A thin copper ring of radius a is charged with q units of electricity . An elctron is placed at the centre of the of electricity . An electron is placed at the centre of the copper ring . If the electron is displaced of a little , it will have frequency :

Options

A. `1/(2pi) sqrt((eq)/(4piepsilon_(0)ma^(3)))`
B. `1/(2pi) sqrt(q/(4piepsilon_(0)ema^(2)))`
C. `sqrt((eq)/(4piepsilon_(0)ma))`
D. `sqrt(q/(4piepsilon_(0)ema^(3)))`

Answer

A

Explanation

Field at distance x on axis is
`E = 1/(4piepsilon_(0)) (qx)/((a^(2)+x^(2))^(3//2)) = (q.x)/(4piepsilon_(0)a^(3))`
Force on electron at centre
`F = eE = (qx)/(4piepsilon_(0)a^(3)) (-e)`
` :. m (d^(2)x)/(dt^(2)) = - 1/(4piepsilon_(0)) (qex)/(a^(3))` .
` :. (d^(2)x)/(dt^(2)) = - 1/(4piepsilon_(0)) (qex)/("ma"^(3))`
So motion is S.H.M
`omega^(2) = 1/(4piepsilon_(0)) (qe)/(ma^(3)) , omega = sqrt((eq)/(4piepsilon_(0),ma^(3)))`
`n = 1/(2pi) sqrt((eq)/(4piepsilon_(0)ma^(3)))`