Oscillation of a point charge perpendicular to field lines

[CrazyNeutrino]

Homework Statement

Two particles A and B each carry a charge Q and are separated by a fixed distance D. A particle c with charge q and mass m is kept at the midpoint of A and B. If C is displaced perpendicular to AB by a distance x where x<<<D,
find the time period of the oscillation of the charge.

Homework Equations

$$F=\frac{Qq}{4 \pi \epsilon_0 x^3} \cdot \hat{x}$$
$$a=-\omega^2 x$$

The Attempt at a Solution

I started off by trying to find an expression for the force acting on the particle. Since the components of the force parallel to AB cancel out because the charges at A & B are the same, I took the vector sum of the perpendicular force in the $\hat{x}$ direction.
$$ma=-\frac{2Qq}{4\pi\epsilon_0 x^3}\cdot \hat{x}$$

I then solved for $a$ and plugged the coefficients into $a=\omega^2 x$.
This gives:
$$\omega^2=\frac{Qq}{4 \pi \epsilon_0 x^3}$$
$$ \frac{2\pi}{T}=\sqrt{\frac{Qq}{4 \pi \epsilon_0 x^3}} $$
$$ T = \sqrt{\frac{8\pi^3\epsilon_0 x^3 m}{qQ}}$$

This is wrong.

 

[kuruman]

Your equation $$F=\frac{Qq}{4 \pi \epsilon_0 x^3} \cdot \hat{x}$$ is incorrect. There should be dependence on D, the separation between the fixed charges. It's also dimensionally incorrect and predicts that F is infinite when x = 0. Actually, F = 0 when x = 0.

 

[CrazyNeutrino]

Your equation $$F=\frac{Qq}{4 \pi \epsilon_0 x^3} \cdot \hat{x}$$ is incorrect. There should be dependence on D, the separation between the fixed charges. It's also dimensionally incorrect and predicts that F is infinite when x = 0. Actually, F = 0 when x = 0.

I understand the dimensionality argument. How and why does the force depend on the charge separation though? Wouldn’t those components cancel each other out?

 

[haruspex]

How and why does the force depend on the charge separation though?

What is the magnitude of the force exerted by one charge Q? What direction is it in? What component is left after cancellation by the force from the other charge?