Question

Two identical pith balls are charged by rubbing one against the other. They are suspended from a horizontal rod through two strings of length 20 cm each, the separation between the suspension points being 5 cm. In equilibrium, the separation between the balls is 3 cm. Find the mass of each ball and the tension in the strings. The charge on each ball has a magnitude 2.0 × 10⁻⁸ C.

Answer 1

Let the tension in the string be T and the force of attraction between the two balls be F.
From the free-body diagram of the balls, we get
Tcosθ = mg ...(1)
Tsinθ = F ...(2)
From ∆ABC,
$$\begin{gathered} \sin \theta =\frac{1}{20} \\ \cos \theta =\sqrt{1-{\left(\frac{1}{20}\right)}^2} \end{gathered}$$
From equation (1) and (2),
$$\begin{gathered} \tan \theta =\frac{F}{mg} \\ \therefore m=\frac{F}{g\tan \theta } \\ \Rightarrow m=\left(\frac{1}{4\mathrm{\pi }{\epsilon }_0}\frac{q^2}{r^2}\right)\times \left(\frac{1}{g\tan \theta }\right) \\ \Rightarrow m=\frac{9\times {10}^9\times {\left(2\times {10}^{-8}\right)}^2}{9\times {10}^{-4}\times 0.98\times \tan \theta } \end{gathered}$$

$\Rightarrow m=0.0082\mathrm{kg}=8.2\mathrm{g}$
$$\begin{gathered} T=\frac{F}{\sin \theta }=\frac{9\times {10}^9\times {\left(2\times {10}^{-8}\right)}^2}{9\times {10}^{-4}\times {\displaystyle \frac{1}{20}}} \\ \Rightarrow T=8.2\times {10}^2\mathrm{N} \end{gathered}$$