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)

\[\Rightarrow \tan\theta = \frac{F}{mg}\] 

\[ \Rightarrow m = \frac{F}{g\tan\theta}        \ldots\left( 3 \right)\]

Here, separation between the two charges,  

\[r = 2l\sin\theta\]

\[F = \frac{1}{4\pi \epsilon_0}\frac{q^2}{r^2} = \frac{1}{4\pi \epsilon_0} \times \frac{q^2}{\left( 2l\sin\theta \right)^2}\]

Substituting the value of F in equation (3), we get

\[m = \frac{q^2 \cot\theta}{16\pi \epsilon_0 g l^2 \sin^2 \theta}\]