A charge q is distributed uniformly on a ring of radius a. A sphere of equal radius a is constructed with its centre at the periphery of the ring.Calculate flux of the electric field through the surface of the sphere.

By gauss law we know that flux is 1/E times the charged enclosed within the closed surface, where E is epsilon nought. if we  imagine the diagram in a 2-D plane , we see we have 2 intersecting circles of same radius, each passing through the  others centre. joining their centres and any one pt of intersection we see thet the triangle has all 3 sides equal to a , and so each angle is 60 degrees. now angle subtended at the centre of 1st circle i.e, the centre of the sphere by the length of intercepted length of second circle, i.e, the part of ring inside the sphere is 120 degrees. thus the length is {(120/360 )* 2*pi*a} ; thus (length * linear charge distribution) gives the charge enclosed which is q/3 and the answer is 1/3E , where E is epsilon nought.