Two  identical metallic balls are placed at a certain distance apart. They carry like charges and repel each other with a certain force. They have moved apart to a distance equal to half their initial separation after bringing them in contact. In comparison with the initial value, the force of repulsion between them increases 4. 5 times. Find the ratio of the initial charges on the balls. 

Force between two charges kept at a distance of r is given by,

$\Rightarrow \mathrm{F}=\frac{{\mathrm{kq}}_1{\mathrm{q}}_2}{{\mathrm{r}}^2}$

Let balls have charges Q₁ and Q₂ respectively

$\Rightarrow F=\frac{k{Q}_1{Q}_2}{r^2}$

when two charges are brought in contact they redistribute their charges and after moving apart to half of their initial separation charge on each of them becomes $\frac{Q_1+Q_2}{2}$, therefore in second case force between them is

${\mathrm{F}}^{\mathrm{\prime }}=\frac{\mathrm{k}{\left(\frac{{\mathrm{Q}}_1+{\mathrm{Q}}_2}{2}\right)}^2}{{\left(\frac{\mathrm{r}}{2}\right)}^2}=\frac{\mathrm{k}{\left({\mathrm{Q}}_1+{\mathrm{Q}}_2\right)}^2}{{\mathrm{r}}^2}$

According to question,

$\Rightarrow {\mathrm{F}}^{\mathrm{\prime }}=4.5\mathrm{F}$

$\Rightarrow \frac{\mathrm{k}{\left({\mathrm{Q}}_1+{\mathrm{Q}}_2\right)}^2}{{\mathrm{r}}^2}=4.5\times \frac{{\mathrm{kQ}}_1{\mathrm{Q}}_2}{{\mathrm{r}}^2}$

$\Rightarrow {Q_1}^2+{Q_2}^2+2Q_1{Q}_2=4$

$.5Q_1{Q}_2$

$\Rightarrow Q_1^2+Q_2^2-2.5Q_1{Q}_2=0$

Divide whole equation by ${Q_2}^2$ and let $\frac{Q_1}{Q_2}$ be x, equation becomes.

$\Rightarrow x^2-2.5x+1=0$

$\Rightarrow x^2-2x-0.5x+1=0$

$\Rightarrow x(x-2)-0.5(x-2)=0$

$\Rightarrow (x-2)(x-0.5)=0$

$\text{ hence, }\frac{Q_1}{Q_2}=2\text{ or }\frac{Q_1}{Q_2}=0.5$