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

Answer 1

Answer: (C)

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. The net force will be towards left.

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

Thus, force on q is directly proportional to displacement x.

When charge q is negative:

When 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 the charge will accelerate towards right.