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(B) Let the distance between the two stationary charges be $2d = 2 \, m$,so $d = 1 \, m$. The mass of the particle is $m = 1 \, mg = 10^{-6} \, kg$.Th

Vedclass · Accepted Answer

$6$

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(B) Let the distance between the two stationary charges be $2d = 2 \, m$,so $d = 1 \, m$. The mass of the particle is $m = 1 \, mg = 10^{-6} \, kg$.The net force on the free charged particle when displaced by $x$ is:$F = \frac{kq^{2}}{(d-x)^{2}} - \frac{kq^{2}}{(d+x)^{2}}$$F = kq^{2} \left[ \frac{(d+x)^{2} - (d-x)^{2}}{(d^{2}-x^{2})^{2}} ight] = kq^{2} \left[ \frac{4dx}{(d^{2}-x^{2})^{2}} ight]$Since $x \ll d$,we can approximate $(d^{2}-x^{2})^{2} \approx d^{4}$:$F \approx \frac{4kq^{2}dx}{d^{4}} = \frac{4kq^{2}}{d^{3}} x$Since the force is directed towards the equilibrium position,$F = -kx_{spring} = -m\omega^{2}x$. Thus:$m\omega^{2} = \frac{4kq^{2}}{d^{3}}$$\omega = \sqrt{\frac{4kq^{2}}{md^{3}}}$Substituting the values $k = 9 	imes 10^{9} \, N \cdot m^{2}/C^{2}$,$q^{2} = 10 \, C^{2}$,$m = 10^{-6} \, kg$,and $d = 1 \, m$:$\omega = \sqrt{\frac{4 	imes 9 	imes 10^{9} 	imes 10}{10^{-6} 	imes 1^{3}}} = \sqrt{36 	imes 10^{16}} = 6 	imes 10^{8} \, rad/s$.Thus,the angular frequency is $6 	imes 10^{8} \, rad/s$.

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