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A particle of mass $M$ and charge $q$ is at rest at the mid point between two other fixed similar charges each of magnitude $Q$ placed a distance $2 d$ apart. The system is collinear as shown in the figure. The particle is now displaced by a small amount $x(x< < d)$ along the joining the two charges and is left to itself. It will now oscillate about the mean position with a time period $\left(\varepsilon_{0}=\right.$ permittivity of free space)
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The correct answer is:
$2 \sqrt{\frac{\pi^{3} M \varepsilon_{0} d^{3}}{Q q}}$
Restoring force on displacement of $x$
$\begin{aligned} F &=K\left[\frac{q}{(d-x)^{2}}-\frac{Q q}{(d+x)^{2}}\right] \\ &=K Q q\left[\frac{1}{(d-x)^{2}}-\frac{1}{(d+x)^{2}}\right] \\ &=K Q q\left[\frac{4 d x}{\left(d^{2}-x^{2}\right)^{2}}\right] \\ &=K Q q\left[\frac{4 d x}{d^{4}}\right] \text { If }(d>>x) \\ F &=K Q q\left[\frac{4 x}{d^{3}}\right] \end{aligned}$
Acceleration, $a=\frac{F}{m}=\frac{4 K Q q x}{M d^{3}}$
or $\omega^{2}=\frac{4 K Q q}{M d^{3}}$
$\therefore \quad T=\frac{2 \pi}{\omega}=2 \pi \sqrt{\frac{M d^{3}}{4 K O q}}$
$=2 \sqrt{\frac{\pi^{3} M d^{3} \varepsilon_{0}}{Q q}}$
$\begin{aligned} F &=K\left[\frac{q}{(d-x)^{2}}-\frac{Q q}{(d+x)^{2}}\right] \\ &=K Q q\left[\frac{1}{(d-x)^{2}}-\frac{1}{(d+x)^{2}}\right] \\ &=K Q q\left[\frac{4 d x}{\left(d^{2}-x^{2}\right)^{2}}\right] \\ &=K Q q\left[\frac{4 d x}{d^{4}}\right] \text { If }(d>>x) \\ F &=K Q q\left[\frac{4 x}{d^{3}}\right] \end{aligned}$
Acceleration, $a=\frac{F}{m}=\frac{4 K Q q x}{M d^{3}}$
or $\omega^{2}=\frac{4 K Q q}{M d^{3}}$
$\therefore \quad T=\frac{2 \pi}{\omega}=2 \pi \sqrt{\frac{M d^{3}}{4 K O q}}$
$=2 \sqrt{\frac{\pi^{3} M d^{3} \varepsilon_{0}}{Q q}}$
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