Two electrons each are fixed at a distance 2d. A third charge proton placed at the midpoint is displaced slightly by a distance x(x <<d) perpendicular to the line joining the two fixed charges. Proton will execute simple harmonic motion having angular frequency:

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${(\frac{π ε_{0} m d^{3}}{2 q^{2}})}^{\frac{1}{2}}$

${(\frac{2 π ε_{0} m d^{3}}{q^{2}})}^{\frac{1}{2}}$

${(\frac{2 q^{2}}{π ε_{0} m d^{3}})}^{\frac{1}{2}}$

${(\frac{q^{2}}{2 π ε_{0} m d^{3}})}^{\frac{1}{2}}$

answer is D.

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Detailed Solution

${Restoring force, F}_{R} = - 2 (F c o s θ)$

$\begin{array}{l} \Rightarrow F_{R} = - 2 (\frac{k e^{2}}{{(\sqrt{d^{2} + x^{2}})}^{2}} \times \frac{x}{\sqrt{d^{2} + x^{2}}}) \\ \Rightarrow m a = - \frac{2 k e^{2}}{{(d^{2} + x^{2})}^{\frac{3}{2}}} x \\ Since, x << d \\ \Rightarrow m a \approx - \frac{2 k e^{2}}{d^{3}} x \\ \Rightarrow a = - [\frac{2 k e^{2}}{m d^{3}}] x \end{array}$

$\begin{array}{l} C o n p a r i n g w i t h a = - ω^{2} x \\ \Rightarrow ω = {(\frac{e^{2}}{2 π ε_{0} m d^{3}})}^{\frac{1}{2}} \\ o r \\ ω = {(\frac{q^{2}}{2 π ε_{0} m d^{3}})}^{\frac{1}{2}} \end{array}$