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Unformatted text preview: fierro (jmf2547) HW 07 berk (60290) 1 This printout should have 10 questions. Multiplechoice questions may continue on the next column or page find all choices before answering. 001 10.0 points A charge of 7 pC is uniformly distributed throughout the volume between concentric spherical surfaces having radii of 1 . 7 cm and 3 . 7 cm. Let: k e = 8 . 98755 10 9 N m 2 / C 2 . What is the magnitude of the electric field 2 . 9 cm from the center of the surfaces? Correct answer: 31 . 8528 N / C. Explanation: Let : q tot = 7 pC = 7 10 12 C , r 1 = 1 . 7 cm , r 2 = 3 . 7 cm , and r = 2 . 9 cm = 0 . 029 m . By Gauss law, c = contintegraldisplay vector E d vector A = q in The tricky part of this question is to deter mine the charge enclosed by our Gaussian surface, which by symmetry considerations is chosen to be a concentric sphere with radius r . Since the charge q is distributed uniformly within the solid, we have the relation q in q tot = V in V tot where q in and V in are the charge and volume enclosed by the Gaussian surface. Therefore q in = q tot parenleftbigg V in V tot parenrightbigg = q tot bracketleftbigg r 3 r 3 1 r 3 2 r 3 1 bracketrightbigg = (7 pC) bracketleftbigg (2 . 9 cm) 3 (1 . 7 cm) 3 (3 . 7 cm) 3 (1 . 7 cm) 3 bracketrightbigg = 2 . 98059 pC = 2 . 98059 10 12 C . And by Gausss Law, E = k e q in r 2 = 8 . 98755 10 9 N m 2 / C 2 2 . 98059 10 12 C (0 . 029 m) 2 = 31 . 8528 N / C . 002 (part 1 of 2) 10.0 points A uniformly charged, straight filament 5 m in length has a total positive charge of 5 C. An uncharged cardboard cylinder 2 cm in length and 10 cm in radius surrounds the filament using the filament as its axis of symmetry, with the filament as the central axis of the cylinder. Find the total electric flux through the cylinder....
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This note was uploaded on 09/19/2008 for the course PHY 303L taught by Professor Turner during the Spring '08 term at University of Texas at Austin.
 Spring '08
 Turner
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