A particle of mass M and charge q is at rest the midpoint between two other fixed similar charges each of magnitude Q placed a distance 2d apart. The system is collinear as shown in the figure. The particle is now displaced by a small amount $x (x < < d)$ along the line joining the two charges and is left to itself. It will now oscillate about the mean position with a time period $(ε_{0} =$ permittivity of free space)

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$2 \sqrt[]{\frac{π^{2} M ε_{0} d^{3}}{Qq}}$

$2 \sqrt[]{\frac{π^{3} M ε_{0} d}{Qq}}$

$2 \sqrt[]{\frac{π^{3} M ε_{0}}{{Qqd}^{3}}}$

$2 \sqrt[]{\frac{π^{3} M ε_{0} d^{3}}{Qq}}$

answer is C.

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Detailed Solution

Restoring force on displacement of x,

$\begin{array}{l} F = K [\frac{Qq}{{(d-x)}^{2}} - \frac{Qq}{{(d+x)}^{2}}] \\ = KQq [\frac{1}{{(d - x)}^{2}} - \frac{1}{{(d + x)}^{2}}] = KQq [\frac{4 d x}{{(d^{2} - x^{2})}^{2}}] \\ = KQq [\frac{4 d x}{d^{4}}] = KQq [\frac{4 x}{d^{3}}] i f (d > > x) \\ a c c e l e r a t i o n a= \frac{F}{M} = \frac{4KQq}{{Md}^{3}} x \Rightarrow ω^{2} = \frac{4KQq}{^{3}} \\ T= \frac{2π}{ω} =2 π \sqrt[]{\frac{{Md}^{3}}{4KQq}} =2 \sqrt[]{\frac{π^{3} {Md}^{3} ε_{0}}{Qq}} \end{array}$