Arc in a dipole field

Let $\vec{p} = -ql\hat{i}$ , the positive charge at $(-0.0005,0)$ , and negative charge at $(0.0005,0)$

The motion would be symmetric about $X$ -axis with the extreme ends as $(0,R)$ , and $(0,-R)$ .

Let us keep the test charge at $0,R$ , as the electric field at whole of $Y$ axis is towards right, the particle would initially go towards right.

Let the particle has covered an angle $\theta$ .

Potential energy at $(0,R)$ = 0,

New potential energy = $\frac{kQ\vec{p}.\vec{r}}{R^3} = -\frac{kqlQsin\theta}{R^2}$

Now, using conservation of mechanical energy,

$\frac{1}{2}mR^2w^2 = \frac{kqlQsin\theta}{R^2}$

$\Rightarrow w = \frac{d\theta}{dt} = \sqrt{\frac{2kqlQsin\theta}{mR^4}}$

Hence, $\displaystyle \int_{0}^{\frac{\pi}{2}} \frac{d\theta}{\sqrt{sin\theta}} = \int_{0}^{\frac{T}{4}} \sqrt{\frac{2kqlQ}{mR^4}}dt$

Now, $\displaystyle \int_{0}^{\frac{\pi}{2}} \frac{d\theta}{\sqrt{sin\theta}} = \int_{0}^{\frac{\pi}{2}} \frac{d\theta}{\sqrt{sin(\frac{\pi}{2} - \theta)}} = \int_{0}^{\frac{\pi}{2}} \frac{d\theta}{\sqrt{cos\theta}} = 2.622$

$\therefore \sqrt{\frac{2kqlQ}{mR^4}}\frac{T}{4} = 2.622$

$\Rightarrow T = 4 \times 2.622\sqrt{\frac{mR^4}{2kqlQ}} = \boxed{1.11 ms}$

Suppose that the dipole, of strength $p = q_+\ell$ , is directed along the $y$ -axis. The electric field due to the dipole is $\mathbf{E} \; = \; \frac{k}{r^3}\left[\frac{3(\mathbf{p}\cdot\mathbf{r})}{r^2}\mathbf{r} - \mathbf{p}\right]$ The desired oscillation can take place entirely in the $xy$ -plane. Suppose that $\theta$ is the angle between $\mathbf{r}$ and $\mathbf{p}$ , and that $|\mathbf{r}| = R$ is constant. Then $\begin{array}{rcl} \mathbf{r} & = & R{\sin\theta \choose \cos\theta} \\ \dot{\mathbf{r}} & = & R{\cos\theta \choose -\sin\theta}\dot{\theta} \\ \ddot{\mathbf{r}} & = & R{\cos\theta \choose -\sin\theta}\ddot{\theta} - R{\sin\theta \choose \cos\theta}\dot{\theta}^2 \\ \mathbf{E} & = & \frac{kp}{R^3}\left[3\cos\theta{\cos\theta \choose \sin\theta} - {0 \choose 1}\right] \\ & = & \frac{kp}{R^3}\left[2\cos\theta{\sin\theta \choose \cos\theta} + \sin\theta{\cos\theta \choose -\sin\theta}\right] \end{array}$ and so a test particle of mass $m$ and charge $Q$ will satisfy the differential equation $m\ddot{\mathbf{r}} \; = \; Q\mathbf{E}$ provided that $mR \ddot{\theta} \; = \; \tfrac{kpQ}{R^3}\sin\theta \qquad -mR\dot{\theta}^2 \; = \; \tfrac{2kpQ}{R^3}\cos\theta$ These two equations are consistent. Putting $\theta = \pi + \alpha$ , we have $\dot{\alpha}^2 \; = \; \tfrac{2kpQ}{mR^4}\cos\alpha$ which represents a semicircular oscillation from $\alpha=-\tfrac12\pi$ to $\alpha = \tfrac12\pi$ of period $T \; = \; 4\sqrt{\tfrac{mR^4}{2kpQ}}\int_0^{\frac12\pi}\frac{d\alpha}{\sqrt{\cos\alpha}}$ With $m=1$ , $R=1$ , $p = 1\times10^{-3}$ , $Q=5$ , we obtain $T = 1.11$ ms.

N.B. This orbit is not stable.