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 separation between the balls now becomes-

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

Answer 1

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


From the FBD of the pith ball,

$T\cos \theta =mg$

$T\sin \theta =\frac{K{q}^2}{r^2}$

$\tan \theta =\frac{K{q}^2}{r^{2}mg}$

$\frac{\left(\frac{r}{2}\right)}{y}=\frac{k{q}^2}{r^{2}mg}$

$y=\frac{r^{3}mg}{2k{q}^2}$

$\Rightarrow y\propto r^3$

$\therefore {\left(\frac{r^{\prime }}{r}\right)}^3=\frac{\left(\frac{y}{2}\right)}{y}$

$\Rightarrow r^{\prime }=r{\left(\frac{1}{2}\right)}^{1/3}=\frac{r}{\sqrt[3]{32}}$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, $\left(C\right)$ is the correct answer.