Harmonic motion of negative charge between two positive ones I have to prove that a negative charge will execute an harmonic motion when placed in the line that separates two positive charges $Q$. (The negative charge $-q$ will not be placed on the center).
I clearly understand that it will be an harmonic motion, but i can't show it mathematically: i've tried the traditional way:
Let the two positive charges be $Q_1$ and $Q_2$, and the position of the $-q$ charge is at an $X$ distance from the center:
$$F_1 - F_2 = ma$$
Where
$$F_1 = \frac{kQ_1q}{(d/2 -x)^2}$$
And
$$F_2 = \frac{kQ_2q}{(d/2 + x)^2}$$
The problem is that im not being able to change this equation in order to get the traditional simple harmonic motion differential equation.
Any help would be great.
Thanks!
 A: The motion cannot be along the line joining the +ve charges because such motion is unstable : the -ve charge would accelerate away from the centre point and towards the nearest +ve charge.
I think it must be assumed that the -ve charge moves along the perpendicular bisector of the two +ve charges, and is constrained to remain on this line.It must be constrained because this motion is again unstable : the smallest deviation either side will result in acceleration towards one of the charges.
Suppose the two +ve charges are $2d$ apart and the -ve charge is distance $y$ from their midpoint along the perpendicular bisector. I am using $y$ to distinguish transverse displacement from longitudinal displacement $x$ along the line joining the +ve charges. 
Then the distance $r$ between the -ve and +ve charges is given by $r^2=d^2+y^2$ and the force of attraction to either charge is $kQq/r^2$. The total of the components of these forces along the bisector is the restoring force
$F=2(kQq/r^2)(y/r)=2kQq(y/r^3)$.
For small values of $y<<d$ we have $r^3 \approx d^3$ so the equation of motion is :
$m\ddot y = -F = -2kQq(y/d^3)$
$\ddot y + \omega^2 y = 0$
where $\omega^2=2kQq/md^3$.
The oscillation is only harmonic for small displacements $y<<d$.
