Question

A particle of mass $M$ and charge $q$ is at rest at 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 ($\epsilon_0$ = permittivity of free space

• A. $2\sqrt{\frac{\pi^{2}M\varepsilon_{0}d}{Qq}}$
• B. $2\sqrt{\frac{\pi^{2}M\varepsilon_{0}d^{3}}{Qq}}$
• C. $2\sqrt{\frac{\pi^{3}M\varepsilon_{0}d^{3}}{Qq}}$
• D. $2\sqrt{\frac{\pi^{3}M\varepsilon_{0}}{Qqd^{3}}}$

Answer 1

Answer: (C)

Restoring force on displacement of $x$
$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) $
$\Rightarrow F=K Q q\left[\frac{4 x}{d^{3}}\right]$
Acceleration, $a=\frac{F}{m}=\frac{4 K Q q x}{M d^{3}}$
or $\omega^{2}=\frac{4 K Q q}{M d^{3}}$
$\therefore T=\frac{2 \pi}{\omega}=2 \pi \sqrt{\frac{M d^{3}}{4 K Q q}}$
$=2 \sqrt{\frac{\pi^{3} M d^{3} \varepsilon_{0}}{Q q}}$