Two positive point charges each of magnitude `10 C` are fixed at positions `A` &`B` at a seperation `2d= 6m`. A negatively charged particle of mass `m= 90gm` and charge of magnitude `10xx10^(-6)C` is revolving in a circular path of radius `4m` in the plane perpendicular to the line `AB` and bisecting the line `AB`. Neglect the effect of gravity. Find the angular velocity of the particle

Options

A. `200 rad//s`
B. `400 rad//s`
C. `250 rad//s`
D. `100rad//s`

Answer

b

Explanation

Net force on `-q` towards the centre,

`F=(2F_(1).sin theta)`
`=2.(KQ.q)/(d^(2)+d^(2))xx(R )/(sqrt(d^(2)+d^(2)))`
For parctice to move in circle
`F= m omega^(2)R`
`implies (2KQqr)/((d^(2)+R^(2)))m omega^(2)R`
`implies omega=sqrt((2KQq)/(m(d^(2)+R^(2))^(3//2)))`
`=sqrt((2 xx9 xx 10^(9) xx 10 xx 10 xx 10^(-6))/(90 xx 10^(-3) xx(3^(2) + 4^(2))^(3//2))) =400 rad//s`