Four point `+ve` charges of same magnitude`(Q)` are placed at four corners of a rigid square frame as shown in figure. The plane of the frame is perpendicular to `z`-axis. If a `-ve` point charge is placed at a distance `z` away from the above frame `(z lt lt L)` then

Options

A. `-ve` charge oscillates along the `Z`-axis
B. it moves away from the frame
C. it moves slowly towards the frame and stays in the plane of the frame
D. it passes through the frame only once

Answer

1

Explanation

Force on `-ve` charge
`F = (1)/(4 pi in_(0)) . (Qq)/([((L)/(sqrt(2)))^(2) + z^(2)])`
Net force on `-q`,
`F' = - 4 F sin theta = -4 . (1)/(4 pi in_(0)) . (Qq)/((L^(2)/(2) +z^(2))) .(z)/(((L^(2))/(2) +z^(2))^((1)/(2)))`
`F' = -(Qq)/(pi in_(0)) (z)/(((L^(2))/(2) +z^(2))^((3)/(2)))`
Since `z lt lt L`
`F' = - (Qq)/(pi in_(0) (L^(3))/(2 sqrt(2))).z`
`F' prop -z` (case of `S.H.M.`)
`-ve` sign , because force is directed towards centre.
Note : Here only electric force is considered.