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 repulsive force between \( P \) and \( S \) is given by \(F_{PS} \propto Q^2\), with \(F_{PS} = 16 \, \text{N}\). 1. Upon bringing the uncharged sphere \( R \) into contact with \( P \), the charge on \( P \) redistributes equally, as they are identical. The charge on each sphere \( P \) and \( R \) becomes \( \frac{Q}{2} \). 2. Subsequently, \( R \) is brought into contact with \( S \). The charge then redistributes equally between \( R \) and \( S \). After this contact, the charge on each sphere \( S \) and \( R \) becomes \( \frac{3Q}{4} \). At this stage, \( P \) possesses a charge of \( \frac{Q}{2} \) and \( S \) has a charge of \( \frac{3Q}{4} \). The new repulsive force between \( P \) and \( S \) is proportional to \( \frac{Q}{2} \times \frac{3Q}{4} = \frac{3Q^2}{8} \). Given that the initial force \( F_{PS} \) was 16 N, the new force is calculated as \(F_{PS} = \frac{3}{8} \times 16 = 6 \, \text{N}\).