JEE Mains · Physics · STD 12 - 1. Electric charges and fields
A particle of mass \(1 \,{mg}\) and charge \(q\) is lying at the mid-point of two stationary particles kept at a distance \('2 \,{m}^{\prime}\) when each is carrying same charge \('q'.\) If the free charged particle is displaced from its equilibrium position through distance \('x'\) \((x\,< \,1\, {m})\). The particle executes \(SHM.\) Its angular frequency of oscillation will be \(....\,\times 10^{8}\, {rad} / {s}\) if \({q}^{2}=10\, {C}^{2}\)
- A \(60\)
- B \(6\)
- C \(76\)
- D \(760\)
Answer & Solution
Correct Answer
(B) \(6\)
Step-by-step Solution
Detailed explanation
Net force on free charged particle \(F =\frac{ kq ^{2}}{( d + x )^{2}}-\frac{ kq ^{2}}{( d - x )^{2}}\) \(F =- kq ^{2}\left[\frac{4 dx }{\left( d ^{2}- x ^{2}\right)^{2}}\right]\) \(a =-\frac{4 kq ^{2} d }{ m }\left(\frac{ x }{ d ^{4}}\right)\)…
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