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Two identical positive charges $\mathrm{ Q}$ each are fixed at a distance of $\mathrm{ ' 2 \mathrm{a} '}$ apart from each other. Another point charge $\mathrm{ \mathrm{q}_{0}}$ with mass $\mathrm{ ' \mathrm{m} '}$ is placed at midpoint between two fixed charges. For a small displacement along the line joining the fixed charges, the charge $\mathrm{ \mathrm{q}_{0}}$ executes SHM. The time period of oscillation of charge $\mathrm{q}_{0}$ will be :
Option: 1
$\mathrm{\sqrt{\frac{4 \pi^{3} \varepsilon_{0} m a^{3}}{q_{0} Q}}}$

Option: 2
$\mathrm{\sqrt{\frac{q_{0} Q}{4 \pi^{3} \varepsilon_{0} m a^{3}}}}$

Option: 3
$\mathrm{\sqrt{\frac{2 \pi^{2} \varepsilon_{0} m a^{3}}{q_{0} Q}}}$

Option: 4
$\mathrm{\sqrt{\frac{8 \pi^{3} \varepsilon_{0} m a^{3}}{q_{0} Q}}}$

Answers (1)

best_answer

If change $\mathrm{q_{o}}$ is displaced by $\mathrm{x}$ from point A to B

$\mathrm{F_{2}=\frac{KQq_{o}}{\left (a-x \right )^{2}}}$

$\mathrm{F_{1}=\frac{KQq_{o}}{\left (a+x \right )^{2}}}$

$\mathrm{F_{net}=F_{2}-F_{1}}$

$\mathrm{=KQq_{0}\left [ \frac{1}{\left ( a-x \right )^{2}}-\frac{1}{\left ( a+x \right )^{2}} \right ]}$

$\mathrm{F_{\text {net }}=k Qq_0\left[\frac{4 a x}{\left(a^2-x^2\right)^2}\right]}$

$\mathrm{a_{\text {net }}=\frac{4 K Q q_0 a x}{m\left(a^2-x^2\right)^2}}$

$\mathrm{x<< a}$

$\mathrm{\therefore {a_{\text {net }}}=-\left(\frac{4 K Q q_0 a}{m a^4}\right){\mathrm{x}}}$

$\mathrm{\omega^2=\frac{4 K Q q_0}{m a^3}}$

$\mathrm{T=\frac{2 \pi}{\omega}=2 \pi \sqrt{\frac{m a^3}{4 k Q q_0}}}$

$\mathrm{T=2 \pi \sqrt{\frac{m a^3 \times 4 \pi \varepsilon_0}{4 Q q_0}}}$

$\mathrm{T=\sqrt{\frac{4 \pi^3 m a^3 \varepsilon_0}{Q q_0}}}$

Hence 1 is correct option

Posted by

Kuldeep Maurya

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