694_Physics ProblemsTechnical Physics

694_Physics ProblemsTechnical Physics - 34 Gausss Law...

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34 Gauss’s Law P24.20 Φ E e kQ R r , .. . . hole hole 22 Nm C C m m =⋅ = F H G I K J = ×⋅ × F H G G I K J J × EA 2 2 96 2 3 2 899 10 100 10 0100 ππ ej e j af Φ E , . hole 2 28 2 P24.21 Φ E q = = × in 2 C 8.85 10 C N m 0 6 12 7 170 10 192 10 . (a) ΦΦ EE bg one face 2 == 1 6 6 7 . Φ E one face 2 MN m C 320 . (b) Φ E 19 2 . MNm C 2 (c) The answer to (a) would change because the flux through each face of the cube would not be equal with an asymmetric charge distribution. The sides of the cube nearer the charge would have more flux and the ones further away would have less. The answer to (b) would remain the same, since the overall flux would remain the same. P24.22 No charge is inside the cube. The net flux through the cube is zero. Positive flux comes out through the three faces meeting at g . These three faces together fill solid angle equal to one-eighth of a sphere as seen from
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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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