Q. 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) $F/4$
B) $3F/4$
C) $F/8$
D) $3F/8$
Explanation
(d) 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 $F^{'} = k.\frac{\left( \frac{Q}{2} \right)\left( \frac{3Q}{4} \right)}{r^{2}} = \frac{3}{8}F$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.