A small, electrically charged bead can slide on a circular, frictionless, thin, insulating ring. Charge on the bead is Q and its mass is m. A small electric dipole, having dipole moment P is fixed at the centre of the circle with the dipole’s axis lying in the plane of the circle. Initially, the bead is held on the perpendicular bisector of the dipole (see fig.). Ignore gravity. The magnitude of normal force exerted by the ring on the bead at position $\theta$ is $(K=\frac{1}{4\pi {\epsilon }_0})$

 

(a) Let us apply the law of conservation of energy for the bead having mass m and charge Q
$\frac{1}{2}m{v}^2+QK\frac{P\cos \theta }{r^2}=\frac{1}{2}m{v}_0^2+\frac{QKP\cos (\pi /2)}{r^2}=0$ 
We can then express the velocity of the bead at angle $\theta$ as
$v=\sqrt{-2\frac{QKP\cos \theta }{m{r}^2}}(\frac{\pi }{2}\le \theta \le \pi )$…………………(1)
 
(b) The circular motion needs a radial force equal to $m{v}^2/r.$
The radial component of electric field due to the dipole $E_r=2\frac{KP\cos \theta }{r^3}$  
Using the expression of the velocity (equation 1), we notice that the radial force on the bead $(=Q{E}_r)$ is just equal to $-m{v}^2/r,$ the required centripetal force. Thus the ring need not exert any force on the bead to sustain circular motion.