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QJEEM22Phb43d: JEE Main 2022 Physics - Class 12 JEEM22Phb43d Solution: Two identical positive charges each are fixed at a distance of…

Two identical positive charges $Q$ each are fixed at a distance of '2a' apart from each other. Another point charge $q_{0}$ with mass 'm' is placed at midpoint between two fixed charges. For a small displacement along the line joining the fixed charges, the charge $\mathrm{q}_{0}$ executes $\mathrm{SHM}$ . The time period of oscillation of charge $\mathrm{q}_{0}$ will be :

Multiple Choice Options

A $\sqrt{\frac{4 \pi^{3} \varepsilon_{0} m a^{3}}{q_{0} Q}}$
B $\sqrt{\frac{q_{0} Q}{4 \pi^{3} \varepsilon_{0} m a^{3}}}$
C $\sqrt{\frac{2 \pi^{2} \varepsilon_{0} m a^{3}}{q_{0} Q}}$
D $\sqrt{\frac{8 \pi^{3} \varepsilon_{0} m a^{3}}{q_{0} Q}}$

Step-by-Step Verified Solution

JEE Main 2022 (Online) 27th July Morning Shift Physics - Electrostatics Question 92 English Explanation

(x << a) ( $\alpha$ is acceleration)

${F_{net}} = - \left( {{{k{q_0}Q} \over {{{(a - x)}^2}}} - {{kQ{q_0}} \over {{{(a + x)}^2}}}} \right)$

$m\alpha = - {{k{q_0}Q} \over {{a^4}}}4ax$

$\Rightarrow \alpha = - {{4k{q_0}Q} \over {m{a^3}}}x$

So, $T = 2\pi \sqrt {{{4\pi {\varepsilon _0}m{a^3}} \over {4{q_0}Q}}}$

or $T = \sqrt {{{4{\pi ^3}{\varepsilon _0}m{a^3}} \over {{q_0}Q}}}$

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