Question

The figures 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

• A.
  both charges execute simple harmonic motion.
• B.
  both charges will continue moving in the direction of their displacement.
• C.
  charge $+q$ executes simple harmonic motion while charge $-q$ continues moving in the direction of its displacement.
• D.
  charge $-q$ executes simple harmonic motion while charge $+q$ continues moving in the direction of its displacement.

Answer 1

The correct option is C

charge $+q$ executes simple harmonic motion while charge $-q$ continues moving in the direction of its displacement.
$q$ is positive:
Let the distance between the line charges be $d$ and the charge $q$ be displaced by a small distance $x$.
$q$ will be repelled more by the right line charge than the left line charge. Then, the net force will be acting towards left which put charge always at mid point. And because of two opposite forces it always oscillate.

$\overrightarrow{F_q}=\frac{2kq\lambda }{d+x}\hat{i}+\frac{2kq\lambda }{d-x}(-\hat{i})=\frac{4kq\lambda x}{d^2-x^2}(-\hat{i})=\frac{4kq\lambda x}{d^2}(-\hat{i})$ when $x<<d$
& $F\propto \frac{1}{r}$

Charge q is negative:
When charge is displaced by $x$ towards right, the force of attraction due to right line charge will be more than the force of attraction due to left line charge and hence, the charge will accelerate towards right.

So, option C is correct.