Two charges `-q` each are fixed separated by distance 2d. A third charge q of mass m placed at the mid-point is displaced slightly by `x (xltltd)` perpendicular to the line joining the two fixed charges as shown in Fig. Show that q will perform simple harmonic oscillation of time period.
`T = [(8pi^(3) in_(0) md^(3))/(q^(2))]^(1//2)`

Explanation

`F_1=1/(4pie_0). Q^2/(d^2+x^2)^2` [along C to A]…..(i) and `F_2=1/(4pi e_0). (qq)/(BC)^2=1/(4 pi e_0).q/(d^2+x^2)`………(ii)
Now, resolve the component of .`F_1 and F_2`
SO net force on q is `F=F_1 cos theta+F_2 cos theta` ,
`F=2 F_1 cos theta=2,1/(4 pi e_0).q^2/((d^2+x^2)) . cos theta[ because F_1=F_2]`
In `Detla ACO, cos theta=x/sqrt(d^2+x^2), F=1/(4pi e_0)=.(2q^2)/(d^2+x^2)^(3//2)`
This force is attractive in nature, so `=1/(4pie_0).(2q^2.x)/(d^2+x^2)^(3//2)`
According to question `x lt lt d`, so neglect `x^2` as compared to `d^2`
`F=-1/(4pi e_0).(2q^2)/d^3.x`............(iii) Here `F prop -x or F=- omega^2x`........(iv)
So, the motion is simple harmonic, where `omega` is the angular frequency. By comparing Equations (iii) and (iv), we get :
`omega=sqrt((2q^2)/(4pi e_0 d^3m)) or omega =(2pi)/T`
Where, T is the time period of SHM of the charged particle of mass m. `T=((8pi r^3 e_0 m d^3)/q^2)`