A thin circular wire ring of radius r carries a charge q. Find the magnitude of electric field strength on the axis of the ring as a function of distance l from its centre. Investigate the obtained function at l >> r. Find the maximum magnitude of field strength and the corresponding distance l. Also draw the approximate plot of the function E(l).









Hence for points l >> r, the ring charge distribution behaves like a point charge.

Thus the magnitude of electric field vector is maximum on both sides of the ring at a distance l = r/√2 from the ring's centre. To get that maximum value we should substitute this value of l in the expression of E(l).





The approximate plot of E(l) is drawn below.

The approximate plot has been drawn keeping in consideration the following three major points :

The field value at the centre of the ring is zero. The field value is maximum at l = r/√2 on both sides of the ring. As l tends to infinity the field value approaches to zero.

All these three points can easily be concluded from the general expression of E(l).





















Assume the charge distribution on the ring to be uniform. As asked in the question, we have to find the magnitude of the electric field strength at a point (say P) lying on the axis of the ring at a distancefrom the plane containing the ring.Consider an infinitesimally small arc element on the ring holding a small charge dq. As the size (or dimensions) of this element is too small compared to its separation from the point of calculation of field (i.e. the point P), we can approximate it as a point charge. Therefore the field produced by the considered arc element is d(as shown in the above figure) such that |d| = k(dq)/R². Here R is the distance of point P from the arc element.The charge distribution is symmetric with respect to the axis of the ring. Therefore at any point on the axis of the ring, the electric field strength vector would be directed along the axis. Alternatively, choose another small arc element lying diametrically opposite to the first element and draw their fields at point P to observe that their resultant field vector comes parallel to the axis. Same is the case for any small element on the ring and therefore it is concludable that the resultant field at any point on the axis points along the axis. Therefore the field at point P can be found as follows.