A ring of radius a contains a charge q distributed. uniformly ober its length. Find the electric field at a point. on the axis of the ring at a distance x from the centre.

Explanation

Consider a differential element of the ring of length ds. Charge on this element is `dq=((q)/(2piR))ds`.
This element sets up a differential electric field `dvecE` at point P.
The resultant field `vecE` at P is found by integrating the effects of all the elements that make up the ring. From symmetry this resultant field must lie along the right axis. Thus, only the component of `dvecE` parallel to thsi axis contributes to the final result.
To find the total x-component `E_(x)` of the field at P, we integrate this expression over all segments of the ring.
`E_(x)=intdEcostheta=(1)/(4piin_(0))(qx)/(2piR(R^(2)+x^(2))^(3//2))intds`
The integral is simply the circumference of the ring which is equal to `2piR`
`therefore E=(1)/(4piin_(0))(qx)/((R^(2)+x^(2))^(3//2))`
As q is a positive charge, hence field is directed away from the centre of the ring. along its axis.