Question

Two equal negative charges each of charge $-q$ are fixed at points $(0,a)$ and $(0,-a)$ on the $y-$axis. A positive charge $Q$ is released from rest at a point $(2a,0)$ on the $x-$ axis. The charge $Q$ will

• A. Execute simple harmonic motion about the origin.
• B. Move to the origin and remain at rest there.
• C. Move to infinity.
• D. Execute oscillatory but not simple harmonic motion.

Answer 1

The correct option is D Execute oscillatory but not simple harmonic motion.

Let the charge $Q$ be at $P$, with $OP=x$. The resultant force $F$ is along the $x-$ axis directed towards the origin. The charge $Q$ moves to $O$, and acquires kinetic energy. It will cross $O$ and move to -ve $x-$ axis until it comes to rest. It is again attracted towards $O$ and crosses it and this process continues. Therefore, charge $Q$ executes oscillatory motion.


Let $AP=BP=r=\sqrt{a^2+x^2}$.
Then
$F_1=F_2=\frac{qQ}{4\pi {ϵ}_0{r}^2}$

Let $\angle APO=\angle BPO=\theta$
The resultant force on $Q$ is
$F=F_1\cos \theta +F_2\cos \theta =\frac{2qQ}{4\pi {ϵ}_0{r}^2}\cos \theta$
$F=\frac{2qQ}{4\pi {ϵ}_0}\frac{x}{(a^2+x^2{)}^{1.5}}$
Thus, $F$ is not of the form $F=kx$ (where $k=$ constant) and hence the motion is not simple harmonic.
Hence, the correct choice is (d).

$\begin{array}{c}\text{Why this question ?}\\ \text{Caution: A body will perform simple}\\ \text{harmonic oscillation only when the}\\ \text{restoring force is directed toward the}\\ \text{equilibrium or mean position.}\end{array}$