Question

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

Answer 1

Answer: (C)

From the given condition, we have

$F _{\text {netq }} =-\left[2 F _{ q / q } \cos \theta\right]$
$F _{\text {netq }} =-2 \cdot \frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{ q ^{2}}{\left(\sqrt{ d ^{2}+ x ^{2}}\right)^{2}} \cdot \frac{ x }{\sqrt{ d ^{2}+ x ^{2}}}$
$=-\frac{ q ^{2}}{2 \pi \varepsilon_{0}} \frac{ x }{\left( d ^{2}+ x ^{2}\right)^{3 / 2}}$
For $x << d$
$F _{ net q }=-\frac{ q ^{2}}{2 \pi \varepsilon_{0} d ^{3}} x$
$\therefore a =-\frac{ q ^{2}}{2 \pi \varepsilon_{0} \cdot md ^{3}} x$
Comparing with equation of $SHM \left( a =-\omega^{2} x \right)$
$\therefore \omega=\sqrt{\frac{ q ^{2}}{2 \pi \varepsilon_{0} md ^{3}}}$