Q160 / 170 · Manipal (MET) 2017 · Physics

Two equal charges $q$ are kept fixed at $a$ and $+a$ alpng 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
  • Correct answer
    C
    the particle executes, SHM along $x$-axis
  • D
    the particle executes SHM along $y$-axis

Solution

From Coulomb's law $\begin{aligned} F & =\frac{1}{4 \pi \varepsilon_0} \frac{q_1 q_2}{r^2} \\ F & =\frac{1}{4 \pi \varepsilon_0} \frac{q \times \frac{q}{2}}{(a+x)^2}-\frac{1}{4 \pi \varepsilon_0} \cdot \frac{q \times \frac{q}{2}}{(a-x)^2} \\ & =\frac{1}{4 \pi \varepsilon_0} \frac{q^2}{2}\left[\frac{1}{(a+x)^2}-\frac{1}{(a-x)^2}\right] \\ & =\frac{1}{4 \pi \varepsilon_0} \cdot \frac{q^2}{2}\left[-\frac{4 a x}{\left(a^2-x^2\right)^2}\right] \end{aligned}$ When $x\lt \lt a$, then $\begin{aligned} & F=-\frac{2 q^2}{4 \pi \varepsilon_0 a^3} x \\ \Rightarrow & F \propto-x \end{aligned}$ Hence, SHM along $x$-axis.

Hence, the correct answer is option: 'C'.

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