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  • 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 moti

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 : (m = mass of charged particle) 
Option: 1 \begin{array}{l} \left(\frac{2 q^{2}}{\pi \varepsilon_{0} m d^{3}}\right)^{\frac{1}{2}} \\ \end{array}
Option: 2 \left(\frac{q^{2}}{2 \pi \varepsilon_{0} m d^{3}}\right)^{\frac{1}{2}} \\
Option: 3 \left(\frac{\pi \varepsilon_{0} m d^{3}}{2 q^{2}}\right)^{\frac{1}{2}}
Option: 4 \left(\frac{2 \pi \varepsilon_{0} m d^{3}}{q^{2}}\right)^{\frac{1}{2}}

Answers (1)

best_answer

\begin{aligned} &\text { Restoring force on proton:- }\\ &F_{r}=2 F_{1} \sin \theta \text { where } F_{1}=-\frac{k q^{2} }{\left(d^{2}+x^{2}\right)}\\ &\mathrm{F}_{\mathrm{r}}=-\frac{2 \mathrm{kq}^{2} \mathrm{x} }{\left[\mathrm{d}^{2}+\mathrm{x}^{2}\right]^{3 / 2}} \end{aligned}

\begin{array}{l} For \ \ x<<<d \\ F_{r}=-\frac{2 k q^{2} x }{d^{3} }=-\frac{q^{2} x}{2 \pi \varepsilon_{0} d^{3}} \end{array}

And we know that restoring force in SHM is given as

F_r=-\omega ^2x

on comparing we get

\omega=\sqrt{\frac{\mathrm{q}^{2}}{2 \pi \varepsilon_{0} \mathrm{md}^{3}}}

 

 

Posted by

avinash.dongre

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