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 \({q}^{2}=10\, {C}^{2}\)

Pick an option to check your answer.

Verified Solution

Answer & Solution

Correct Answer

(B) \(6\)

Detailed explanation

Net force on free charged particle

\(F =\frac{ kq ^{2}}{( d + x )^{2}}-\frac{ kq ^{2}}{( d - x )^{2}}\)

\(F =- kq ^{2}\left[\frac{4 dx }{\left( d ^{2}- x ^{2}\right)^{2}}\right]\)

\(a =-\frac{4 kq ^{2} d }{ m }\left(\frac{ x }{ d ^{4}}\right)\)

\(a =-\left(\frac{4 kq ^{2}}{ md ^{3}}\right) x\)

So, angular frequency

\(\omega=\sqrt{\frac{4 kq ^{2}}{ md ^{3}}}\)

\(\omega=\sqrt{\frac{4 \times 9 \times 10^{9} \times 10}{1 \times 10^{-6} \times 1^{3}}}\)

\(\omega=6 \times 10^{8}\; rad / sec\)

Same subject

From JEE Mains