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Q.35 Physics ⭐ 360° Option Elimination Active
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NEET 2025 Physics Question 35 Solution

Topic: Electric Charges and Fields · Official NTA Booklet Code 45
NEET 2025 Physics Question 35: Work Done / Change in Potential Energy on Rotating Dipole in Uniform Field

Question

An electric dipole with dipole moment 5×10⁻⁶ C m is aligned with the direction of a uniform electric field of magnitude 4×10⁵ N/C. The dipole is then rotated through an angle of 60° with respect to the electric field. The change in the potential energy of the dipole is:

✓
Verified Against NTA Final Answer Key
Option (2) : 1.0 J
📚 Electric Charges and Fields Easy
📋 Given Parameters & Problem Setup
Dipole Moment (p) 5 × 10⁻⁶ C·m
Electric Field Magnitude (E) 4 × 10⁵ N/C
Initial Orientation (θ₁) 0° (aligned with electric field)
Final Orientation (θ₂) 60°
⚡

The 30-Second Exam Hall Shortcut: ⚡ Dipole Potential Energy ΔU Shortcut

⏱️ Save ~30 seconds

ΔU = p E (cos θ₁ - cos θ₂) = p E (1 - cos 60°) = (p E) / 2 = [(5 × 10⁻⁶) × (4 × 10⁵)] / 2 = 2.0 / 2 = 1.0 J!

1 Potential energy of dipole in electric field: U(θ) = -p E cos θ.
2 Change in potential energy: ΔU = U(θ₂) - U(θ₁) = -p E cos 60° - (-p E cos 0°) = p E (cos 0° - cos 60°).
3 Evaluate cos values: cos 0° = 1, cos 60° = 0.5 ⟹ factor is (1 - 0.5) = 0.5.
4 Calculate product: p E = (5 × 10⁻⁶ C·m) × (4 × 10⁵ N/C) = 20 × 10⁻¹ = 2.0 J.
5 Multiply: ΔU = 0.5 × 2.0 J = 1.0 J.
6 Select Option (2) in under 10 seconds!
🛡️
Why You Can 100% Trust This Shortcut: Electrostatic Potential Energy of Electric Dipole (NCERT Physics · Electrostatic Potential: U = -p · E).
📐

Step-by-Step Rigorous Workflow & Mathematical Proof

Step 1

Formula for Potential Energy of an Electric Dipole

The electrostatic potential energy U of an electric dipole with dipole moment p in a uniform electric field E is given by:

U(θ)=−p⃗⋅E⃗=−pEcos⁡θU(\theta) = -\vec{p} \cdot \vec{E} = -p E \cos\theta
  • • Stable equilibrium occurs at θ = 0° where U is minimum: U(0°) = -p E.
Step 2

Calculate Initial and Final Potential Energies

Evaluate U at initial angle θ₁ = 0° and final angle θ₂ = 60°:

U1=−pEcos⁡0∘=−pE,U2=−pEcos⁡60∘=−12pEU_1 = -p E \cos 0^\circ = -p E, \quad U_2 = -p E \cos 60^\circ = -\frac{1}{2} p E
  • • Change in potential energy is defined as ΔU = U₂ - U₁.
Step 3

Compute ΔU and Substitute Numerical Values

Calculate the difference and substitute p and E:

ΔU=U2−U1=−12pE−(−pE)=12pE=12(5×10−6 C⋅m)(4×105 N/C)=12×2.0 J=1.0 J\Delta U = U_2 - U_1 = -\frac{1}{2} p E - (-p E) = \frac{1}{2} p E = \frac{1}{2} (5 \times 10^{-6}\text{ C}\cdot\text{m}) (4 \times 10^5\text{ N/C}) = \frac{1}{2} \times 2.0\text{ J} = 1.0\text{ J}
  • • Matches Option (2) with complete precision.
⭐

360° Option Elimination Engine Key Feature

Diagnose every choice: Know exactly why the right option works and why distractors fail.

4-Option Deep Scan
1 A
✗ Incorrect
Verdict & Scientific Reason:

Incorrect. Arises from using cos 37° = 0.8 or sin function.

⚠️
Student Fallacy (Why candidates pick this trap): Using torque formula τ = p E sin θ instead of work/energy formula.
2 B
✓ 100% Correct
Verdict & Scientific Reason:

100% correct. ΔU = p E (1 - cos 60°) = 2.0 × 0.5 = 1.0 J.

💡
Why this is airtight: None; direct application of dipole potential energy formula.
3 C
✗ Incorrect
Verdict & Scientific Reason:

Incorrect. Arbitrary distractor.

⚠️
Student Fallacy (Why candidates pick this trap): Arithmetic slip in evaluating p E.
4 D
✗ Incorrect
Verdict & Scientific Reason:

Incorrect. Corresponds to rotation to 120°: p E (1 - cos 120°) = 2.0 × 1.5 / 2 = 1.5 J.

⚠️
Student Fallacy (Why candidates pick this trap): Using 120° angle instead of 60°.
⚠️

NTA Trap Alert: Confusing Torque with Work/Energy (sin θ vs. cos θ)

Torque on a dipole is τ = p E sin θ, whereas potential energy is U = -p E cos θ. Using sin θ yields p E sin 60° = 2 × (√3/2) = √3 ≈ 1.732 J.

🛡️ How to avoid the -1 mark: Always remember: Work/Energy uses COSINE; Torque uses SINE.
📖

Core NEET Syllabus Standard

NEET Physics Core Syllabus · Electric Charges and Fields Standard Benchmark
Core Scientific Principle Tested:

"ΔU = W_{ext} = p E (cos θ₁ - cos θ₂)."

Syllabus Focus: Dipole in a Uniform External Field
✓ Core NEET UG Syllabus Anchor Syllabus & Academic Standards ↗
ΔU = p E (1 - cos 60°) = 2.0 × (1/2) = 1.0 J!

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