Question

Two equal charges $q$ are kept fixed at $a$ and $+$$a$ along the x-axis. A particle of mass $m$ and charge $\frac{q}{2}$ is brought to the origin and given a small displacement along the x-axis, then:

• A. the particle executes oscillatory motion
• B. the particle remains stationary
• C. the particle executes, SHM along x-axis
• D. the particle executes SHM along y-axis

Answer 1

Answer: (C)

the particle executes, SHM along x-axis

Repulsive force acting on the particle due to the farther charge $=\frac{kq\frac{q}{2}}{(a+x{)}^2}$

Repulsive force acting on the particle due to the nearer charge $=\frac{kq\frac{q}{2}}{(a-x{)}^2}$

Thus net force acts on the particle towards the origin(equilibrium position ) and is equal to

$=\frac{k{q}^2}{2}[\frac{1}{(a-x{)}^2}-\frac{1}{(a+x{)}^2}]$

$=\frac{k{q}^2}{2a^2}[\frac{1}{(1-\frac{x}{a}{)}^2}-\frac{1}{(1+\frac{x}{a}{)}^2}]$

$=\frac{k{q}^2}{2a^2}\left[\right(1-\frac{x}{a}{)}^{-2}-(1+\frac{x}{a}{)}^{-2}]$

Since $x<<a$

$F=\frac{k{q}^2}{2a^2}\left[\frac{4x}{a}\right]\propto x$

Hence it is an SHM.