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Unformatted text preview: mcconnell (kam2342) – homework07 – Turner – (60230) 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 5 pC is uniformly distributed throughout the volume between concentric spherical surfaces having radii of 1 . 1 cm and 3 . 1 cm. Let: k e = 8 . 98755 × 10 9 N · m 2 / C 2 . What is the magnitude of the electric field 2 cm from the center of the surfaces? Correct answer: 26 . 3255 N / C. Explanation: Let : q tot = 5 pC = 5 × 10 − 12 C , r 1 = 1 . 1 cm , r 2 = 3 . 1 cm , and r = 2 cm = 0 . 02 m . By Gauss’ law, Φ c = contintegraldisplay vector E · d vector A = q in ǫ 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 = (5 pC) × bracketleftbigg (2 cm) 3 (1 . 1 cm) 3 (3 . 1 cm) 3 (1 . 1 cm) 3 bracketrightbigg = 1 . 17164 pC = 1 . 17164 × 10 − 12 C . And by Gauss’s Law, E = k e q in r 2 = 8 . 98755 × 10 9 N · m 2 / C 2 × 1 . 17164 × 10 − 12 C (0 . 02 m) 2 = 26 . 3255 N / C . 002 (part 1 of 2) 10.0 points A uniformly charged, straight filament 7 m in length has a total positive charge of 3 μ C. An uncharged cardboard cylinder 3 cm in length and 7 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 11/22/2010 for the course PHYS 303 taught by Professor Turner during the Spring '10 term at University of Texas.
 Spring '10
 Turner

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