Quantrex Quantrex AcademyJEE · NEET · NDA PYQs with solutions Open app
JEE Advanced 2026Paper 2
Exam
JEE Advanced
Year
2026
Paper
Paper 2
Subject
Physics
Chapter
Electrostatics

JEE Advanced 2026 · Paper 2 · Physics · Electrostatics

Consider an electric dipole comprising two charges +q and -q each...

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

Options

  1. AThe center of mass of the dipole is deflected towards j^\hat{j} in the presence of the field.
  2. BIf the magnitude of the final angular velocity ωf=2qEmd\omega_f = \sqrt{\dfrac{2qE}{md}}, then θf=π6\theta_f = \dfrac{\pi}{6}.
  3. CIf θf=π/3\theta_f = \pi/3, then the change in kinetic energy of the dipole is given by 23 qEd2\sqrt{3}\, qEd.
  4. DFor θf=π/4\theta_f = \pi/4, the dipole rotates around its center of mass with a constant angular velocity after t>tft > t_f.

Correct answer

B. If the magnitude of the final angular velocity ωf=2qEmd\omega_f = \sqrt{\dfrac{2qE}{md}}, then θf=π6\theta_f = \dfrac{\pi}{6}.

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 (qEj^qE\hat{j} and −qEj^-qE\hat{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(d2)2+m(d2)2=md22I = m\left(\dfrac{d}{2}\right)^2 + m\left(\dfrac{d}{2}\right)^2 = \dfrac{md^2}{2}.

The torque on the dipole is τ⃗=p⃗×E⃗\vec{\tau} = \vec{p} \times \vec{E}. The angle between the dipole moment p⃗\vec{p} and the electric field E⃗\vec{E} is 90∘−θ90^{\circ} - \theta. The magnitude of the torque is τ=pEsin⁡(90∘−θ)=qdEcos⁡θ\tau = pE \sin(90^{\circ} - \theta) = qdE \cos\theta.

The work done by the electric field as the dipole rotates from θ=0\theta = 0 to θ=θf\theta = \theta_f is equal to the change in kinetic energy:

ΔK=∫0θfτdθ=∫0θfqdEcos⁡θdθ=qdEsin⁡θf\Delta K = \int_{0}^{\theta_f} \tau d\theta = \int_{0}^{\theta_f} qdE \cos\theta d\theta = qdE \sin\theta_f

Equating this to the final kinetic energy 12Iωf2\dfrac{1}{2}I\omega_f^2:

12(md22)ωf2=qdEsin⁡θf\dfrac{1}{2} \left(\dfrac{md^2}{2}\right) \omega_f^2 = qdE \sin\theta_f

md24ωf2=qdEsin⁡θf\dfrac{md^2}{4} \omega_f^2 = qdE \sin\theta_f

For option (B), substituting ωf=2qEmd\omega_f = \sqrt{\dfrac{2qE}{md}}:

md24(2qEmd)=qdEsin⁡θf\dfrac{md^2}{4} \left(\dfrac{2qE}{md}\right) = qdE \sin\theta_f

qdE2=qdEsin⁡θf⇒sin⁡θf=12⇒θf=π6\dfrac{qdE}{2} = qdE \sin\theta_f \Rightarrow \sin\theta_f = \dfrac{1}{2} \Rightarrow \theta_f = \dfrac{\pi}{6}

Thus, option (B) is correct.

For option (C), if θf=π3\theta_f = \dfrac{\pi}{3}, the change in kinetic energy is ΔK=qdEsin⁡(π3)=32qEd\Delta K = qdE \sin\left(\dfrac{\pi}{3}\right) = \dfrac{\sqrt{3}}{2} qEd. Thus, option (C) is incorrect.

For option (D), after t>tft > 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 ωf=2qEmd\omega_f = \sqrt{\dfrac{2qE}{md}}, then θf=π6\theta_f = \dfrac{\pi}{6}.; For θf=π/4\theta_f = \pi/4, the dipole rotates around its center of mass with a constant angular velocity after t>tft > t_f.
Practice Electrostatics on Quantrex Academy →

More from Electrostatics

Two charges Q_1 = q and Q_2 = mq are placed at the points P_1(a, b) and P_2(ma, mb)… JEE Advanced 2026Two point charges q_1=3\,μ C and q_2=-4\,μ C are placed at points (2 i +3 j +3 k ) and (… JEE Main 2026The electric potential as a function of x, y is given by V = 5(x^2 - y^2) V. The electric… JEE Main 2026A thin half ring of radius 35 cm is uniformly charged with a total charge of Q coulomb.… JEE Main 2026A three coulomb charge moves from the point (0, -2, -5) to the point (5, 1, 2) in an… JEE Main 2026A particle of charge q and mass m is projected from origin with an initial velocity v = (… JEE Main 2026A rigid dipole undergoes a simple harmonic motion about its centre in the presence of an… JEE Main 2026Given below are two statements: one is labelled as Assertion A and the other is labelled… JEE Main 2026 Full Electrostatics list All JEE Advanced PYQs