Question
A charge is distributed uniformly over a ring of radius 'a'. Obtain an expression for the electric intensity E at a point on the axis of the ring. Hence, show that for points at large distance from the ring, it behaves like a point charge.
Solution
Consider a ring of radius 'a' which carries uniformly distributed positive total charge Q. 
To find: electric field due to a ring at a point P lying at a distance x from its centre along the central axis perpendicular to the plane of the ring.
As the charge is distributed uniformly over the ring, the charge density over the ring is,
The perpendicular component of electric field due to charge on the ring along the x-axis cancels each other out.
As there is same charge on both sides of the ring, the magnitude of the electric field at P due to the segment of charge dQ is given by,
dE = ke 
Ex =
1. At the centre (X = 0) , electric field is zero.
2. When x>> a, a can be neglected in the denominator.
Therefore,
Ex =
1. At the centre (X = 0) , electric field is zero.
2. When x>> a, a can be neglected in the denominator.
Therefore,
