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 ${{q}_{A}}={{q}_{C}}=\frac{{{q}_{A}}+{{q}_{B}}}{2}$ $=\frac{(q/2)+q}{2}$ Hence, electric force acting between $B$ and $C$ is $F=\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{{{q}_{B}}{{q}_{C}}}{{{r}^{2}}}$ $=\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{(q/2)(3q/4)}{{{r}^{2}}}$ $=\frac{3}{8}\left[ \frac{1}{4\pi {{\varepsilon }_{0}}}\frac{{{q}^{2}}}{{{r}^{2}}} \right]=\frac{3F}{8}$