Two equal negative charges - q. are fixed at points (0, a) and (0, - a) on the y-axis. A positive charge Q is released from rest at a point (2a, 0) on the x-axis. The charge Q will 

Let the charge Q be at P, with OP = x. The resultant force F is along the x-axis directed towards the origin. The charge Q moves to O, and acquires kinetic energy. It will cross O and move to -ve axis until it comes to rest. It is again attracted towards O and crosses it and this process continues. Therefore charge Q execute periodic motion (see fig.)

Let AP = BP = r,  then

$F_1=F_2=\frac{qQ}{4\pi {\epsilon }_0{r}^2}$The resultant force on Q is

$F=F_1\cos \theta +F_2\cos \theta =\frac{2qQ}{4\pi {\epsilon }_0{r}^2}\cos \theta$

$F=\frac{2qQx}{4\pi {\epsilon }_0{r}^3}=\frac{2qQ}{4\pi {\epsilon }_0}\frac{x}{{\left(a^2+x^2\right)}^{3/2}}$

Thus F is not of the form F = kx (where k = constant) and hence the motion is not simple harmonic.