Interactions between charges

Let the initial distance between the two charges be $x$ meters. Since the force between the two charges follows the inverse square law, that is, it is inversely proportional to the square of distance between them, then we have:

$\begin{aligned} \frac{(x+5)^2}{x^2} & = \frac{36}{25} \\ \Rightarrow \frac{x+5}{x} & = \frac{6}{5} \\ 5x + 25 & = 6 x \\ \Rightarrow x = \boxed{25} \end{aligned}$

We know that electrostatic force of attraction is given by, $\vec{F}=k\dfrac{q_1q_2}{r^2}\\\therefore k \dfrac{q_1q_2}{r^2} = 36\mu \implies m\times q_1q_2=r^236\mu$

Now, due to increase in the distance, $r'=r+5$ ; $\therefore k \dfrac{q_1q_2}{(r+5)^2}=25\mu\implies m\times q_1q_2=(r+5)^225\mu$ $\therefore r^236\mu=(r+5)^225\mu\\6r=(r+5)5\implies 6r-5r=25\\\boxed{r=25m}$