic energy: K = _ 0 ^ _f d = _ 0 ^ _f qdE d = qdE _f Equating this to the final kinetic energy 1 2 I _f^2: 1 2 ( md^2 2 ) _f^2 = qdE _f md^2 4 _f^2 = qdE _f For option (B), substituting _f = 2qE md : md^2 4 ( 2qE md ) = qdE _f qdE 2 = qdE _f _f = 1 2 _f = 6 Thus, option (B) is correct. For option (C), if _f = 3 , the change in kinetic energy is K = qdE ( 3 ) = 3 2 qEd. Thus, option (C) is incorrect. For option (D), after t > t_f, the electric field is turned off. The net torque on the dipole becomes zero. By Newton's first law of rotational motion, the dipole will continue to rotate about its center of mass with a constant angular velocity. Thus, option (D) is correct. Answer: If the magnitude of the final angular velocity _f = 2qE md , then _f = 6 .; For _f = /4, the dipole rotates around its center of mass with a constant angular velocity after t > t_f.

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