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

Answer: (B)

6 N

Answer 2

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:
$F_{PS} = \frac{3}{8} \times 16 = 6 \, \text{N}$