A Charge Q is distributed uniformly on a ring of radius r. A sphere of equal r is constructed with its centre at the periphery of the ring (figure 30.12) Find the flux of the electric field through the surface of the sphere.

Explanation

From the geometry of the fig `OA=OO_(1)` and `O_(1)A=O_(1)O`. Thus `OAO_(1)` is an equilateraltriangle.
Hence `/_A OO_(1)=60^(@)/_AOB=120^(@)`
The arc `AO_(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 `=Q/3`. Thus flux out of sphere `Q/(3 epsilon_(0))`