Two identical tennis balls each having mass 'm' and charge 'q' are… | JEE Main 2021 (27 Jul, Shift 1)

Q60JEE Main 2021 · 27 JUL · SHIFT 1Medium

Electric Charge & Coulomb's Law

Two identical tennis balls each having mass 'm' and charge 'q' are suspended from a fixed point by threads of length 'l'. What is the equilibrium separation when each thread makes a small angle ' $\theta$ ' with the vertical?

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Answer: (b)

For each ball $\tan\theta = \dfrac{F}{mg}$ ; for small $\theta$ , $\tan\theta \approx \sin\theta = \dfrac{x/2}{l}$ .
$\dfrac{x}{2l} = \dfrac{q^2}{4\pi\varepsilon_0 x^2 mg}$
$x^3 = \dfrac{2lq^2}{4\pi\varepsilon_0 mg} = \dfrac{q^2 l}{2\pi\varepsilon_0 mg}$
So the answer is (b) $x = \left(\dfrac{q^2 l}{2\pi\varepsilon_0 mg}\right)^{1/3}$ .

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