ch22_p58 - ² at distant points). (c) Differentiating Eq....

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58. (a) It is clear from symmetry (also from Eq. 22-16) that the field vanishes at the center. (b) The result ( E = 0) for points infinitely far away can be reasoned directly from Eq. 22- 16 (it goes as 1/ z ² as z ) or by recalling the starting point of its derivation (Eq. 22-11, which makes it clearer that the field strength decreases as 1/ r
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Unformatted text preview: ² at distant points). (c) Differentiating Eq. 22-16 and setting equal to zero (to obtain the location where it is maximum) leads to dE dz = q ( R 2 − 2 z 2 ) 4 πε o ( R 2 + z 2 ) 5/2 = 0 ¡ z = + R 2 = 0.707 R . (d) Plugging this value back into Eq. 22-16 with the values stated in the problem, we find E max = 3.46 × 10 7 N/C....
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This note was uploaded on 01/25/2011 for the course PHYSICS 17029 taught by Professor Rebello,nobels during the Fall '10 term at Kansas State University.

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