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A particle of mass m and charge -q moves diametrically through a uniformly charged sphere of radius R with total charge Q. The angular frequency of the particle's simple harmonic motion, if its amplitude <R, is given by

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$\sqrt{\frac{1}{4 π ε_{0}} \frac{q Q}{m R}}$

$\sqrt{\frac{1}{4 π ε_{0}} \frac{q Q}{m R^{2}}}$

$\sqrt{\frac{1}{4 π ε_{0}} \frac{q Q}{m R^{3}}}$

$\sqrt{\frac{1}{4 π ε_{0}} \frac{m}{q Q}}$

answer is C.

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Detailed Solution

$\begin{array}{l} F = \frac{k Q r q}{R^{3}} = - m a \\ a = - (\frac{Q q}{4 π ε_{0} m R^{3}}) r = - ω^{2} r or ω = \sqrt{\frac{Q q}{4 π ε_{0} m R^{3}}} \end{array}$

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