Two spherical conductors B and C having equal | vedclass

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(D) Initially,the force between $B$ and $C$ is given by Coulomb's law: $F = k \frac{Q^2}{r^2}$.When the uncharged conductor (let's call it $D$) is bro

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$3F/8$

Vedclass · Answer

(D) Initially,the force between $B$ and $C$ is given by Coulomb's law: $F = k \frac{Q^2}{r^2}$.When the uncharged conductor (let's call it $D$) is brought in contact with $B$,the charge $Q$ is shared equally between them because they have equal radii. Thus,the charge on $B$ becomes $Q_B = Q/2$.Next,the conductor $D$ (now carrying charge $Q/2$) is brought in contact with $C$ (which has charge $Q$). The total charge $(Q/2 + Q) = 3Q/2$ is shared equally between $C$ and $D$. Thus,the new charge on $C$ becomes $Q_C = (3Q/2) / 2 = 3Q/4$.The new force of repulsion between $B$ and $C$ is $F' = k \frac{Q_B \cdot Q_C}{r^2} = k \frac{(Q/2) \cdot (3Q/4)}{r^2} = \frac{3}{8} \left( k \frac{Q^2}{r^2} \right) = \frac{3}{8} F$.

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