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A thin fixed ring of radius 1m has a pos...

A thin fixed ring of radius 1m has a positive charge `1xx10^(-5)C` uniformly distributed over it. A particle of mass 0.9 g and having a negative charge of `1xx10^(-6)C` is placed on the axis at a distance of 1cm from the centre of the ring. The motion of the negatively charged particle is approximately simple harmonic. The time period of oscillations is `(api)/(10)`, then value of a is

Text Solution

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The correct Answer is:
2

`E=(1)/(4piepsilon_(0))(q_(1)x)/((R^(2)+x^(2))^(3//2))`
`R^(2)+x^(2)~~R^(2)`
`E=(q_(1)x)/(4piepsilon_(0)R^(3))`
Force on the particle at `P=-q_(2)E`
Force on particle `=(-q_(1)q_(2)x)/(4piepsilon_(0)R^(3))` towards O.
`a=(F)/(m)=(-q_(1)q_(2))/(4piepsilon_(0)MR^(3))impliesa=-omega^(2)x`
`T=2pisqrt(|x/a|)=2pisqrt((4piepsilon_(0)MR^(3))/(q_(1)q_(2)))=(2pi)/(10)`
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