Two identical small equally charged conducting balls are suspended from long threads secured at one point. The charges and masses of the balls are such that they are in equilibrium when the distance between them is a (the length of thread L >> a).One of the balls is then discharged. Again for the certain value of distance b (b << l) between the balls the equilibrium is restored, The value of $\frac{a^3}{b^3}$ is _______________.

$\begin{array}{l}T\sin \theta =F=\frac{1}{4\pi {\in }_0}\frac{q^2}{a^2} .....\left(i\right)\\ T\cos \theta =mg .......\left(ii\right)\\ \text{ From (i) and (ii), }\tan \theta =\frac{q^2}{4\pi {\in }_0{a}^{2}mg}\\ \frac{a/2}{L}=\frac{q^2}{4\pi {\in }_0{a}^{2}mg}(\because \text{ for small }\theta ,\tan \theta \approx a/2L)\\ \Rightarrow \frac{a^3}{L}=\frac{q^2}{2\pi {\in }_{0}mg} .......\left(iii\right)\end{array}$

when one of the ball is discharged, the balls come closes and touch each other and again separate due to repulsion. The charge on each ball after touching each other = q/2.Replacing g with q/2 in (iii), we get

$\frac{b^3}{L}=\frac{(q/2{)}^2}{2\pi {\in }_{0}mg} .....\left(iv\right)$

$\text{ From (iii) and (iv) }\frac{b^3}{a^3}=\frac{1}{4}\Rightarrow \frac{a^3}{b^3}=4$