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58. (a) It is clear from symmetry (also from Eq. 2216) 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. 2211,
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. 2216 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. 2216 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.
 Fall '10
 Rebello,NobelS

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