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Unformatted text preview: benavides (jjb2356) – homework 07 – Turner – (59130) 1 This printout should have 11 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. 001 10.0 points A charge of 6 pC is uniformly distributed throughout the volume between concentric spherical surfaces having radii of 1 cm and 3 cm. Let: k e = 8 . 98755 × 10 9 N · m 2 / C 2 . What is the magnitude of the electric field 2 . 3 cm from the center of the surfaces? Correct answer: 43 . 7825 N / C. Explanation: Let : q tot = 6 pC = 6 × 10 − 12 C , r 1 = 1 cm , r 2 = 3 cm , and r = 2 . 3 cm = 0 . 023 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 = (6 pC) × bracketleftbigg (2 . 3 cm) 3 (1 cm) 3 (3 cm) 3 (1 cm) 3 bracketrightbigg = 2 . 577 pC = 2 . 577 × 10 − 12 C . And by Gauss’s Law, E = k e q in r 2 = 8 . 98755 × 10 9 N · m 2 / C 2 × 2 . 577 × 10 − 12 C (0 . 023 m) 2 = 43 . 7825 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 8 μ C. An uncharged cardboard cylinder 2 cm in length and 8 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. Correct answer: 2581 . 56 N · m 2 / C. Explanation: Let : L = 7 m , r = 8 cm = 0 . 08 m , l = 2 cm = 0 . 02 m , and Q = 8 μ C = 8 × 10 − 6 C . Calculate the flux through the cylinder us ing Gauss’ law. The flux through a closed surface is Φ = q enclosed ǫ . Our Gaussian surface will be the cardboard tube, but we will close off the ends of the cylinder for our imaginary surface. We will assume that the filament is long enough (com pared tp the cylinder) that field lines emerge from the filament only radially — they do not penetrate the caps on our Gaussian surface....
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This note was uploaded on 03/26/2012 for the course PHY 303L taught by Professor Turner during the Spring '08 term at University of Texas.
 Spring '08
 Turner
 Charge, Work

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