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.

Given 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, we have to find the equilibrium separation when the strings are clamped at half the height. We can see from the figure that under equilibrium, \(\tan \theta=\frac{\mathrm{F}}{\mathrm{mg}}\)...(i) and also, \(\tan \theta=\frac{\mathrm{r} / 2}{\mathrm{y}} \ldots\) (ii) Equating (i) and (ii), we get \(\frac{\mathrm{F}}{\mathrm{mg}}=\frac{\mathrm{r}}{2 \mathrm{y}}\) also, Force, \(\mathrm{F}=\frac{\mathrm{kq}^2}{\mathrm{r}^2}\) where \(\mathrm{k}=\) constant \(=9 \times 10^9\) So, we get \(\mathrm{r}^3=\left(\frac{\mathrm{kq}^2}{\mathrm{mg}}\right) \mathrm{y} \ldots\) (iii) Let \(r^{\prime}\) be the equilibrium separation when \(y \rightarrow \frac{y}{2}\) So, equation (iii) then reduces to: \(\mathrm{r}^{\prime 3}=\left(\frac{\mathrm{kq}^2}{\mathrm{mg}}\right) \frac{\mathrm{y}}{2} \ldots \text { (iv) }\) Dividing (iv) by (iii) we get, So, \(\frac{\mathrm{r}^{\prime 3}}{\mathrm{r}^3}=\frac{1}{2}\) \(r^{\prime}=r\left(\frac{1}{2}\right)^{1 / 3}\)