problem22_33

University Physics with Modern Physics with Mastering Physics (11th Edition)

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22.33: To find the charge enclosed, we need the flux through the parallelepiped: C m N 5 . 37 60 cos ) C N 10 50 . 2 )( m 0600 . 0 )( m 0500 . 0 ( 60 cos 2 4 1 1 = ° × = ° = Φ AE C m N 105 60 cos ) C N 10 00 . 7 )( m 0600 . 0 )( m 0500 . 0 ( 120 cos 2 4 2 2 - = ° × = ° = Φ AE So the total flux is , C m N 5 . 67 C m N ) 105 5 . 37 ( 2 2 2 1 - = - = Φ + Φ = Φ and . C 10 97 . 5 ) C m N 5 . 67 ( 10 0 2 0 - × - = - = Φ = ε ε q b) There must be a net charge (negative) in the parallelepiped since there is a net
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Unformatted text preview: flux flowing into the surface. Also, there must be an external field or all lines would point toward the slab....
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This document was uploaded on 02/06/2008.

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