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Topic: Electric charge, conservation, Coulomb's law, superposition· Electrostatics

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 sph…

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 :

1. \(\frac{3F}{5}\)

2. \(\frac{2F}{3}\)

3. \(\frac{F}{2}\)

4. \(\frac{3F}{8}\)

1 answer

Best answer

Answer: 4. \(\frac{3F}{8}\)

Key idea: identical conducting spheres share charge equally on contact.

  • Third sphere touches A: A and C each have \(q/2\).
  • Then C (\(q/2\)) touches B (\(q\)): each has \(\dfrac{q/2+q}{2}=\dfrac{3q}{4}\).

Now A has \(q/2\) and B has \(3q/4\). Since \(F=k\dfrac{q^2}{r^2}\):

$$F'=k\frac{(q/2)(3q/4)}{r^2}=\frac38F$$
Doubtly AI · 30 Oct 2026 AI-generated · not yet verified Report
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