# Irodov's Solution

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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 Xis 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 perpendicular to B'C'. Hence the total electric field strength is given by,

Irodov Problem 3.12

Consider an infinitesimally small section of the ring that subtends an angle and is located at angle. The length of this infinitesimally small section of the ring is. Since the linear charge density of the ring depends on as , the total charge contained in this infinitesimally small section of the ring is given by,

The electric field generated by this infinitesimally small section of the ring at point O, as depicted in the figure, is then given by,

As seen from the Figure, using elementary trigonometry we See that,

From (1), (2), (3a) and (3b) we obtain,

The field at the center of the ring can be obtained by setting x=0 as,

Irodov Problem 3.16

The easiest way to solve this problem is to choose a coordinate system where the x-axis is oriented in the direction of the vector. In such a coordinate system the vector can be simply written as. The charge density of the

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