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Two identical balls having like charges and placed at a certain distance apart repel each other with a certain force. They are brought in contact and then moved apart to distance equal to half their initial separation. The force of repulsion between them increases $4.5\,times$ in comparison with the initial value. The ratio of the initial charges of the balls is
$2$
$3$
$4$
$6$
Solution
Suppose the balls having charge $Q_{1}$ and $Q_{2}$ Intially$:$
$F^{\prime}=\frac{\left(k \frac{Q_{1}+Q_{2}}{2}\right)^{2}}{\left(\frac{r}{2}\right)^{2}}=\frac{k\left(Q_{1}+Q_{2}\right)^{2}}{r^{2}}$
It is given that $F=4.5 F$ so $\frac{\left(Q_{1}+Q_{2}\right)^{2}}{r^{2}}=4.5 k \frac{Q_{1} Q_{2}}{r^{2}}$
$\Rightarrow\left(Q_{1}+Q_{2}\right)^{2}=4.5 Q_{1} Q_{2}$
On solving it gives $\frac{Q_{1}}{Q_{2}}=\frac{2}{1}$
