Q.34 Physics ⭐ 360° Option Elimination Active

NEET 2026 Physics Question 34 Solution

Topic: Electrostatics & Potential · Official NTA Booklet Code 50

Question

Consider a fixed uniformly charged insulating sphere with radius R and total charge +Q. A point charge −q (q << Q) with mass m is released from rest at a distance of 3R from the centre of the charged sphere. When the point charge reaches the surface of the sphere, its speed is: (ε₀ is the permittivity of vacuum, neglect gravitational forces).

✓

Verified Against NTA Final Answer Key

Option (3) : v = sqrt(Qq / (3 * pi * epsilon_0 * m * R))

📚 Electrostatics & Potential Medium

📋 Given Parameters & Problem Setup

Charged Sphere Fixed insulating sphere, radius R, total charge +Q

Point Charge Charge -q, mass m (q << Q), released from rest

Initial Distance r_i = 3R from the centre

Final Distance r_f = R (at the surface of the sphere)

Target Velocity Speed v upon reaching the surface

⚡

The 30-Second Exam Hall Shortcut: ⚡ Potential Energy Difference Shortcut

⏱️ Saves 2 minutes of calculus force integrals.

Equate kinetic energy gain to electrostatic potential energy loss

1 Outside the sphere (r >= R), potential is V(r) = Q / (4pi * epsilon_0 * r).

2 Potential difference from 3R to R: Delta V = V(R) - V(3R) = [Q / (4pi * epsilon_0)] * (1/R - 1/3R) = [Q / (4pi * epsilon_0 * R)] * (2/3) = Q / (6pi * epsilon_0 * R).

3 Kinetic energy gained by charge q: (1/2)*m*v^2 = q * Delta V = Q*q / (6pi * epsilon_0 * R).

4 Multiply by 2/m: v^2 = 2*Q*q / (6pi * epsilon_0 * m * R) = Q*q / (3pi * epsilon_0 * m * R).

5 Take square root: v = sqrt( Q*q / (3pi * epsilon_0 * m * R) ). Solved in 20 seconds!

🛡️

Why You Can 100% Trust This Shortcut: Conservation of mechanical energy: Delta K + Delta U = 0. Since the point charge never penetrates the sphere interior, only the exterior Coulomb potential 1/r is required.

📐

Step-by-Step Rigorous Workflow & Mathematical Proof

Step 1

Formulate Conservation of Mechanical Energy

Because the insulating sphere is fixed and electrostatic forces are conservative:

K_i + U_i = K_f + U_f \implies 0 + U(3R) = \frac{1}{2} m v^2 + U(R)

• Initial kinetic energy K_i = 0 (released from rest).
• Initial distance r_i = 3R; final distance r_f = R.

Step 2

Evaluate Electrostatic Potential Energy at Both Positions

Outside a uniformly charged sphere, electrostatic potential energy between +Q and -q is:

U(r) = -\frac{1}{4\pi \epsilon_0} \frac{Qq}{r}

• At r = 3R: U(3R) = -\frac{Qq}{12\pi \epsilon_0 R}.
• At r = R: U(R) = -\frac{Qq}{4\pi \epsilon_0 R}.

Step 3

Solve for Kinetic Energy and Velocity v

Isolate (1/2)*m*v^2 and simplify:

\frac{1}{2} m v^2 = U(3R) - U(R) = -\frac{Qq}{12\pi \epsilon_0 R} - \left(-\frac{Qq}{4\pi \epsilon_0 R}\right) = \frac{Qq}{4\pi \epsilon_0 R}\left(1 - \frac{1}{3}\right) = \frac{Qq}{6\pi \epsilon_0 R}

• Multiply by 2/m: v^2 = \frac{2 Qq}{6\pi \epsilon_0 m R} = \frac{Qq}{3\pi \epsilon_0 m R}.
• Taking the square root gives: v = \sqrt{\frac{Qq}{3\pi \epsilon_0 m R}}.
• Matches Option (3) exactly.

⭐

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 Option (1)

✗ Incorrect

Verdict & Scientific Reason:

Inverted fraction factor.

⚠️

Student Fallacy (Why candidates pick this trap): Multiplies by 3/2 instead of 2/3 during work calculation.

Note: 1 - 1/3 = 2/3, not 3/2.

2 Option (2)

✗ Incorrect

Verdict & Scientific Reason:

Forgot factor of 1/2 in kinetic energy.

⚠️

Student Fallacy (Why candidates pick this trap): Equates m*v^2 directly to work without dividing by 2.

Note: Kinetic energy is (1/2)*m*v^2, so v^2 = 2*W/m.

3 Option (3)

✓ 100% Correct

Verdict & Scientific Reason:

100% mathematically exact; confirmed by Official NTA Final Answer Key.

💡

Why this is airtight: Flawless execution of potential energy difference and kinetic energy balance.

Note: Exact physics.

4 Option (4)

✗ Incorrect

Verdict & Scientific Reason:

Assumes released from infinity.

⚠️

Student Fallacy (Why candidates pick this trap): Takes initial potential energy as 0 (as if released from r = infinity instead of 3R).

Note: Point charge starts at 3R, not infinity.

⚠️

NTA Trap Alert: Assuming Release from Infinity (The 1/4 Factor Trap)

Students who use v = sqrt(2*q*V_surface/m) assume the particle started at infinity, picking Option (4).

🛡️ How to avoid the -1 mark: Check the problem statement carefully: the particle starts at 3R! The potential factor is (1/R - 1/3R) = 2/(3R), NOT 1/R.

📖

Core NEET Syllabus Standard

NEET UG Physics Core Standards · Electrostatic Potential and Capacitance Standard Benchmark

Core Scientific Principle Tested:

"Kinetic energy gained by a particle in an electric field: (1/2)*m*v^2 = q*[V(r_final) - V(r_initial)]."

Syllabus Focus: Potential Energy of Charge Configurations & Work-Energy Conservation

✓ Core NEET UG Syllabus Anchor Syllabus & Academic Standards ↗

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NEET 2026 Physics Question 34 Solution

Question

The 30-Second Exam Hall Shortcut: ⚡ Potential Energy Difference Shortcut

Step-by-Step Rigorous Workflow & Mathematical Proof

Formulate Conservation of Mechanical Energy

Evaluate Electrostatic Potential Energy at Both Positions

Solve for Kinetic Energy and Velocity v

360° Option Elimination Engine Key Feature

NTA Trap Alert: Assuming Release from Infinity (The 1/4 Factor Trap)

Core NEET Syllabus Standard

Ready to attempt under real exam pressure?

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