Question

Two pith balls carrying identical charges are suspended from a common point by strings of equal lengths, the equilibrium separation between them is $r$. Now the strings are rigidly clamped at half the height. The equilibrium separation between the balls now will be :

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

Answer 1

The correct option is A $\frac{r}{2^{1/3}}$


Let $m$ be the mass of each ball and $q$ be the charge on each ball.

Force of repulsion, $F=\frac{1}{4\pi {ϵ}_0}\frac{q^2}{r^2}$

In equilibrium,

$T\cos \theta =mg\dots \dots \left(1\right)$
$T\sin \theta =F\dots \dots \left(2\right)$

From $\left(1\right)$ and $\left(2\right)$ , we get,

$\tan \theta =\frac{F}{mg}=\frac{\frac{1}{4\pi {ϵ}_0}\frac{q^2}{r^2}}{mg}$

From figure (a),

$\frac{\left(\frac{r}{2}\right)}{y}=\frac{\frac{1}{4\pi {ϵ}_0}\frac{q^2}{r^2}}{mg}$ $\dots \dots \left(3\right)$

From figure (b)


$\frac{\left(\frac{r^{{}^{\prime }}}{2}\right)}{\frac{y}{2}}=\frac{\frac{1}{4\pi {ϵ}_0}\frac{q^2}{r^{{}^{\prime }{}^2}}}{mg}$$\dots \dots \left(4\right)$

From $\left(3\right)$ and $\left(4\right)$, we get

$\frac{2r^{{}^{\prime }}}{r}=\frac{r^2}{r^{{}^{\prime 2}}}\Rightarrow r^{{}^{\prime 3}}=\frac{r^3}{2}\Rightarrow r^{{}^{\prime }}=\frac{r}{2^{1/3}}$

Hence, option (d) is the correct answer.