Physics - Electrostatics Question with Solution | TestHub

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, $\tan \theta =\frac{\mathrm{r}/2}{\mathrm{y}}\dots$ (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 ${\mathrm{r}}^3=\left(\frac{{\mathrm{kq}}^2}{\mathrm{mg}}\right)\mathrm{y}\dots$ (iii) Let $r^{\prime }$ be the equilibrium separation when $y\to \frac{y}{2}$ So, equation (iii) then reduces to: ${\mathrm{r}}^{\prime 3}=\left(\frac{{\mathrm{kq}}^2}{\mathrm{mg}}\right)\frac{\mathrm{y}}{2}\dots \text{ (iv) }$ Dividing (iv) by (iii) we get, So, $\frac{{\mathrm{r}}^{\prime 3}}{{\mathrm{r}}^3}=\frac{1}{2}$ $r^{\prime }=r{\left(\frac{1}{2}\right)}^{1/3}$