A thin non conducting ring  of radius R has linear charge density has given above where the given angle is azimithul angle find the magnitude of electric field strength on the axis of the ring as a function of the distance x from the centre (x<<R)

Its 2 pie * R *   * X  /  [ 4 pie ebsilon knot  ( X ² + R²) ] 

.........for X >>>R ..it becomes 2 pie * R *   / 4 pie epsilon knot  X²

^.........for X<<<R ..it becomes 2 pie *   / 4 pie epsilon knot R 

Mast question hai!!

The symmetry of this distribution is such that half ring becomes negatively charged,, other positively charged...vector E at the centre of the ring is directed to right..

we will have to integrate this projection and we obtain

this is the electric field at center.

Let X and Y axes be on the plane of the ring and Z be perpendicular to the ring.

Consider a small arc on the ring that makes an angle of  with the X axis and subtebds an angle of d at the ring's center. Then the electric field due to this element at a point located at a diatance on the Z axis is

Integrate the equation from 0 to 2 to get the answer

Finally as its given x << R, apply binomial approximation