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Unformatted text preview: lew (dl9564) hw 3 Opyrchal (121006) 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 An electric field of magnitude 11000 N / C and directed upward perpendicular to the Earths surface exists on a day when a thunderstorm is brewing. A truck that can be approximated as a rectangle 3 . 3 m by 2 . 3 m is traveling along a road that is inclined 10 relative to the ground. Determine the electric flux through the bot tom of the truck. Correct answer: 82221 . 6 N m 2 / C. Explanation: Let : E = 11000 N / C , = 3 . 3 m , w = 2 . 3 m , and = 10 . By Gauss law, = vector E vector A The flux through the bottom of the car is = E A cos = E w cos = (11000 N / C) (3 . 3 m) (2 . 3 m) cos(10 ) = 82221 . 6 N m 2 / C . 002 10.0 points A nonconducting wall carries a uniform charge density of 7 . 64 C / cm 2 . What is the electric field 10 . 5 cm in front of the wall? Correct answer: 4 . 31434 10 9 N / C. Explanation: Let : = 7 . 64 C / cm 2 and r = 10 . 5 cm . The electric field of an infinite plane of surface charge density is E = 2 = 7 . 64 C / cm 2 2 (8 . 85419 10 12 C 2 / N m 2 ) parenleftbigg 1 C 10 6 C parenrightbigg parenleftbigg 100 cm 1 m parenrightbigg 2 = 4 . 31434 10 9 N / C . 003 (part 1 of 2) 10.0 points A thin spherical shell of radius 6 . 85 m has a total charge of 1 . 84 C distributed uniformly over its surface. Let : k e = 8 . 988 10 9 N m 2 / C 2 . 6 . 8 5 m vector E + + + + + + + + + + + + + + + + + + + + Find the electric field vector E 8 . 3 m from the center of the shell (outside the shell). Correct answer: 2 . 40063 10 8 N / C. Explanation: Let : a = 6 . 85 m , Q = 1 . 84 C , and r = 8 . 3 m . If we construct a spherical Gaussian surface of radius r > a , concentric with the shell, then the charge inside this surface is Q . Therefore the field at a point outside the shell is equiva lent to that of a point charge Q at the center....
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 Fall '08
 Opyrchal

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