P24_008 - V = 15 m 3 , and recognizing that we are dealing...

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8. (a) The total surface area bounding the bathroom is A =2(2 . 5 × 3 . 0) + 2 (3 . 0 × 2 . 0) + 2 (2 . 0 × 2 . 5) = 37 m 2 . The absolute value of the total electric flux, with the assumptions stated in the problem, is | Φ | = | ~ E · ~ A | = | ~ E | A = (600)(37) = 22 × 10 3 N · m 2 /C. By Gauss’ law, we conclude that the enclosed charge (in absolute value) is | q enc | = ε 0 | Φ | =2 . 0 × 10 7 C. Therefore, with volume
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Unformatted text preview: V = 15 m 3 , and recognizing that we are dealing with negative charges (see problem), we Fnd the charge density is q enc /V = − 1 . 3 × 10 − 8 C/m 3 . (b) We Fnd ( | q enc | /e ) /V = (2 . × 10 − 7 / 1 . 6 × 10 − 19 ) / 15 = 8 . 2 × 10 10 excess electrons per cubic meter....
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This note was uploaded on 11/12/2011 for the course PHYS 2001 taught by Professor Sprunger during the Fall '08 term at LSU.

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