Two pith balls carrying equal charges are suspended from a common point by strings of equal length, the equilibrium separation between them is $r$. Now the strings are rigidly clamped at half the height. The equilibrium separation between the balls now become

Let $m$ be mass of each ball and $q$be charge on each ball.

Force of repulsion, $F=\frac{1}{4{\mathrm{\pi \epsilon }}_0}\frac{q^2}{r^2}$

In equilibrium

       $$\begin{gathered} T \cos \theta =mg ....\left(i\right) \\ T \sin \theta =F ....\left(ii\right) \end{gathered}$$

Divide (ii) by (i), we get

$\tan \theta =\frac{F}{mg}=\frac{{\displaystyle \frac{1}{4{\mathrm{\pi \epsilon }}_0}\frac{q^2}{r^2}}}{mg}$

From figure (a),

$\frac{r/2}{y}=\frac{{\displaystyle \frac{1}{4{\mathrm{\pi \epsilon }}_0}\frac{q^2}{r^2}}}{mg} ...\left(iii\right)$

Similarly, $\tan \theta \text{'}=\frac{{\displaystyle \frac{1}{4{\mathrm{\pi \epsilon }}_0}\frac{q^2}{r{\text{'}}^2}}}{mg}$

From figure (b)

$\frac{r\text{'}/2}{y/2}=\frac{{\displaystyle \frac{1}{4{\mathrm{\pi \epsilon }}_0}\frac{q^2}{r{\text{'}}^2}}}{mg} ....\left(iv\right)$

On dividing (iv) by (iii), we get, $\frac{2r\text{'}}{r}=\frac{r^2}{r{\text{'}}^2}$  $\Rightarrow$ $r\text{'}=\frac{r}{\sqrt[3]{2}}$