672_Physics ProblemsTechnical Physics

# 672_Physics ProblemsTechnical Physics - 12 Electric Fields...

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12 Electric Fields P23.30 Ek x xR e =− + F H G I K J 21 22 πσ E x x x x ×− + F H G G I K J J + F H G I K J 2 8 99 10 7 90 10 1 0350 446 10 1 0123 93 2 2 8 2 π .. . . . ej e j af (a) At x = 00500 . m , E = 383 10 383 8 N C M N C (b) At x = 0100 m , E = 324 10 324 8 N C M N C (c) At x = 0500 m , E = 807 10 807 7 NC MNC (d) At x = 200 m , E = 668 10 668 8 P23.31 (a) From Example 23.9: x e + F H G I K J σ == × = × = Q R E 2 3 87 184 10 1 04 10 0 900 9 36 10 93 6 . . . Cm 2 a f appx: e 0 4 MN C about 11% high bg (b) E + F H G I K J = 104 10 1 30 0 300 1 04 10 0 004 96 0 516 8 2 8 . . . . cm 30.0 cm 2 appx: Q r e × × = 2 9 6 2 899 10 520 10 030 0519 . . . . a f MN C about 0.6% high P23.32 The electric field at a distance x is x xe + L N M M O Q P P This is equivalent to Rx + L N M M O Q P P 1 1 For large x , R x 2 2 1 << and 11 2 2 2 2 2 +≈ + R x R x so k e + F H G G I K J J = +− + 1 12 2 1 Substitute = Q R 2 , E kQ x R x
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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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