A particle of mass 1 ,{mg} and charge q is ly | vedclass

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Question

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 ^

Vedclass · Accepted Answer

$6$

Vedclass · Answer

Net force on free charged particle

$F =\frac{ kq ^{2}}{( d + x )^{2}}-\frac{ kq ^{2}}{( d - x )^{2}}$

$F =- kq ^{2}\left[\frac{4 dx }{\left( d ^{2}- x ^{2}ight)^{2}}ight]$

$a =-\frac{4 kq ^{2} d }{ m }\left(\frac{ x }{ d ^{4}}ight)$

$a =-\left(\frac{4 kq ^{2}}{ md ^{3}}ight) x$

So, angular frequency

$\omega=\sqrt{\frac{4 kq ^{2}}{ md ^{3}}}$

$\omega=\sqrt{\frac{4 	imes 9 	imes 10^{9} 	imes 10}{1 	imes 10^{-6} 	imes 1^{3}}}$

$\omega=6 	imes 10^{8}\; rad / sec$

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