Two pith balls carrying equal charges are suspended from a common...

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})