Question

Two identical pith balls, each carrying a charge q, are suspended from a common point by two strings of equal length l. Find the mass of each ball if the angle between the strings is 2θ in equilibrium.

Answer 1

Let the tension in the string be T and force of attraction between the two balls be F.
From the free-body diagram of the ball, we get
Tcos θ = mg ...(1)
Tsin θ = F ...(2)
$$\begin{gathered} \Rightarrow \tan \theta =\frac{F}{mg} \\ \Rightarrow m=\frac{F}{g\tan \theta }\dots \left(3\right) \end{gathered}$$
Here, separation between the two charges, $r=2l\sin \theta$
$F=\frac{1}{4\mathrm{\pi }{\epsilon }_0}\frac{q^2}{r^2}=\frac{1}{4\mathrm{\pi }{\epsilon }_0}\times \frac{q^2}{{\left(2l\sin \theta \right)}^2}$
Substituting the value of F in equation (3), we get
$m=\frac{q^2\cot \theta }{16\mathrm{\pi }{\epsilon }_{0}g{l}^2{\sin }^2\theta }$