A charge Q is distributed uniformly on a ring of radius r. A sphere of equal radius r is constructed with its centre at the periphery of the ring (see figure). Find the flux of the electric field through the surface of the sphere.
 

 

From the geometry of the figure, $OA=O{O}_1\text{\hspace{0.17em}\hspace{0.17em}and\hspace{0.17em}}{O}_{1}A=O_{1}O.$ 

Thus, $OA{O}_1$  is an equilateral triangle. 

Hence $\angle AO{O}_1=60°$ or  $\angle AOB=120°$. 

The arc $A{O}_{1}B$ of the ring subtends an angle $120°$ at the centre O. 

Thus, one third of the ring is inside the sphere. 

The charge enclosed by the sphere $=\frac{Q}{3}$.

From Gauss’s law, the flux of the electric field through the surface of the sphere is $=\frac{Q}{3{\epsilon }_0}$.