JEE MAIN - Physics (2022 - 27th July Morning Shift - No. 10)
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 :
$$\sqrt{\frac{4 \pi^{3} \varepsilon_{0} m a^{3}}{q_{0} Q}}$$
$$\sqrt{\frac{q_{0} Q}{4 \pi^{3} \varepsilon_{0} m a^{3}}}$$
$$\sqrt{\frac{2 \pi^{2} \varepsilon_{0} m a^{3}}{q_{0} Q}}$$
$$\sqrt{\frac{8 \pi^{3} \varepsilon_{0} m a^{3}}{q_{0} Q}}$$
Explanation
_27th_July_Morning_Shift_en_10_1.png)
(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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