F=r2kQ2 on touching; F′=r2k(Q/2)(3Q/4) F′=83F 
After touching → (A)&(C)→ Total charge →q+0=q QA′=2q,QC′=2q (B) & (C) Total charge =q+2q=23qQB′=43qQC′=43q
NEET UG 2025 — Physics Electromagnetism
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 :
Held on 30 Apr 2025 · Verified 9 Jul 2026.
53F
32F
2F
83F
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