Question

A particle of mass $1\text{mg}$ and charge $q$ is lying at the mid-point of two stationary particles kept at a distance $2\text{m}$, when each is carrying same charge ${}^{\prime }{q}^{\prime }.$ If the free charged particle is displaced from its equilibrium position through distance ${)}^{\prime }{x}^{\prime }(x<<1\text{m}).$ The particle executes SHM. Its angular frequency of oscillation will be____ $\times {10}^8\text{rad/s}$ if $q^2=10{\text{C}}^2$.

Answer 1

Net force on free charged particle is,
$F=\frac{k{q}^2}{(d+x{)}^2}-\frac{k{q}^2}{(d-x{)}^2}$
$F=-k{q}^2\left[\frac{4xd}{(d^2-x^2{)}^2}\right]$
As $x<<<d\Rightarrow (d^2-x^2)\approx d^2$
$a=-\frac{4k{q}^2}{m}\left(\frac{x}{d^3}\right)$
$a=-\left(\frac{4k{q}^2}{m{d}^3}\right)x$
Comparing the above equation with, $a=-{\omega }^{2}x$
we get the angular frequency as,
$\omega =\sqrt{\frac{4k{q}^2}{m{d}^3}}$
$\omega =\sqrt{\frac{4\times 9\times {10}^9\times 10}{1\times {10}^{-6}\times 1^3}}$
$\omega =6\times {10}^8\text{rad/sec}$