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)

Let the spherical conductors $B$ and $C$ have same charge as $q$ . The electric force between them is $F=\frac{1}{4\pi {\epsilon }_0}\frac{q^2}{r^2}$ $r$ , 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 $q_A=q_C=\frac{q_A+q_B}{2}$ $=\frac{(q/2)+q}{2}$ Hence, electric force acting between $B$ and $C$ is $F=\frac{1}{4\pi {\epsilon }_0}\frac{q_B{q}_C}{r^2}$ $=\frac{1}{4\pi {\epsilon }_0}\frac{(q/2)(3q/4)}{r^2}$ $=\frac{3}{8}\left[\frac{1}{4\pi {\epsilon }_0}\frac{q^2}{r^2}\right]=\frac{3F}{8}$