Two charges `-q` each are fixed separated by distance 2d. A third charge q of mass m placed at the mid-point is displaced at the mid-point is placed 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

Let us elaborate the figure first.
given two charge -q at A and B
`AB=AO+OB=2d`
x=small distance perpendicular to O.
i.e., xltd mass of charge q is So, force attraction at P towards A and B are each `F=(q(q))/(4piepsi_(0)r^(2))`, where AP=BP=r
Horizontal components of these forces `F_(n)` are cancel out. Vertical components along PO add.
If `angleAPO=O`, the net force on q along PO is `F'=2FcosQ` ltBrgt `=(2q^(2))/(4piepsi_(0)r^(2))((x)/(r))`
`=(2q^(2)x)/(4piepsi_(0)(d^(2)+x^(2))^(3//2))`
When, `x lt lt d, F'=(2q^(2)x)/(4piepsi_(0)d^(3))=kX`
Where, `K=(2q^(2))/(4piepsi_(0)d^(3))`
I.e., force on charge q is proportional to its displacement from the centre O and it is directed towards O.
Hence, motion of charge q would be simple harmonic, where
`omega=sqrt((K)/(m))`
and `T=(2pi)/(omega)=2pisqrt((m)/(K))`
`=2pisqrt((m.4piepsi_(0)d^(3))/(2q^(2)))=[(8pi^(3)epsi_(0)md^(3))/(q^(2))]^(1//2)`