Q.28 Physics ⭐ 360° Option Elimination Active

NEET 2025 Physics Question 28 Solution

Topic: Electric Charges and Fields · Official NTA Booklet Code 45

Question

Two identical charged conducting spheres A and B have their centres separated by a certain distance. Charge on each sphere is q and the force of repulsion between them is F. A third identical uncharged conducting sphere is brought in contact with sphere A first and then with B and finally removed from both. New force of repulsion between spheres A and B (Radii of A and B are negligible compared to the distance of separation so that for calculating force between them they can be considered as point charges) is best given as:

✓

Verified Against NTA Final Answer Key

Option (4) : \frac{3F}{8}

📚 Electric Charges and Fields Easy

📋 Given Parameters & Problem Setup

Initial Configuration Two identical conducting spheres A and B, each with charge q, separated by distance r

Initial Force F = k q² / r²

Successive Operations Third identical uncharged sphere C contacts A first, then contacts B, then is removed

⚡

The 30-Second Exam Hall Shortcut: ⚡ Successive Conduction Fraction Shortcut

⏱️ Save ~40 seconds

Sphere C takes half of A (q/2). Then C (q/2) shares with B (q): (q + q/2)/2 = 3q/4 on B! New Force F' = (1/2) × (3/4) F = 3/8 F!

1 Initially: Sphere A has q, Sphere B has q ⟹ F = k q² / r².

2 Contact 1 (C with A): Equal sharing gives q_A' = q_C' = (q + 0) / 2 = q / 2.

3 Contact 2 (C with B): Equal sharing gives q_B' = (q + q/2) / 2 = (3q/2) / 2 = 3q / 4.

4 Final charges on A and B: q_A' = q / 2 and q_B' = 3q / 4.

5 New Coulomb force: F' ∝ q_A' × q_B' = (1/2) × (3/4) = 3/8 F.

6 Select Option (4) in under 15 seconds!

🛡️

Why You Can 100% Trust This Shortcut: Coulomb’s Law of Electrostatics (NCERT Physics · Electric Charges and Fields: F = k|q₁q₂|/r²).

📐

Step-by-Step Rigorous Workflow & Mathematical Proof

Step 1

Initial Electrostatic Force

By Coulomb’s law, the initial repulsive force between the two identical spheres A and B carrying charge q separated by distance r is:

F = \frac{k q^2}{r^2}

• Spheres are treated as point charges because radii are negligible compared to distance r.

Step 2

Step 1: Contact Between Sphere C and Sphere A

An identical uncharged conducting sphere C (initial charge = 0) is brought into physical contact with sphere A (charge q). Due to identical capacitance and symmetry, charge redistributes equally:

q_A' = q_C' = \frac{q + 0}{2} = \frac{q}{2}

• Sphere A now retains charge q/2, and sphere C carries charge q/2.

Step 3

Step 2: Contact Between Sphere C and Sphere B & Final Force

Sphere C (charge q/2) is now brought into contact with sphere B (charge q). The total charge divides equally between the two identical spheres:

q_B' = q_C'' = \frac{q + \frac{q}{2}}{2} = \frac{3q}{4}

• Sphere C is now removed.
• The new force between A (charge q/2) and B (charge 3q/4) is:
• F' = k (q/2)(3q/4) / r² = (3/8) [k q² / r²] = (3/8) F.
• Corresponds uniquely to Option (4).

⭐

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. Denominator 5 is physically impossible in binary equal sharing.

⚠️

Student Fallacy (Why candidates pick this trap): Random fraction guess.

2 B

✗ Incorrect

Verdict & Scientific Reason:

Incorrect. Arithmetic slip.

⚠️

Student Fallacy (Why candidates pick this trap): Dividing by 3 instead of successive halves.

3 C

✗ Incorrect

Verdict & Scientific Reason:

Incorrect. Considers only the halving of charge on A while ignoring the charge reduction on B.

⚠️

Student Fallacy (Why candidates pick this trap): Assuming charge on B remained untouched at q.

4 D

✓ 100% Correct

Verdict & Scientific Reason:

100% correct. q_A = q/2, q_B = 3q/4 ⟹ F' = (1/2) × (3/4) F = 3F / 8.

💡

Why this is airtight: None; quintessential NCERT problem.

⚠️

NTA Trap Alert: Ignoring Charge Carried Over by Sphere C

Students frequently assume sphere C is still uncharged when it touches sphere B! But sphere C brings charge q/2 from its contact with A, so B shares (q + q/2), leaving 3q/4 on B.

🛡️ How to avoid the -1 mark: Track the charge of sphere C: 0 ⟹ q/2 ⟹ touches B with q/2!

📖

Core NEET Syllabus Standard

NEET Physics Core Syllabus · Electric Charges and Fields Standard Benchmark

Core Scientific Principle Tested:

"Identical conducting spheres in contact share charge equally: q_shared = (q₁ + q₂) / 2."

Syllabus Focus: Coulomb’s Law and Charge by Conduction

✓ Core NEET UG Syllabus Anchor Syllabus & Academic Standards ↗

Half from A (1/2) × Three-Quarters on B (3/4) = 3/8 F!

Ready to attempt under real exam pressure?

Test your speed and accuracy in the authentic 180-minute timed NTA simulator.

Launch NEET 2025 CBT Test →

← Previous (Q.27) All Physics Questions Grid Next (Q.29) →

Spotted a mistake in this solution? Report an error (support@mycompetitiveexams.com)

⚖️ Legal & Fair Dealing Notice: Sourced from publicly released NEET 2025 question papers and NEET answer keys for private study, research, and non-commercial educational practice under Fair Dealing (Section 52(1)(a) of the Indian Copyright Act, 1957). Not affiliated with or endorsed by NTA or NMC.

NEET 2025 Physics Question 28 Solution

Question

The 30-Second Exam Hall Shortcut: ⚡ Successive Conduction Fraction Shortcut

Step-by-Step Rigorous Workflow & Mathematical Proof

Initial Electrostatic Force

Step 1: Contact Between Sphere C and Sphere A

Step 2: Contact Between Sphere C and Sphere B & Final Force

360° Option Elimination Engine Key Feature

NTA Trap Alert: Ignoring Charge Carried Over by Sphere C

Core NEET Syllabus Standard

Ready to attempt under real exam pressure?

🔗 Related Questions — Electric Charges and Fields

Electric Potential on the Axial Line of a Dipole

Torque on a Dipole and Determination of Charge Magnitude

Flux Balance and Determination of Enclosed Charge

Electric Dipole in Non-Uniform Electric Field and Potential Energy

Solve 4 NEET Questions a Day — One per Subject