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 {{\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} $