JEE Advanced 2026 Electric Dipole Question: Full Solution & Correct Answers

eSaral › JEE › JEE Main ›JEE Advanced 2026 Electric Dipole Question

🚀 Checkout eSaral Courses

What Is the JEE Advanced 2026 Electric Dipole Question? 

This multiple-choice question from JEE Advanced 2026 is based on the behaviour of an electric dipole placed in a uniform electric field. It tests four chapters simultaneously — a hallmark of JEE Advanced's design philosophy.

Here is the exact setup:

The question asks which of the four options (A, B, C, D) is correct.

To see how this question fits into the complete exam, refer to the JEE Advanced exam paper and explore the full paper from which it was taken.

Key Concepts Involved in This Question 

This single question requires mastery of four major chapters:

1. Electrostatics — Electric Dipole in a Field

When an electric dipole of moment p is placed in a uniform electric field E, it experiences:

• Torque: τ = p × E (magnitude = pE sinθ)
• Potential Energy: U = −p · E = −pE cosθ
• Net Force = 0 (in a uniform field)

2. Work, Power & Energy — Energy Conservation

To find the angular velocity ω at any angle, we apply conservation of mechanical energy:

Initial KE + Initial PE = Final KE + Final PE

3. Rotational Mechanics — Moment of Inertia

The dipole rotates about its centre of mass. Since both charges have equal mass m and each is at a distance d/2 from the centre:

I = 2 × m(d/2)² = md²/2

4. Centre of Mass — System Motion

In a uniform electric field, the net force on the dipole system is zero (+qE and −qE cancel). Therefore, the centre of mass does not move — it only rotates about its own position.

How to Read a JEE Advanced Question the Right Way

Reading a JEE Advanced question is itself a skill. Here is a structured approach used in eSaral's live sessions:

Step 1 — Read up to the physical setup, then stop. After reading "dipole of +q, −q, mass m, separation d, initially at rest, along î direction", stop. Draw the figure. The dipole moment p = qd î.

Step 2 — Note what changes with time. At t = 0: field E ĵ is switched on. Between t = 0 and t = t_f: torque acts, dipole rotates. At t = t_f: field is switched off. After t_f: no torque.

Step 3 — Evaluate each option as a mini-problem. Never read all four options together. Each option is a separate sub-question. Treat it that way.

This is how eSaral trains students to approach every JEE Advanced paragraph-based or multi-correct question in every class — not just for this problem.

Step-by-Step Solution: Option A — Centre of Mass Displacement 

Option A states: The centre of mass of the dipole is deflected towards ĵ (y-direction) in the presence of the electric field.

Is This Correct?

Ask one simple question: What is the net force on the dipole system?

• Force on +q: F₁ = +qEĵ (upward)
• Force on −q: F₂ = −qEĵ (downward)
• Net force = F₁ + F₂ = 0

By Newton's second law for the centre of mass: F_net = Ma_cm

Since F_net = 0, acceleration of centre of mass = 0. The centre of mass does not move — it stays at the origin.

Option A is INCORRECT. ✗

Step-by-Step Solution: Option B — Finding Angular Velocity ω 

Option B states: If the magnitude of the final angular velocity is a specific expression, then θ_f = π/6.

Setting Up Energy Conservation

At t = t_f, the dipole has rotated from θ = 0 (along x-axis) to θ = θ_f. Apply energy conservation:

Initial State (t = 0):

• KE_initial = 0 (dipole at rest)
• PE_initial = −pE cos(angle between p and E)

Here, p is along î and E is along ĵ. The angle between them = 90°.

PE_initial = −pE cos 90° = 0

Final State (t = t_f):

The angle between p and E at time t_f is (90° − θ_f), not θ_f. This is a common error students make.

🚨 Critical mistake to avoid: Students often write cos θ_f for the final PE. The correct angle between the dipole vector and E-field vector is (90° − θ_f). So cos(90° − θ_f) = sin θ_f.

• PE_final = −pE sin θ_f = −qd · E · sin θ_f
• KE_final = ½Iω² = ½ · (md²/2) · ω²

Energy Conservation Equation

0 + 0 = −qEdsinθ_f + ½ · (md²/2) · ω²

½ · (md²/2) · ω² = qEd sinθ_f

ω² = 4qE sinθ_f / (md)

ω = √(4qE sinθ_f / md)

Checking Option B

Option B gives a specific value of ω and says θ_f = π/6.

Substituting θ_f = π/6: sin(π/6) = 1/2

ω = √(4qE · (1/2) / md) = √(2qE / md)

This exactly matches the expression given in Option B.

Option B is CORRECT. ✓

Step-by-Step Solution: Option C — Kinetic Energy at θ_f = π/3 

Option C states: When θ_f = π/3, the kinetic energy equals a specific value (given in the options).

Calculating KE at θ_f = π/3

From the energy conservation derived above:

KE = qEd sinθ_f

At θ_f = π/3: sin(π/3) = √3/2

KE = qEd · (√3/2) = (√3/2) · qEd

Now check whether Option C gives this value. The value given in Option C does not equal (√3/2)qEd.

Option C is INCORRECT. ✗

Step-by-Step Solution: Option D — Motion After Electric Field Is Switched Off 

Option D states: If θ_f = π/4, the dipole rotates about its centre of mass with constant angular velocity after t > t_f.

What Happens When E is Switched Off?

At t = t_f:

• The electric field is set to zero: E = 0
• Torque = pE sinθ = p · 0 = 0
• Angular acceleration α = τ/I = 0

Since α = 0, the angular velocity ω that existed at t = t_f remains constant for all t > t_f.

The dipole continues to rotate about its centre of mass with the same ω it had at the moment the field was switched off.

This is exactly what Option D describes.

Option D is CORRECT. ✓

Final Answer and Concept Map 

✅ Correct Options: B and D

Chapters Used in This Question

1. Electrostatics — dipole moment, torque, PE in electric field
2. Work, Power & Energy — energy conservation for rotation
3. Rotational Mechanics — moment of inertia, angular velocity, angular acceleration
4. Centre of Mass — COM motion requires a net external force

This question is a masterclass in multi-concept integration — the defining trait of JEE Advanced. If you have been preparing with eSaral, every one of these concepts has been covered at this exact level in class.

🚀 Checkout eSaral Courses