Question

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 $2d$ apart. The system is colinear as shown in 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:

(${\epsilon }_0=$ permittivity of free psace)

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

$2\sqrt{\frac{{\pi }^{3}M{\epsilon }_0{d}^3}{Qq}}$

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