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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

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F/4

3F/8

3F/4

F/8

answer is D.

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Detailed Solution

Initially, $F = k \frac{Q^{2}}{r^{2}}$ (Fig. A). When a third spherical conductor comes in contact alternately with B and C and then removed, the charges on B and C are Q/2 and 3Q/4, respectively (Fig. B)

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

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