## Sunday, January 24, 2010

### 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,

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ReplyDeleteIt is better to use vector form of Coulomb's law when in a hurry :-

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U can consider two dipoles with axis on equatorial and then considering force due to dipole is anti parallel on equatorial it's a 1 line answer

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