Question

Two pith-balls of same weight are suspended from the same point by means of silk threads of equal length. On charging the pith-balls equally, they are found to repel each other to a certain distance and remain at equilibrium. Identify the forces responsible for them to remain under equilibrium.

Answer 1

Given that two pith balls of same weight are suspended from same point by means of silk threads of equal length. On charging the pith balls equally, they repel each other. We have to find forces responsible to remain under equilibrium.

According to the figure,

At equilibrium, force of repulsion is balanced by the $\sin$ component of tension in the string i.e, $F=T\sin \theta$ ...(i)

Weight is balanced by the $\cos$ component of the tension $mg=T\cos \theta$ ...(ii)

where $m=$ mass of each pith ball

$T=$ tension in the string

$F=$ force of repulsion

$g=$ acceleration due to gravity

Dividing (i) and (ii), $\tan \theta =\frac{F}{mg}$

$\Rightarrow F=mg\tan \theta$

This is the force under equilibrium.