Two charges, each equal to q, aer 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 `q_0` is given a small displacement (ylt lt a) along the y-axis, the net force acting on the particle is proportional to

Options

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

Answer

A

Explanation

In Fig, let `OP = y lt lt q`.

Net force on charge `(-q//2)` at `P` due to the two given charges is
`F_(net) = 2 F cos theta`, where `/_APO = /_BPO = theta`
`= (2kq (q//2))/((a^(2) + y^(2))) (y)/(sqrt(a^(2) + y^(2))) = (kq^(2))/((a^(2) + y^(2))^(3//2))`
`F_(net) = (kq^(2) y)/(a^(3))`, where `y lt lt a :. F_("net") prop y`