CBSE Class 12-science Answered

We determine the field at point P on the axis of the ring. It should be apparent from symmetry that the field is along the axis. The field dE due to a charge element dq is shown, and the total field is just the superposition of all such fields due to all charge elements around the ring. The perpendicular fields sum to zero, while the differential x-component of the field is

We now integrate, noting that r and x are constant for all points on the ring:

This gives the predicted result. Note that for x much larger than a (the radius of the ring), this reduces to a simple Coulomb field. This must happen since the ring looks like a point as we go far away from it. Also, as was the case for the gravitational field, this field has extrema at x = +/-a.