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 charge q0 is given a small displacement (y a) along the y-axis, the net force acting on the particle is proportional to:

Options

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

Answer

B

Explanation

`F_(net)=2F cos theta therefore F_(net) =- (2kq (q/2))/ (sqrt(y^2+a^2))^2. y/sqrt(y^2+a^2)`
[negative sign indicates that net force is acting towards the mean position]
`F_(net) =- (2kq (q/2)y)/(y^2+a^2)^(3//2) implies (kq^2y)/a^3 prop -y`