# Question #b6b7c

May 3, 2017

There is a specific formula for the dipole moment and torque but we can sorta derive that as we go along:

We start with Newton's 2nd Law for circular motion: $m a t h b f \tau = I m a t h b f \alpha q \quad \star$

We place the origin at the mid point and exploit the symmetry and for minimum time we arrange the positive and negative charges as shown. The torque is:

$m a t h b f \tau = 2 \times \left(m a t h b f r \times m a t h b f F\right)$

$m a t h b f r = \left(\begin{matrix}\frac{L}{2} \cos \theta \\ \frac{L}{2} \sin \theta\end{matrix}\right)$ and $m a t h b f F = \left(\begin{matrix}E q \\ 0\end{matrix}\right)$

$\implies m a t h b f \tau = - L E q \sin \theta \setminus m a t h b f {e}_{z}$

The inertia for 2 point masses is $2 \times M {\left(\frac{L}{2}\right)}^{2} = \frac{M {L}^{2}}{2}$

And so from $\star$:

$- L E q \sin \theta \setminus m a t h b f {e}_{z} = \frac{M {L}^{2}}{2} \ddot{\theta} \setminus m a t h b f {e}^{z}$

We make the usual linearisation assumption, $\theta \approx \sin \theta$ for small $\theta$ and we get this DE:

$\ddot{\theta} + \frac{2 E q}{M L} \theta = 0$ which is in usual form: $\ddot{\theta} + {\omega}^{2} \theta = 0$

So period $T = \frac{2 \pi}{\omega} = 2 \pi \sqrt{\frac{M L}{2 E q}}$

This has been linearised to simple harmonic motion (we have a linear restorative torque), and the minimum time $\tau$ for the arrangement to first go through the equilibrium point (ie where $\theta = 0$) is $\tau = \frac{T}{4} = \textcolor{b l u e}{\frac{\pi}{2} \sqrt{\frac{M L}{2 E q}}}$

I'd roll with (3) ... unless you see a mistake :)