Question bank/NEET UG/Physics

NEET UGPhysicsModerateNEET 2025

Two identical charged conducting spheres A and B have their centres separated by a certain distance. Charge on each sphere is qq and the force of repulsion between them is FF. 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:

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Sol.
<image_1>
F=Kqqr2F = \frac{Kqq}{r^2}
<image_2>
F′=Kq 3q2 4r2F' = \frac{\frac{Kq \ 3q}{2 \ 4}}{r^2}
F′=3F8F' = \frac{3F}{8}

Source: geekyrakshit/indian-exam-questions / neet2025-code45

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