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A particle of mass $1 \,{mg}$ and charge $q$ is lying at the mid-point of two stationary particles kept at a distance $'2 \,{m}^{\prime}$ when each is carrying same charge $'q'.$ If the free charged particle is displaced from its equilibrium position through distance $'x'$ $(x\,< \,1\, {m})$. The particle executes $SHM.$ Its angular frequency of oscillation will be $....\,\times 10^{8}\, {rad} / {s}$ if ${q}^{2}=10\, {C}^{2}$
$60$
$6$
$76$
$760$
Solution
Net force on free charged particle
$F =\frac{ kq ^{2}}{( d + x )^{2}}-\frac{ kq ^{2}}{( d – x )^{2}}$
$F =- kq ^{2}\left[\frac{4 dx }{\left( d ^{2}- x ^{2}\right)^{2}}\right]$
$a =-\frac{4 kq ^{2} d }{ m }\left(\frac{ x }{ d ^{4}}\right)$
$a =-\left(\frac{4 kq ^{2}}{ md ^{3}}\right) x$
So, angular frequency
$\omega=\sqrt{\frac{4 kq ^{2}}{ md ^{3}}}$
$\omega=\sqrt{\frac{4 \times 9 \times 10^{9} \times 10}{1 \times 10^{-6} \times 1^{3}}}$
$\omega=6 \times 10^{8}\; rad / sec$
