121HW3 - lew(dl9564 hw 3 Opyrchal(121006 This print-out...

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lew (dl9564) – hw 3 – Opyrchal – (121006) 1 This print-out should have 11 questions. Multiple-choice 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 Earth’s 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 θ = 10 . By Gauss’ law, Φ = vector E · vector A The flux through the bottom of the car is Φ = E A cos θ = E ℓ w cos θ = (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 μ 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 : σ = 7 . 64 μ C / cm 2 and r = 10 . 5 cm . The electric field of an infinite plane of surface charge density σ is E = σ 2 ǫ 0 = 7 . 64 μ C / cm 2 2 (8 . 85419 × 10 12 C 2 / N · m 2 ) × parenleftbigg 1 C 10 6 μ 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 . 85 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. For r > a , E 1 = k Q r 2 = (8 . 988 × 10 9 N · m 2 / C 2 )(1 . 84 C) (8 . 3 m) 2 = 2 . 40063 × 10 8 N / C .
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lew (dl9564) – hw 3 – Opyrchal – (121006) 2 004 (part 2 of 2) 10.0 points Find the electric field E 2 3 . 94 m from the center of the shell (inside the shell).
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