Question

Two identical conducting spheres P and S with charge Q on each, repel each other with a force 16N. A third identical uncharged conducting sphere R is successively brought in contact with the two spheres. The new force of repulsion between P and S is :

• A. 4 N
• B. 6 N
• C. 1 N
• D. 12 N

Answer 1

The Correct Option is B

The initial force of repulsion between \( P \) and \( S \) is:
\[F_{PS} \propto Q^2\]
\[F_{PS} = 16 \, \text{N}\]
1. When the uncharged sphere \( R \) is brought in contact with \( P \), the charge on \( P \) and \( R \) redistributes equally because they are identical spheres. Thus, after contact with \( P \):
\[\text{Charge on each of } P \text{ and } R = \frac{Q}{2}\]
2. Next, \( R \) is brought in contact with \( S \), and the charge will again redistribute equally between \( R \) and \( S \). After this contact:
\[\text{Charge on each of } S \text{ and } R = \frac{3Q}{4}\]
Now, \( P \) has a charge of \( \frac{Q}{2} \) and \( S \) has a charge of \( \frac{3Q}{4} \). The new force of repulsion between \( P \) and \( S \) is:
\[F_{PS} \propto \frac{Q}{2} \times \frac{3Q}{4} = \frac{3Q^2}{8}\]
Since the initial force \( F_{PS} \) was 16 N, we have:
\[F_{PS} = \frac{3}{8} \times 16 = 6 \, \text{N}\]