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(D) Initial condition: Two spheres $A$ and $B$ each have charge $q$ and are separated by distance $r$. The initial force is $F = \frac{k q^2}{r^2}$.St

Vedclass · Accepted Answer

$\frac{3F}{8}$

Vedclass · Answer

(D) Initial condition: Two spheres $A$ and $B$ each have charge $q$ and are separated by distance $r$. The initial force is $F = \frac{k q^2}{r^2}$.Step $1$: When the uncharged sphere $C$ is touched to sphere $A$,the total charge $q + 0 = q$ is shared equally between them because they are identical. Thus,the charge on $A$ becomes $q_A = \frac{q}{2}$ and the charge on $C$ becomes $q_C = \frac{q}{2}$.Step $2$: Now,sphere $C$ (with charge $\frac{q}{2}$) is touched to sphere $B$ (with charge $q$). The total charge is $q + \frac{q}{2} = \frac{3q}{2}$. Since the spheres are identical,this charge is shared equally between $B$ and $C$. Therefore,the new charge on $B$ is $q_B = \frac{1}{2} 	imes \frac{3q}{2} = \frac{3q}{4}$.Step $3$: The final force $F^{\prime}$ between sphere $A$ (charge $\frac{q}{2}$) and sphere $B$ (charge $\frac{3q}{4}$) at the same distance $r$ is:$F^{\prime} = \frac{k q_A q_B}{r^2} = \frac{k (q/2) (3q/4)}{r^2} = \frac{3}{8} \frac{k q^2}{r^2}$.Since $F = \frac{k q^2}{r^2}$,we have $F^{\prime} = \frac{3}{8} F$.

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