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(C) In the first case,by Coulomb's law,the force between the spheres is:$F_1 = \frac{k Q_1 Q_2}{d^2}$When the two identical spheres are made to touch,

Vedclass · Accepted Answer

$\frac{4Q_2}{Q_1}$

Vedclass · Answer

(C) In the first case,by Coulomb's law,the force between the spheres is:$F_1 = \frac{k Q_1 Q_2}{d^2}$When the two identical spheres are made to touch,the total charge is redistributed equally between them. The new charge on each sphere is:$q = \frac{Q_1 + Q_2}{2}$Since $Q_1 >> Q_2$,we can approximate $q \approx \frac{Q_1}{2}$.After separating them to the original distance $d$,the new force $F_2$ is:$F_2 = \frac{k q^2}{d^2} = \frac{k (Q_1/2)^2}{d^2} = \frac{k Q_1^2}{4d^2}$Now,calculating the ratio $F_1/F_2$:$\frac{F_1}{F_2} = \frac{(k Q_1 Q_2 / d^2)}{(k Q_1^2 / 4d^2)} = \frac{4 Q_1 Q_2}{Q_1^2} = \frac{4 Q_2}{Q_1}$

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