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)

Pick an option to check your answer.

Verified Solution

Answer & Solution

Correct Answer

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

Step-by-step Solution

Detailed explanation

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 \(SHM \left( a =-\omega^{2} x \right)\)

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

Same subject

Explore more questions on app

From JEE Mains

Explore more questions on app