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NEET Physics Electrostatics 2025 NEET 2025

NEET Physics Question (2025) — Solution

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 :

Options

  1. A. 3 F 5
  2. B. 2 F 3
  3. C. F 2
  4. D. 3 F 8

Answer

D. 3 F 8

Step-by-step solution

F = kQ ^2 r ^2 on touching; F^ = k(Q / 2)(3 Q / 4) r^2 F^ = 3 F 8 After touching ( A ) \&( C ) Total charge q +0= q Q _ A ^ = q 2 , Q _ C ^ = q 2 (B) & (C) aligned & Total charge =q+ q 2 = 3 2 q \\ & Q_B^ = 3 4 q Q_C^ = 3 4 q aligned

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