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 C is brought in contact with sphere A first and then with sphere B and finally removed from both. New force of repulsion between spheres A and B (Radius of A and B are negligible compared to the distance of separation so that they can be considered as point charges) is best given as:

• A. \( \frac{2F}{3} \)
• B. \( \frac{F}{2} \)
• C. \( \frac{3F}{8} \)
• D. \( \frac{3F}{4} \)

Hint:

When two identical conducting spheres with charges \(q_1\) and \(q_2\) are brought into contact, the charge on each sphere after separation becomes \( \frac{q_1 + q_2}{2} \) due to the redistribution of charge.

Answer 1

The Correct Option is C

We will resolve the problem by systematically examining the interactions when sphere C contacts sphere A, followed by sphere B. Both spheres A and B initially possess a charge of \(q\). The initial repulsive force between them is defined as:

\( F = k \frac{q^2}{d^2} \)

where \(k\) represents Coulomb's constant and \(d\) denotes the separation between the sphere centers.

1. Interaction with A: Sphere C begins with no charge. Upon contact with sphere A, charge distribution occurs equally due to their identical nature. Consequently, the charge on each sphere post-contact becomes \( \frac{q + 0}{2} = \frac{q}{2} \).

Sphere A now holds a charge of \( \frac{q}{2} \), and sphere C also carries a charge of \( \frac{q}{2} \).

2. Interaction with B: Sphere C, now charged at \( \frac{q}{2} \), makes contact with sphere B, which has a charge of \( q \). Again, charges will be distributed equally. The combined charge is \( \frac{q}{2} + q = \frac{3q}{2} \).

The charge on each sphere after this contact is \( \frac{\frac{3q}{2}}{2} = \frac{3q}{4} \).

After sphere C is separated, sphere A retains a charge of \( \frac{q}{2} \) and sphere B has a charge of \( \frac{3q}{4} \).

3. Recalculation of Force: The new repulsive force between spheres A and B is calculated as:

\( F' = k \frac{\left( \frac{q}{2} \right) \left( \frac{3q}{4} \right)}{d^2} = k \frac{\frac{3q^2}{8}}{d^2} = \frac{3F}{8} \)

Therefore, the updated repulsive force between spheres A and B is \( \frac{3F}{8} \).