JEE Mains · Physics · STD 12 - 1. Electric charges and fields
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 & 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 \(F _{\text {net}}=-\left[2 F _{ q / q } \cos \theta\right]\) \(F _{\text {net}}=-2 \cdot \frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{ q ^{2}}{\left(\sqrt{ d ^{2}+ x ^{2}}\right)^{2}} \cdot \frac{ x }{\sqrt{ d ^{2}+ x ^{2}}}\)…
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