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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 ?
Option: 1 x=\left(\frac{q^{2} l}{2 \pi \varepsilon_{0} m g}\right)^{\frac{1}{2}}
Option: 2 x=\left(\frac{q^{2} l}{2 \pi \varepsilon_{0} m g}\right)^{\frac{1}{3}}
Option: 3 x=\left(\frac{q^{2} l^{2}}{2 \pi \varepsilon_{0} m^{2} g}\right)^{\frac{1}{3}}
Option: 4 \mathrm{x}=\left(\frac{\mathrm{q}^{2} l^{2}}{2 \pi \varepsilon_{0} \mathrm{~m}^{2} \mathrm{~g}^{2}}\right)^{\frac{1}{3}}

Answers (1)

best_answer

\begin{aligned} &\tan \theta=\frac{F}{m g}=\frac{k q^{2} / x^{2}}{m g} \\ &\frac{(x / 2)}{\sqrt{l^{2}-\left(\frac{x}{2}\right)^{2}}}=\frac{k q^{2}}{x^{2} m g} \end{aligned}

\begin{aligned} &x^{3}=\frac{2 k q^{2}}{m g} \sqrt{(\ell)^{2}-\left(\frac{x}{2}\right)^{2}} \\ &\because \theta \rightarrow \text { small } \\ &\therefore x<<l \end{aligned}

\begin{aligned} &x^{3}=\frac{2 \mathrm{kq}^{2} l}{m g}=\frac{2 q^{2} l}{4 \pi \varepsilon_{0} m g} \\ &x=\left(\frac{q^{2} l}{2 \pi \varepsilon_{0} m g}\right)^{1 / 3} \end{aligned}

The correct option is (2)

Posted by

vishal kumar

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