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Unformatted text preview: Version 003 – TEST01 – Tsoi – (59090) 1 This printout should have 14 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 30000 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 9 . 5 m by 4 . 8 m is traveling along a road that is inclined 8 ◦ relative to the ground. Determine the electric flux through the bot tom of the truck. 1. 262159.0 2. 251035.0 3. 144657.0 4. 92604.1 5. 1354690.0 6. 396236.0 7. 170259.0 8. 506672.0 9. 146112.0 10. 151979.0 Correct answer: 1 . 35469 × 10 6 N · m 2 / C. Explanation: Let : E = 30000 N / C , ℓ = 9 . 5 m , w = 4 . 8 m , and θ = 8 ◦ . 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 = (30000 N / C) (9 . 5 m) × (4 . 8 m) cos(8 ◦ ) = 1 . 35469 × 10 6 N · m 2 / C . 002 10.0 points An insulating sphere of radius 12 cm has a uniform charge density throughout its vol ume. 12 cm 4 . 2065 cm 26 . 3749 cm p If the magnitude of the electric field at a distance of 4 . 2065 cm from the center is 36352 N / C, what is the magnitude of the electric field at 26 . 3749 cm from the center? 1. 123690.0 2. 33211.1 3. 17024.7 4. 11456.8 5. 32792.1 6. 10481.2 7. 42930.6 8. 21466.9 9. 28655.3 10. 2698.74 Correct answer: 21466 . 9 N / C. Explanation: Let : R = 12 cm , E 1 = 36352 N / C , r 1 = 4 . 2065 cm , r 2 = 26 . 3749 cm , V = 4 3 π R 3 , and ρ = Q V . R r 1 r 2 p Method 1: We know the magnitude of the electric field at a radius r 1 = 4 . 2065 cm Version 003 – TEST01 – Tsoi – (59090) 2 (corresponding to a smaller sphere with sur face area A 1 and volume V 1 ): the magnitude is E 1 = 36352 N / C. We want to find the magnitude E 2 at a radius r 2 = 26 . 3749 cm, corresponding to a sphere with surface area A 2 and volume V 2 that is larger than the insulating sphere. From Gauss’s Law, we know that since the flux is constant over the sphere, E 1 A 1 = Φ 1 = Q 1 ǫ relating the flux through the Gaussian sphere of radius r 1 to the charge enclosed, Q 1 . We also know Q 1 = ρ V 1 . For the outer sphere (radius r 2 = 26 . 3749 cm), E 2 A 2 = Φ 2 = Q ǫ with Q = ρ V (Not ρ V 2 , as the Gaussian sur face is larger than the actual physical sphere, and no charge is outside of the sphere.), so E 2 A 2 E 1 A 1 = ρ V ǫ ◦ ρ V 1 ǫ ◦ = V V 1 . We know the surface area of a sphere is pro portional to the radius squared, and the vol ume is proportional to the cube of the radius (in particular: A = 4 π R 2 and V = 4 3 π R 3 ), so E 2 A 2 E 1 A 1 = E 2 r 2 2 E 1 r 2 1 and V V 1 = R 3 r 3 1 ....
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This note was uploaded on 10/21/2009 for the course PHY 59090 taught by Professor Tsoi during the Fall '09 term at University of Texas at Austin.
 Fall '09
 TSOI
 Physics

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