Physics · Electrostatics

NEET (UG) 2025 — Question 26

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 :

Option A:
$\frac{3 F}{5}$
Option B:
$\frac{2 F}{3}$
Option C:
$\frac{F}{2}$
Option D:
$\frac{3 F}{8}$
Correct

Answer: D

Step-by-step solution

$F=\frac{k Q^{2}}{r^{2}}$

on touching; $F^{\prime}=\frac{k(Q / 2)(3 Q / 4)}{r^{2}}$

$F^{\prime}=\frac{3 F}{8}$

After touching $\rightarrow$

$(\mathrm{A}) \&(\mathrm{C}) \rightarrow$

Total charge $\rightarrow \mathrm{q}+0=\mathrm{q}$

$\mathrm{Q}_{\mathrm{A}}^{\prime}=\frac{\mathrm{q}}{2}, \mathrm{Q}_{\mathrm{C}}^{\prime}=\frac{\mathrm{q}}{2}$

(B) & (C)

Total charge $=\mathrm{q}+\frac{\mathrm{q}}{2}=\frac{3}{2} \mathrm{q}$

$\mathrm{Q}_{\mathrm{B}}^{\prime}=\frac{3}{4} \mathrm{q} \quad \mathrm{Q}_{\mathrm{C}}^{\prime}=\frac{3}{4} \mathrm{q}$

Answer key and solution verified before publishing.

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Exam: NEET (UG) 2025
Paper: 2025 paper
Subject: Physics
Chapter: Electrostatics
Topic: Properties of Conductors and Redistribution of Charge

Answer: D

Step-by-step solution

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