Two spherical conductors B and C having equal | vedclass

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(D) Let the initial charge on spherical conductors $B$ and $C$ be $q$.The initial electric force between them is given by Coulomb's law:$F = \frac{1}{

Vedclass · Accepted Answer

$\frac{3F}{8}$

Vedclass · Answer

(D) Let the initial charge on spherical conductors $B$ and $C$ be $q$.The initial electric force between them is given by Coulomb's law:$F = \frac{1}{4 \pi \varepsilon_{0}} \frac{q^2}{r^2}$When the uncharged conductor $A$ is brought in contact with $B$,the total charge $q$ is shared equally between them because they have equal radii:$q_B = q_A = \frac{q + 0}{2} = \frac{q}{2}$Now,conductor $A$ (carrying charge $q/2$) is brought in contact with $C$ (carrying charge $q$):$q_A = q_C = \frac{(q/2) + q}{2} = \frac{3q/2}{2} = \frac{3q}{4}$After removing $A$,the new charges on $B$ and $C$ are $q_B = q/2$ and $q_C = 3q/4$.The new force of repulsion $F'$ between $B$ and $C$ is:$F' = \frac{1}{4 \pi \varepsilon_{0}} \frac{q_B q_C}{r^2} = \frac{1}{4 \pi \varepsilon_{0}} \frac{(q/2)(3q/4)}{r^2}$$F' = \frac{3}{8} \left( \frac{1}{4 \pi \varepsilon_{0}} \frac{q^2}{r^2} \right) = \frac{3F}{8}$

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