No Springs Attached - Part 2

This was a fun one. The system Lagrangian is (where $k$ is the Coulomb constant):

$\mathcal{L} = \frac{1}{2} m \dot{y}^2 + \frac{2 k Q^2}{\sqrt{a^2 + y^2}}$

Equation of motion:

$\frac{d}{dt} \frac{\partial{\mathcal{L}}}{\partial{\dot{y}}} = \frac{\partial{\mathcal{L}}}{\partial{y}}$

Crunching this out results in:

$m \ddot{y} = -k Q^2 (a^2 + y^2)^{-3/2} (2 y)$

Now we can make use of the fact that $y^2 << a^2$ :

$m \ddot{y} \approx -k Q^2 a^{-3} (2 y)$

Simplifying:

$\ddot{y} = - \frac{2 k Q^2}{m a^3} y$

This corresponds to simple harmonic motion with angular frequency $\omega$ :

$\omega = \sqrt{\frac{2 k Q^2}{m a^3}}$

Plugging in $k = \frac{1}{4 \pi \epsilon_0}$ and further simplifying results in:

$T = \frac{2 \pi}{\omega} = \sqrt{\frac{8 \pi^3 \, \epsilon_0 \, m \, a^3}{Q^2}}$

Force acting on the charge is approximately $\dfrac{Q^2y}{2π\epsilon_0a^3}$ acting in the direction opposite to displacement. Hence the acceleration of the charge is $-\dfrac{Q^2}{2π\epsilon_0ma^3}\times y$ . So the time period of the oscillation is $2π\sqrt {\dfrac{2π\epsilon_0ma^3}{Q^2}}=\sqrt {\dfrac{8π^3\epsilon_0ma^3}{Q^2}}$ . Therefore $A=8, B=3, C=1, D=2$ and so $A+B+C+D=\boxed {14}$ .