The diagram shows a small bead of mass m carrying charge q. The bead can freely move on the smooth fixed ring placed on a smooth horizontal plane. In the same plane a charge +Q has also been fixed as shown. The potential at the point P due to +Q is V. The velocity with which the bead should projected from the point P so that it can complete a circle should be greater than 

If we send the bead diametrically opposite, it will complete full circle as resultant of repulsive force between the charges and normal reaction between the ring and bead will provide force in tangential direction of ring. Hence at point 2, $v_2=0.$ 

Given $\frac{1}{4\pi {\epsilon }_0}\frac{Q}{4a}=V$

or $\frac{1}{4\pi {\epsilon }_0}\frac{Q}{a}=4V\mathrm{..............}\left(i\right)$

Let velocity given at 1 be $v_1,$ then 

$\frac{1}{2}m{v}_1^2+\frac{1}{4\pi {\epsilon }_0}\frac{Qq}{4a}=0+\frac{1}{4\pi {\epsilon }_0}\frac{Qq}{a}$

or $\frac{1}{2}m{v}_1^2+qV=q.4V$

or $\frac{1}{2}m{v}_1^2=3Vqor{V}_1=\sqrt{\frac{6Vq}{m}}$