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 (${ϵ}_0$ = permittivity of free space

• A. $2\sqrt{\frac{{\pi }^{2}M{\epsilon }_{0}d}{Qq}}$
• B. $2\sqrt{\frac{{\pi }^{2}M{\epsilon }_0{d}^3}{Qq}}$
• C. $2\sqrt{\frac{{\pi }^{3}M{\epsilon }_0{d}^3}{Qq}}$
• D. $2\sqrt{\frac{{\pi }^{3}M{\epsilon }_0}{Qq{d}^3}}$

Answer 1

Answer: (C)

Restoring force on displacement of $x$
$F=K\left[\frac{q}{(d-x{)}^2}-\frac{Qq}{(d+x{)}^2}\right]$
$=KQq\left[\frac{1}{(d-x{)}^2}-\frac{1}{(d+x{)}^2}\right]$
$=KQq\left[\frac{4dx}{{\left(d^2-x^2\right)}^2}\right]$
$=KQq\left[\frac{4dx}{d^4}\right]\text{ If }(d>x)$
$\Rightarrow F=KQq\left[\frac{4x}{d^3}\right]$
Acceleration, $a=\frac{F}{m}=\frac{4KQqx}{M{d}^3}$
or ${\omega }^2=\frac{4KQq}{M{d}^3}$
$\therefore T=\frac{2\pi }{\omega }=2\pi \sqrt{\frac{M{d}^3}{4KQq}}$
$=2\sqrt{\frac{{\pi }^{3}M{d}^3{\epsilon }_0}{Qq}}$