Question

Two infinitely long static line charges 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 the $x-$ axis from 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 continues to move in the direction of its displacment
• C. Charge takes circular path
• D. Charge takes parabolic path

Answer 1

The correct option is B Charge continues to move in the direction of its displacment

Initially, the net force acting on the charge $+q$ is zero.
So, the 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

$F_{net}=F_2-F_1$

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

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

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

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

From above solution, if the $-q$ charge displaced by a small distance $x$ to the right, then the net force acts in the direction of displacement act. Hence, the net force on the charged particle will be towards right, so the particle keeps going towards right.

Similar is the case when charge displaced to the left, net force acts towards left