Question

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

Answer 1

Given:$q=2.0\times {10}^{-8}$c,$\sin \theta =\frac{1}{20}$

Force between the charges

$F=\frac{k{q}_1{q}_2}{r^2}=\frac{9\times {10}^9\times 2\times {10}^{-8}\times 2\times {10}^{-8}}{{(3\times {10}^{-2})}^2}$

$\frac{9\times 4\times {10}^{-7}}{9\times {10}^{-4}}=4\times {10}^{-3}$N

$mg\sin \theta =F\Rightarrow m=\frac{F}{g\sin \theta }=\frac{4\times {10}^{-3}}{10\times \frac{1}{20}}=8\times {10}^{-3}=8$gm.

$\cos \theta =\sqrt{1-{\sin }^2\theta }=\sqrt{1-\frac{1}{400}}=\sqrt{\frac{400-1}{400}}=0.99=1$(approx.)

So, $T=mg\cos \theta$

or $T=8\times {10}^{-3}\times 10\times 0.99=8\times {10}^{-2}$N