NEET UG 2013 — Physics Electromagnetism

Given two pith balls carrying equal charges are suspended from a common point by strings of equal length. The equilibrium separation between them is $r$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, $\tan \theta=\frac{\mathrm{F}}{\mathrm{mg}}$tanθ=mgF​...(i) and also, $\tan \theta=\frac{\mathrm{r} / 2}{\mathrm{y}} \ldots$tanθ=yr/2​… (ii) Equating (i) and (ii), we get $\frac{\mathrm{F}}{\mathrm{mg}}=\frac{\mathrm{r}}{2 \mathrm{y}}$mgF​=2yr​ also, Force, $\mathrm{F}=\frac{\mathrm{kq}^2}{\mathrm{r}^2}$F=r2kq2​ where $\mathrm{k}=$k= constant $=9 \times 10^9$=9×109 So, we get $\mathrm{r}^3=\left(\frac{\mathrm{kq}^2}{\mathrm{mg}}\right) \mathrm{y} \ldots$r3=(mgkq2​)y… (iii) Let $r^{\prime}$r′ be the equilibrium separation when $y \rightarrow \frac{y}{2}$y→2y​ So, equation (iii) then reduces to: $\mathrm{r}^{\prime 3}=\left(\frac{\mathrm{kq}^2}{\mathrm{mg}}\right) \frac{\mathrm{y}}{2} \ldots \text { (iv) }$r′3=(mgkq2​)2y​… (iv)  Dividing (iv) by (iii) we get, So, $\frac{\mathrm{r}^{\prime 3}}{\mathrm{r}^3}=\frac{1}{2}$r3r′3​=21​ $r^{\prime}=r\left(\frac{1}{2}\right)^{1 / 3}$r′=r(21​)1/3