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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 a 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
$\mathrm{F}_{1}=\frac{\mathrm{k} \mathrm{Q}_{1} \mathrm{Q}_{2}}{\mathrm{r}^{2}} \quad \mathrm{F}_{2}=\frac{\mathrm{k}\left(\mathrm{Q}_{1}+\mathrm{Q}_{2}\right)^{2}}{\mathrm{r}^{2}}$
Given $\quad \mathrm{F}_{2}=4.5 \mathrm{F}_{1}$
${\frac{\mathrm{k}\left(\mathrm{Q}_{1}+\mathrm{Q}_{2}\right)^{2}}{\mathrm{r}^{2}}=\frac{\mathrm{k} \mathrm{Q}_{1} \mathrm{Q}_{2}}{\mathrm{r}^{2}} 4.5} $
${\mathrm{Q}_{1}^{2}+\mathrm{Q}_{2}^{2}=2.5 \mathrm{Q}_{1} \mathrm{Q}_{2}}$
${\left(\frac{\mathrm{Q}_{1}}{\mathrm{Q}_{2}}\right)^{2}+1=\frac{2.5 \mathrm{Q}_{1}}{\mathrm{Q}_{2}}}$
Let $\frac{Q_{1}}{Q_{2}}=x$
$\Rightarrow x^{2}+1-2.5 x=0$
$\Rightarrow x^{2}+1-2.5 x=0$
$\quad x^{2}-2 x-0.5 x+1=0$
$(X-2)(x-0.5)=0$
$x=2$ or $\frac{1}{2}$
$\Rightarrow \frac{Q_{1}}{Q_{2}}=2$
