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

Let's solve the problem step by step by analyzing the effects of bringing sphere C in contact with sphere A and then sphere B. Initially, the charge on both spheres A and B is \(q\). The force of repulsion between them is given by:

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

where \(k\) is the Coulomb's constant and \(d\) is the distance between the centers of the spheres.

1. Contacting C with A: Sphere C is initially uncharged. When C is touched to A, the charges will redistribute equally because they are identical spheres. Therefore, the charge on each sphere after contact becomes \( \frac{q + 0}{2} = \frac{q}{2} \).

Now, sphere A has a charge of \( \frac{q}{2} \), and sphere C also has a charge of \( \frac{q}{2} \).

2. Contacting C with B: Sphere C with charge \( \frac{q}{2} \) is now touched to sphere B with charge \( q \). Again, the charges will distribute equally. The total charge is \( \frac{q}{2} + q = \frac{3q}{2} \).

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

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

3. New Force Calculation: The new force of repulsion between A and B is:

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

Thus, the new force of repulsion between spheres A and B is \( \frac{3F}{8} \).