Question

Two charges each of magnitude $Q$ are fixed apart at a distance of '$2a$'. A third charge ($-q$ of mass ${}^{\prime }{m}^{\prime }$) is placed at the midpoint of the two charges. Now if $-q$ charge is slightly displaced perpendicular to the line joining the charges then, find its time period.

Answer 1

For a small displacement $x$ perpendicular to the line joining the charges

the potential energy , U ( r ) = $\frac{Qq}{4\pi {ϵ}_{0}r}$ , of the displaced charge is now :

$U\left(x\right)=2\times -\frac{Qq}{4\pi {ϵ}_0\sqrt{a^2+x^2}}$

The force vector acting on the particle is in the negative derivative w.r.t $x$ of this conservative potential function.

Form symmetry we know that it will act "perpendicular to the line joining the charges".

$\overrightarrow{F}$ = - $\frac{du}{dx}$

= $\frac{Qq}{2\pi {ϵ}_0}\frac{x}{(a^2+x^2{)}^{\frac{1}{2}}}$

By taylor expansion, for small $x$ we have:

$\frac{x}{(a^2+x^2{)}^{\frac{3}{2}}}$ = $\frac{x}{a^3}(1+\frac{x^2}{a^2}{)}^{-\frac{3}{2}}$

In other words, $\overrightarrow{F}$ linearises as:

$\overrightarrow{F}$ = $-\frac{Qq}{2a^3\pi {ϵ}_0}x$ ,

and it is this linearise restorative force that characterises the harmonic oscillaator.

i.e. $x+\frac{Qq}{2\pi m{a}^3{ϵ}_o}x=0$ can be written in the standrad SHM wave form $x+{\omega }^{2}x=0$

Therefore;

$T=\frac{2\pi }{w}$

= $2\pi \sqrt{\frac{2m{a}^3\pi {ϵ}_0}{Qq}}$