671_Physics ProblemsTechnical Physics

# 671_Physics ProblemsTechnical Physics - Chapter 23 Section...

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Chapter 23 11 Section 23.5 Electric Field of a Continuous Charge Distribution P23.25 E k dd kQ e e e = + = + = + = ×× + λ A A AA af bg afaf ej e j a f 899 10 220 10 0 290 0 140 0 290 96 .. . E 159 10 6 ., N C directed toward the rod. FIG. P23.25 P23.26 E kdq x e = z 2 , where dq dx = 0 Ek dx x k x k x e x e x e == F H G I K J = z λλ 0 2 0 0 0 0 0 1 The direction is or left for −> ± i 0 0 P23.27 E kxQ xa x x x x e = + = + = × + 22 32 5 2 899 10 750 10 0100 674 10 00100 e j e j . . . (a) At x = . m , Ei i = 664 10 664 6 . ± . ± NC MNC (b) At x = 00500 m , i = 241 10 241 7 . ± . ± (c) At x = 0300 m , i = 640 10 640 6 . ± . ± (d) At x = 100 m , i = 0664 5 . ± . ± P23.28 EE i ii i L N M M M O Q P P P =− F H G I K J zz z d kx d x x kxx d xkx x k x e x e x e x e 00 3 3 2 0 0 0 1 2 2 ± ±± ± P23.29 E kQx e = + For a maximum, dE dx Qk x e = + + L N M M M O Q P
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## This note was uploaded on 12/14/2011 for the course PHY 203 taught by Professor Staff during the Fall '11 term at Indiana State University .

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