Question

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 seperation between the balls now become

• A. ${\left(\frac{1}{\sqrt{2}}\right)}^2$
• B. $\left(\frac{r}{\sqrt[3]{32}}\right)$
• C. $\left(\frac{2r}{\sqrt{3}}\right)$
• D. $\left(\frac{2r}{3}\right)$

Answer 1

The correct option is B $\left(\frac{r}{\sqrt[3]{32}}\right)$

Let the length of the strings be $L$ and mass of the ball be $m$ and charge be $q$.

At equilibrium, $\sum F_x=0$ and $\sum F_y=0$

$\therefore$ $Tsin\theta =mg$ ..........(1)

Also $Tcos\theta =F_e$ $⟹Tcos\theta =\frac{K{q}^2}{r^2}$ .......(2) where $K=\frac{1}{4\pi {ϵ}_o}$

Dividing (1) and (2) gives $r^2=\frac{mg}{K{q}^2}tan\theta$

Now as $\frac{mg}{K{q}^2}=constant=C$ and $tan\theta =\frac{y}{\frac{r}{2}}$

$\therefore$ $r^2=C\times \frac{2y}{r}$ $⟹r\propto (y{)}^{\frac{1}{3}}$

Thus $\frac{r^{\prime }}{r}=\frac{(y^{\prime }{)}^{\frac{1}{3}}}{y^{\frac{1}{3}}}$

Now $y^{\prime }=\frac{y}{2}$ $⟹r^{\prime }=\frac{r}{2^{\frac{1}{3}}}$