The figure below depict two situations in which two infinitely long static line charges of constant positive line charge density `lambda` are kept parallel to each other. In their resulting electric field, point charges q and `-q` are kept in equilibrium between them. The point charges are confined to move in the x direction only. If they are given a small displacement about their equilibrium positions, then the correct statement(s) is(are)

Options

A. (a) Both charges execute simple harmonic motion
B. (b) Both charges will continue moving in the direction of their displacement
C. (c) Charge `+q` executes simple harmonic motion while charge `-q` continues moving in the direction of its displacement
D. (d) Charge `-q` executes simple harmonic motion while charge `+q` continues moving in the direction of its displacement

Answer

C

Explanation

Force on change q when it is given a small displacement x is `F_(n et)=F_1-F_2`
`F_(n et)=(1)/(2piepsilon_0)(lambda)/(d-x)-(1)/(2piepsilon_0)(lambda)/(d+x)`
`:.` `F_(n et)=(lambda)/(2piepsilon_0)[(d+x-d+x)/(d^2-x^2)]`
`:.` `F_(n et)=(lambda)/(2piepsilon_0)(2x)/(d^2-x^2)`
When `x lt lt d` then

`F_(n et)=(lambda)/(piepsilon_0)x` and is directed towards the mean position therefore the charge +q will execute SHM.

In case of charge `(-q)`
`F_2gtF_1` therefore the charge `-q` continues to move in the direction of its displacement.
[C] is the correct option.