Irodov Problem 3.10

The net electric field at a point O is the sum of the electric field Ering due to the ring and Eq due to the charge located at the center. In Problem 3.9 we have already derived Ering (in Eqn 4) as,




The electric field due to the point charge at the center of the ring is given by,





The net electric field is given by,





We can rewrite (3) as,

Irodov Problem 3.9

Consider an infinitesimally thin section of the ring that subtends and angle at the center of the circle. The point in question, O, is located exactly above the center at a height l from the plane of the ring. The charge in the infinitesimally small section is





The electric field due to the infinitesimally small section of ring will be equal to,













From the figure we can see that,




Hence, the electric field is given by,





When l much larger than r, (4) simply becomes,





In other words it acts like a point charge.

Clearly the electric field is zero at the center of the ring (l=0) and also zero when l is very very large. Hence, there must be a value of l for which the electric field is maximum. This value of l can be obtained by setting the first derivative of E to zero as follows,










From (5) and (4) we have the maximum possible value of E as,

Irodov Problem 3.8

Consider an infinitesimally small section of the half-ring that subtends and angle from the center. The charge contained in this infinitesimally thin section of the ring is given by,




The electric field due to this infinitesimally small section of the half-ring is given by,

Irodov Problem 3.7

As shown in Figure 1, the charges are located in at the four corners of the square ABCD whose diagonal is of length 2l. Since the point X is located at a height of x units from the plane of ABCD along its central axis, the distance of X from any of the corners A,B,C and D is . The electric field strengths due to each of the four charges located at corners A,B,C and D are given by,













The vertical components of EC and EB will cancel each other. Similarly the vertical components of EA and ED will cancel each other. Hence, only the horizontal components , , , and will remain as shown in Figure 2 (which depicts the top view).



















From the geometry in Figure 1 it is clear that,





Since, A'B'C'D' is a square, the diagonals are perpendicular and hence A'D' is perpendicaular to B'C'. Hence the total electric field strnegth is given by,