Two electrons each are fixed at a distance '2d'. A third charge p… - Vidyadip

MCQ

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)

A
$\left(\frac{2 q^{2}}{\pi \varepsilon_{0} m d^{3}}\right)^{\frac{1}{2}}$
B
$\left(\frac{\pi \varepsilon_{0} md ^{3}}{2 q ^{2}}\right)^{\frac{1}{2}}$
✓
$\left(\frac{ q ^{2}}{2 \pi \varepsilon_{0} md ^{3}}\right)^{\frac{1}{2}}$
D
$\left(\frac{2 \pi \varepsilon_{0} md ^{3}}{ q ^{2}}\right)^{\frac{1}{2}}$

✓

Answer

Correct option: C.

$\left(\frac{ q ^{2}}{2 \pi \varepsilon_{0} md ^{3}}\right)^{\frac{1}{2}}$

c
From the given condition, we have

$F _{\text {net}}=-\left[2 F _{ q / q } \cos \theta\right]$

$F _{\text {net}}=-2 \cdot \frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{ q ^{2}}{\left(\sqrt{ d ^{2}+ x ^{2}}\right)^{2}} \cdot \frac{ x }{\sqrt{ d ^{2}+ x ^{2}}}$

$=-\frac{q^{2}}{2 \pi \varepsilon_{0}} \frac{x}{\left(d^{2}+x^{2}\right)^{3 / 2}}$

For $x ( x << d )$

$F_{\text {net }}=-\frac{q^{2}}{2 \pi \varepsilon_{0} d^{3}} x$

$\therefore a =-\frac{ q ^{2}}{2 \pi \varepsilon_{0} \cdot md ^{3}} x$

Comparing with equation of $SHM \left( a =-\omega^{2} x \right)$

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

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