Question

Two spherical conductors $B$ and $C$ having equal radii and carrying equal charges in them repel each other with a force $F$ when kept apart at some distance. A third spherical conductor having same radius as that of $B$ but uncharged, is brought in contact with $B$, then brought in contact with $C$ and finally removed away from both. The new force of repulsion between $B$ and $C$ is

• A. $\frac{F}{4}$
• B. $\frac{3F}{4}$
• C. $\frac{F}{8}$
• D. $\frac{3F}{8}$

Answer 1

Answer: (D)

$\frac{3F}{8}$

Answer 2

Let the spherical conductors $B$ and $C$ have same charge as $q$. The electric force between them is $F=\frac{1}{4 \pi \varepsilon_{0}} \frac{q^{2}}{r^{2}}$ $r$, being the distance between them. When third uncharged conductor $A$ is brought in contact with $B$, then charge on each conductor $q_{A}=q_{B}=\frac{q_{A}+q_{B}}{2}=\frac{0+q}{2}=\frac{q}{2}$ When this conductor $A$ is now brought in contact with $C$, then charge on each conductor $q_{A}=q_{C}=\frac{q_{A}+q_{C}}{2}$ $=\frac{(q / 2)+q}{2}=\frac{3 q}{4}$ Hence, electric force acting between $B$ and $C$ is $F'=\frac{1}{4 \pi \varepsilon_{0}} \frac{q_{B} q_{C}}{r^{2}}$ $=\frac{1}{4 \pi \varepsilon_{0}} \frac{(q / 2)(3 q / 4)}{r^{2}}$ $=\frac{3}{8}\left[\frac{1}{4 \pi \varepsilon_{0}} \frac{q^{2}}{r^{2}}\right]$ $=\frac{3 F}{8}$