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. 100
B. 200
C. 300
D. 400

Answer

D

Explanation

Net force on -q towards the centre.
`F=(2F_(1).sin theta)=2.(KQq)/(d^(2)+d^(2))xxR/(sqrt(d^(2)+d^(2)))`
For particle to move in circle
`F=m omega^(2)R`
`rArr (2KQ qR)/((d^(2)+R^(2))^(3//2))=m omega^(2)R`
`rArr W=sqrt((2KQq)/(m(d^(2)+R^(2))^(3//2)))`
`= sqrt((2xx9xx10^(9)xx10xx10xx10^(-6))/(90xx10^(-3)xx(3^(2)+4^(2))^(3//2))) rad//s` =400 rad/s , no