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 $\hat{\imath}$. A uniform electric field $E \hat{j}$ is turned on at time $t=0$ and it is turned off at $t=t_{f}$, when the dipole moment makes an angle $\theta_{f}$ with $\hat{i}$. Neglecting any sources of energy loss, correct option(s) is/are:

(B) If the magnitude of the final angular velocity $\omega_{f}=\sqrt{\frac{2 q E}{m d}}$ then $\theta_{f}=\frac{\pi}{6}$; (D) For $\theta_{f}=\pi / 4$, the dipole rotates around its center of mass with a constant angular velocity after $t>t_{f}$.

(A) The center of mass of the dipole is deflected towards $\hat{j}$ in the presence of the field.

(C) If $\theta_{f}=\pi / 4$, then the change in kinetic energy of the dipole is given by $2 \sqrt{3} q E d$.

JEE Advanced · 2026 · 17 May 2026 · Paper 2 · Electrostatics

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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

\overset{}{^}

. A uniform electric field

E \hat{j}

is turned on at time

t = 0

and it is turned off at

t = t_{f}

, when the dipole moment makes an angle

θ_{f}

with

\hat{i}

. Neglecting any sources of energy loss, correct option(s) is/are:

(A)The center of mass of the dipole is deflected towards $\hat{j}$ in the presence of the field.
(B)If the magnitude of the final angular velocity $ω_{f} = \frac{2 q E}{m d}$ then $θ_{f} = \frac{π}{6}$ ✓ Correct
(C)If $θ_{f} = π /4$ , then the change in kinetic energy of the dipole is given by $23 q E d$ .
(D)For $θ_{f} = π /4$ , the dipole rotates around its center of mass with a constant angular velocity after $t > t_{f}$ .✓ Correct

Answer: (B), (D)

Solution

(A) Net force on center of mass:

F_{net} = q E - q E = 0

. The center of mass is not deflected. Incorrect.

Torque:

τ = p \times E = q E d cos θ \hat{k}

. Moment of inertia:

I = 2 m (d /2)^{2} = \frac{m d ^{2}}{2}

.

Change in kinetic energy:

Δ K E = q E d sin θ_{f}

, so

\frac{1}{2} I ω_{f}^{2} = q E d sin θ_{f}

.

(B) If

ω_{f} = \frac{2 q E}{m d}

, then

\frac{1}{2} \cdot \frac{m d ^{2}}{2} \cdot \frac{2 q E}{m d} = q E d sin θ_{f}

, giving

sin θ_{f} = \frac{1}{2}

, so

θ_{f} = \frac{π}{6}

. Correct.

(C) If

θ_{f} = \frac{π}{4}

, then

Δ K E = q E d sin (π /4) = \frac{2}{2} q E d

, not

23 q E d

. Incorrect.

(D) For

t > t_{f}

, the electric field is off, so net torque is zero and angular velocity remains constant. Correct.

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JEE Advanced · 2026 · 17 May 2026 · Paper 2 · Electrostatics

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