JEE Mains · Physics · STD 12 - 1. Electric charges and fields
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?
- A \({x}=\left(\frac{{q}^{2} l}{2 \pi \varepsilon_{0} {mg}}\right)^{1 / 2}\)
- B \({x}=\left(\frac{{q}^{2} l^{2}}{2 \pi \varepsilon_{0} {m}^{2} {g}^{2}}\right)^{1 / 3}\)
- C \({x}=\left(\frac{{q}^{2} l}{2 \pi \varepsilon_{0} {mg}}\right)^{1 / 3}\)
- D \({x}=\left(\frac{{q}^{2} l^{2}}{2 \pi \varepsilon_{0} {m}^{2} {g}}\right)^{1 / 3}\)
Answer & Solution
Correct Answer
(C) \({x}=\left(\frac{{q}^{2} l}{2 \pi \varepsilon_{0} {mg}}\right)^{1 / 3}\)
Step-by-step Solution
Detailed explanation
\(T \cos \theta= mg\) \(T \sin \theta=\frac{ kq ^{2}}{ x ^{2}}\) \(\tan \theta=\frac{k q^{2}}{x^{2} m g}\) as \(\tan \theta \approx \sin \theta \approx \frac{x}{2 L}\) \(\frac{ x }{2 L }=\frac{ Kq ^{2}}{ x ^{2} mg }\)…
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