Skip to content
PhysicsNEET 2025Coulomb's law and superposition

Electric Charges and Fields

Two identical charged conducting spheres A and B have their centres separated by a certain distance. The 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. The new force of repulsion between spheres A and B (the radii of A and B are negligible compared to the distance of separation, so they can be considered point charges) is best given as:

  1. A.3F5\dfrac{3F}{5}
  2. B.2F3\dfrac{2F}{3}
  3. C.F2\dfrac{F}{2}
  4. D.3F8\dfrac{3F}{8}✓

Solution

Touching A: the third sphere and A share qq, so A becomes q/2q/2. Touching B: it carries q/2q/2 to B, which then holds q/2+q2=3q4\frac{q/2 + q}{2} = \frac{3q}{4}. New force ∝q2⋅3q4=38q2\propto \frac{q}{2}\cdot\frac{3q}{4} = \frac{3}{8}q^2, i.e. 3F8\frac{3F}{8}.

Save for spaced revision (free account)

More from this chapter in past papers

From Physics › Electric Charges and Fields › Coulomb's law and superposition — lectures, notes and every past question on it.

Last revised . Found a mistake? Tell us — corrections are the fastest thing we act on.

ShareWhatsAppTelegram