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Exam1 - Version 003 TEST01 Tsoi(59090 This print-out should...

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Version 003 – TEST01 – Tsoi – (59090) 1 This print-out should have 14 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 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 , = 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
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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 0 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 0 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 . Combining, we get E 2 r 2 2 E 1 r 2 1 = R 3 r 3 1 E 2 = R 3 r 1 r 2 2 · E 1 = (12 cm) 3 (4 . 2065 cm) (26 . 3749 cm) 2 (36352 N / C) = 21466 . 9 N / C . Note that in this solution, we did not actually need to remember the specific formulae for the surface area and volume of a sphere as the constants cancelled. Method 2: First, calculate the charge density ρ inside the insulating sphere. Select a spherical Gaussian surface with r = 4 . 2065 cm, concentric with the charge distribution. To apply Gauss’ law in this situation, we must know the charge q in within the Gaussian surface of volume V . To calculate q in , we use the fact that q in = ρ V , where ρ is the charge per unit volume and V is the volume enclosed by the Gaussian
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