Two charges, each equal to q , are kept at `x = - a ` and x = a on the x - axis . A particle of mass m and charge `q_(0)=q/2` is placed at the origin . If cahrge `q_(0)` is given a small disaplacement `( y lt lt a)` along the y - axis , the net force acting on the particle is proportional to :

Options

A. `-y`
B. `1/y`
C. `-1/y`
D. y

Answer

D

Explanation

`F_("net") = 2F cos theta `
`= 2 (k.q.q//2)/((sqrt(a^(2)+y^(2)))^(2)) .y/(sqrt(a^(2)+y^(2))) = (2kq^(2))/(2(a^(2))) xx y/a `
`= (kq^(2)y)/(a^(3))`
`F cos theta `
`( :. a gt gt y " therefore .y. can be neglected as compared to y ")`