Question

Two infinitely long static line charge of constant positive line charge density $\lambda$ are kept parallel to each other. If a point charge $+q$ is kept in equilibrium between them and is given small displacement along $x-$ axis about its equilibrium position then the correct statement is
[Charge $+q$ is confined to move in the $x$ direction only]

• A. Charge executes simple harmonic motion
• B. Charge contines to move in the direction of its displacement
• C. Charge takes circular path
• D. Charge takes parabolic path

Answer 1

The correct option is A Charge executes simple harmonic motion

Initially, net force acting on the charge $+q$ is zero.
So, point $\text{P}$ is the equilibrium point.



Now displace $+q$ towards right by a small distance $x$,

$F_1=$ force acting on $+q$ due to rod $1$.

$F_2=$ force acting on $+q$ due to rod $2$.

Applying the formula of electric field due to infinite rod, we get

$F_1=q{E}_1=q\frac{2k\lambda }{r+x}$

$F_2=q{E}_2=q\frac{2k\lambda }{r-x}$

So, net force $\left(F_{net}\right)$ acting on $+q$ in the direction of displacement of charge is given by

$F_{net}=F_1-F_2$

$\Rightarrow F_{net}=q\frac{2k\lambda }{r+x}-q\frac{2k\lambda }{r-x}$

$\Rightarrow F_{net}=2k\lambda q\frac{-2x}{r^2-x^2}$

Since $x<<r$, so $x^2$ can be neglected.

$\Rightarrow F_{net}=-\frac{4k\lambda q}{r^2}x$

Let $m$ be the mass of charge $+q$

$\Rightarrow ma=-\frac{4k\lambda q}{r^2}x$

$\Rightarrow ma=-\frac{4k\lambda q}{m{r}^2}x=\text{constant}\times (-x)$

$\Rightarrow a\propto -x$

This is the condition of SHM and $F_{net}$ is the restoring force. Thus, it performs simple harmonic motion.

Hence, option $\left(a\right)$ is correct.