Question 32 of 311

A particle of mass $1mg$ and charge $q$ is lying at the mid-point of two stationary particles kept at a distance $2m$ when each is carrying same charge $q$. If the free charged particle is displaced from its equilibrium position through distance $x$ $\left(x\lt \lt 1m\right)$. The particle executes SHM. Its angular frequency of oscillation will be _______ $\times {10}^{5}rad{s}^{-1}$ (if ${q}^{2}=10{C}^{2}$)

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Answer: 6000

Net force on free charged particle,
$F=\dfrac{k{q}^{2}}{{\left(d+x\right)}^{2}}-\dfrac{k{q}^{2}}{{\left(d-x\right)}^{2}}$
$F=-k{q}^{2}\left[\dfrac{4dx}{{\left({d}^{2}-{x}^{2}\right)}^{2}}\right]$
$a=-\dfrac{4k{q}^{2}d}{m}\left(\dfrac{x}{{d}^{4}}\right)$
$a=-\left(\dfrac{4k{q}^{2}}{m{d}^{3}}\right)x$
So, angular frequency
$\omega =\sqrt{\dfrac{4k{q}^{2}}{m{d}^{3}}}$
$\omega =\sqrt{\dfrac{4\times 9\times {10}^{9}\times 10}{1\times {10}^{-6}\times {1}^{3}}}$
$\omega =6\times {10}^{8}rad{\sec }^{-1}$