Question

Two equal negative charges each $-q$, are fixed at points $(0,a)$ and $(0,-a)$ on $Y-$axis. A positive charge $q$ is released from point $(2a,0)$. This charge will be:

• A. executes S.H.M. about the origin.
• B. oscillate but not execute S.H.M.
• C. move towards origin and will become stationary
• D. execute S.H.M. along $Y-$axis

Answer 1

The correct option is B oscillate but not execute S.H.M.


From figure,
$F\to$ force interaction between a pair of $-q$ and $+q$

Using Coulomb's law

$F=\frac{k{q}^2}{r^2}$

So, net force $F_{net}$ on the particle will be

$\Rightarrow F_{net}=F\cos \theta +F\cos \theta$

$\Rightarrow F_{net}=2F\cos \theta$


From above figure,

$\cos \theta =\frac{x}{r}=\frac{x}{\sqrt{x^2+a^2}}$

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

$\Rightarrow F_{net}=2\frac{k{q}^{2}x}{r^3}$

$\Rightarrow F_{net}=\frac{2k{q}^{2}x}{(x^2+r^2{)}^{3/2}}$ (towards $O$)

Here as soon as the particle reaches $O$ its velocity get maximum and as it goes beyond $O$ in $-x$ direction, net force changes its direction again towards origin.
So the particle will oscillate, but since $F_{net}$ is not directly proportional to $x$, it's not SHM.

Why this question? Tip: Every oscillation is not SHM.