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  • A particle of mass and charge is lying at the mid-point of two stationary particles kept at a distance <img alt="' 2 \mathr

A particle of mass 1 \mathrm{mg} and charge \mathrm{q} is lying at the mid-point of two stationary particles kept at a distance ' 2 \mathrm{~m} ' when each is carrying same charge ' q '. If the free charged particle is displaced from its equilibrium position through distance 'x$ ' $(\mathrm{x} \ll<1 \mathrm{~m}). The particle executes SHM. Its angular frequency of oscillation will be_________ \times 10^{5} \mathrm{rad} / \mathrm{s}$ if $\mathrm{q}^{2}=10 \mathrm{C}^{2}.
 

Answers (1)

best_answer

O- Equilibrium position

P- Displaced position

The net force acting on the charge q at the displaced position due to the other two charges is,
F_{net}= \left ( F_{2}-F_{1} \right )
          = \frac{kq^{2}}{\left ( 1-x \right )^{2}}-\frac{kq^{2}}{\left ( 1+x \right )^{2}}
          = kq^{2}\left [ \frac{1+2x+x^{2}-1+2x-x^{2}}{\left ( 1^{2}-x^{2} \right )^{2}} \right ]
F_{net}=\frac{4 kq^{2}x}{\left ( 1-x^{2} \right )^{2}}
a_{net}=\frac{F_{net}}{m}= \frac{4kq^{2}x}{m}= \omega ^{2}x
\bar{a}= -\omega ^{2}\left ( \bar{x} \right )
\omega ^{2}= \frac{4kq^{2}}{m}
\omega= \sqrt{\frac{4kq^{2}}{m}}
   = \sqrt{\frac{4\times 9\times 10^{9}\times 10}{10^{-6}}}
= \sqrt{36\times 10^{16}}= 6\times 10^{8}\, \frac{rad}{s}
 

Posted by

vishal kumar

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