Question

A particle of mass $m$ and charge $q$ is placed between two fixed point charges of charge $q$ and separation $2L$. Find the frequency of oscillation of mass particle, if it is displaced for a small distance along the line joining the charges.

Answer 1

Force on charge at $o$ is same for both charge at point $A$ and $B$ same and in opposite direction. So this two equal force will cancel each other. but when middle charge displace by a small amount $x$, these two force can not remain same. At the displace position forces on charge at middle are,

$F_{AO}=\frac{q^2}{{(l-x)}^2}$ due to point charge at $\overrightarrow{A}$ in $\overrightarrow{AO}$ direction and,

$F_{BO}=\frac{q^2}{{(l-x)}^2}$ due to point charge at $\overrightarrow{B}$ in $\overrightarrow{OB}$ direction.

Resultant of these is $F=F_{AO}-F_{BO}$

$\Rightarrow F=q^2(\frac{1}{{(l-x)}^2}-\frac{1}{{(l-x)}^2})$

$\Rightarrow F=\frac{q}{(l^2-x^2)}\left(\right(l+x{)}^2-(l-x{)}^2)$

$\Rightarrow F=\frac{q}{(l^2-x^2)}.4lx$

Now, as $x<<l$, $l^2-x^2\approx l^2$

So, $F=\frac{q}{l^4}\times 4lx$

$\Rightarrow F=\left(\frac{4q}{l^3}\right)x$

As, the force directed to the mean position we can call is restoring force.

$F_r=-\left(\frac{4q}{l^3}\right)x$

$\Rightarrow m\frac{d^{2}x}{{dt}^2}=-\left(\frac{4q}{l^3}\right)x$

$\Rightarrow \frac{d^{2}x}{{dt}^2}+\frac{4q}{{ml}^3}x=0$

$\Rightarrow \frac{d^{2}x}{{dt}^2}+{\omega }^{2}x=0$

Angular frequency, $\omega =\sqrt{\frac{4q}{{ml}^3}}$

Time period, $T=\frac{2\pi }{\omega }=\pi \sqrt{\frac{{ml}^3}{q}}=\pi l\sqrt{\frac{ml}{q}}$

N.B: Here I have done the calculation considiribg C.G.S unit. The displacement is taken in the direction of $\overrightarrow{OA}$