question_answer

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                                   [AIEEE 2004]

A)                    \[F/4\]                            

B)                    \[3F/4\]

C)                    \[F/8\]                            

D)                    \[3F/8\]

Correct Answer: D

Solution :

             Initially \[F=k.\frac{{{Q}^{2}}}{{{r}^{2}}}\] (fig. A). Finally when a third spherical conductor comes in contact alternately with B and C then removed, so charges on B and C are Q / 2 and 3Q / 4 respectively (fig. B) Now force \[F'=k.\frac{\left( \frac{Q}{2} \right)\,\left( \frac{3Q}{4} \right)}{{{r}^{2}}}=\frac{3}{8}F\]