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 statements(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

If q is displaced slightly along x axis it will come back to original position
$F_{net}$ on q on displacing by small displacement x = restoring force = $\frac{\lambda q}{2\pi\varepsilon_{0}\left(r-x\right)}-\frac{\lambda q}{2\pi \varepsilon _{0}\left(r+x\right)}$
$\Rightarrow\, F = \frac{q\lambda}{2\pi\varepsilon_{0}} \left[\frac{1}{r-x}-\frac{1}{r+x}\right]$
$\Rightarrow\,F = \frac{q\lambda\times2x}{2\pi\varepsilon_{0}\left(r^{2}-x^{2}\right)} \quad$ neglecting x
$F_{restoring} = \frac{q\lambda x}{\pi\varepsilon_{0}r^{2}}$
$a = \frac{q\lambda x}{\pi\varepsilon_{0}r^{2}}$
$a = \frac{q\lambda x}{m\pi\varepsilon_{0}r^{2}}$
$a ? x$ so q will do SHM
- q will not able to oscillation as $F_{net}$ on - q will not send it back to its original equilibrium position. - qcontinue moving in direction of its displacement.