Consider a small part of the ring. All points on the ring are symmetrical to any point on the axis of the ring. Given situation is depicted in the figure. Consider an infinitesimal element at point A on the circumference of the ring. Let charge on this element be dq.
The magnitude of the intensity of electric field d vector E at a point P situated at a distance x from the center on its axis is,
Its direction is from A to P. Now consider two components of d vector E (i) dE sin θ , parallel to the axis of the ring and (ii) dE cos θ , parallel to the axis.
Here it is clear that in the vector sum of intensities due to all such elements taken all over the circumference, the dE sin θ components of the diametrically opposite elements will cancel each other as they are mutually opposite. Hence only dEcos θ components should be considered for integration.
∴ The total intensity of electric field at point P,
