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(x<<d) perpendicular to the line joining the two fixed charged as shown in figure. Show that q will perform simple harmonic oscillation of time period.

T=8π3ε0md3q21/2

Let us elaborate the figure fist.

Given, two charge -q at A and B

AB=AO+OB=2d

x = small distance perpendicular to O.

i.e., x<d mass of charge Q is. So, force of attraction at P towards A and B are each F=q(q)4πε0r2, where AP=BP=r

Horizontal components of these forces Fn are cancelled out. Vertical components along PO add.

If APO=O, the net force on q along PO is 

F'=2F cos Q
=2q24πε0r2(xr)
=2q2x4πε0(d2+x2)3/2

When, x<<d, F'=2q2x4πε0d3=Kx

where, K=2q24πε0d3

Fx

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

ω=Km

and T=2πω=2πmK

=2πm4πε0d32q2=8π3ε0md3q21/2