Question 36 of 311

Two electrons each are fixed at a distance $2d.$ A third charge proton placed at the midpoint is displaced slightly by a distance $x\left(x≪d\right)$ perpendicular to the line joining the two fixed charges. Proton will execute simple harmonic motion having angular frequency:
($m=$ mass of charged particle)

A${\left(\dfrac{\pi {\epsilon }_{0}m{d}^{3}}{2{q}^{2}}\right)}^{\dfrac{1}{2}}$B${\left(\dfrac{2\pi {\epsilon }_{0}m{d}^{3}}{{q}^{2}}\right)}^{\dfrac{1}{2}}$C${\left(\dfrac{{q}^{2}}{2\pi {\epsilon }_{0}m{d}^{3}}\right)}^{\dfrac{1}{2}}$D${\left(\dfrac{2{q}^{2}}{\pi {\epsilon }_{0}m{d}^{3}}\right)}^{\dfrac{1}{2}}$

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Answer: (C) ${\left(\dfrac{{q}^{2}}{2\pi {\epsilon }_{0}m{d}^{3}}\right)}^{\dfrac{1}{2}}$

From the given condition, we have

${F}_{netq}=-\left[2{F}_{q/q}\cos \theta \right]$
${F}_{netq}=-2\cdot \dfrac{1}{4\pi {\epsilon }_{0}}\cdot \dfrac{{q}^{2}}{{\left(\sqrt{{d}^{2}+{x}^{2}}\right)}^{2}}\cdot \dfrac{x}{\sqrt{{d}^{2}+{x}^{2}}}$
$=-\dfrac{{q}^{2}}{2\pi {\epsilon }_{0}}\dfrac{x}{{\left({d}^{2}+{x}^{2}\right)}^{3/2}}$
For $x\lt \lt d,$
${F}_{netq}=-\dfrac{{q}^{2}}{2\pi {\epsilon }_{0}{d}^{3}}x$
$∴a=-\dfrac{{q}^{2}}{2\pi {\epsilon }_{0}\cdot m{d}^{3}}x$
Comparing with equation of SHM $\left(a=-{\omega }^{2}x\right)$
$∴\omega =\sqrt{\dfrac{{q}^{2}}{2\pi {\epsilon }_{0}m{d}^{3}}}$