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(D) Initially,the force between spheres $A$ and $B$ is given by Coulomb's law: $F = \frac{k q^2}{r^2}$.
When the third identical uncharged sphere

Vedclass · Accepted Answer

$\frac{3 F}{8}$

Vedclass · Answer

(D) Initially,the force between spheres $A$ and $B$ is given by Coulomb's law: $F = \frac{k q^2}{r^2}$.
When the third identical uncharged sphere $C$ is brought in contact with sphere $A$,the total charge $q + 0 = q$ is shared equally between them. Thus,the new charge on $A$ is $q_A' = \frac{q}{2}$ and on $C$ is $q_C' = \frac{q}{2}$.
Next,sphere $C$ (now with charge $\frac{q}{2}$) is brought in contact with sphere $B$ (which has charge $q$). The total charge is $q + \frac{q}{2} = \frac{3q}{2}$. This charge is shared equally between $B$ and $C$. Thus,the new charge on $B$ is $q_B' = \frac{3q}{4}$ and on $C$ is $q_C'' = \frac{3q}{4}$.
The new force of repulsion between $A$ and $B$ is $F' = \frac{k q_A' q_B'}{r^2} = \frac{k (q/2) (3q/4)}{r^2} = \frac{3}{8} \frac{k q^2}{r^2}$.
Since $F = \frac{k q^2}{r^2}$,we get $F' = \frac{3 F}{8}$.

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