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. $F/4$
• B. $3F/4$
• C. $F/8$
• D. $3F/8$

Answer 1

Answer: (D)

$3F/8$

Answer 2

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 $F' = k.\frac{\left( \frac{Q}{2} \right)\left( \frac{3Q}{4} \right)}{r^{2}} = \frac{3}{8}F$