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solution_pdft1

# solution_pdft1 - Version 096/ABCAA midterm 01 Turner(59070...

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Version 096/ABCAA – midterm 01 – Turner – (59070) 1 This print-out should have 18 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points A vertical electric field of magnitude 19100 N / C exists above Earth’s surface on a day when a thunderstorm is brewing. A car with a rectangular size of 6 . 02 m by 4 . 09 m is traveling along a roadway sloping downward at 9 . 7 . Determine the electric flux through the bot- tom of the car. 1. 573419.0 2. 514644.0 3. 1322320.0 4. 553942.0 5. 235821.0 6. 472045.0 7. 1064980.0 8. 463560.0 9. 567291.0 10. 206298.0 Correct answer: 4 . 6356 × 10 5 N · m 2 / C. Explanation: Let : a = 6 . 02 m , b = 4 . 09 m , E = 19100 N / C , and θ = 9 . 7 . Flux is Φ = E A cos θ = E a b cos θ = (19100 N / C) (6 . 02 m) (4 . 09 m) cos 9 . 7 = 4 . 6356 × 10 5 N · m 2 / C 002 10.0 points Three identical point charges, each of mass 160 g and charge + q , hang from three strings, as in the figure. The acceleration of gravity is 9 . 8 m / s 2 , and the Coulomb constant is 8 . 98755 × 10 9 N · m 2 / C 2 . 9 . 8 m / s 2 10 . 9 cm 160 g + q 10 . 9 cm 160 g + q 42 160 g + q If the lengths of the left and right strings are each 10 . 9 cm, and each forms an angle of 42 with the vertical, determine the value of q . 1. 0.786265 2. 0.918923 3. 0.622486 4. 0.817624 5. 0.44089 6. 1.14218 7. 0.528588 8. 0.676143 9. 0.738161 10. 0.320938 Correct answer: 0 . 817624 μ C. Explanation: Let : θ = 42 , m = 160 g = 0 . 16 kg , L = 10 . 9 cm = 0 . 109 m , g = 9 . 8 m / s 2 , and k e = 8 . 98755 × 10 9 N · m 2 / C 2 . Newton’s 2nd law: summationdisplay F = m a . Electrostatic force between point charges q 1 and q 2 separated by a distance r F = k e q 1 q 2 r 2 . All three charges are in equilibrium, so for each holds summationdisplay F = 0 . Consider the forces acting on the charge on the right. There must be an electrostatic force F acting on this charge, keeping it balanced

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Version 096/ABCAA – midterm 01 – Turner – (59070) 2 against the force of gravity m g . The electro- static force is due to the other two charges and is therefore horizontal. In the x -direction F T sin θ = 0 . In the y -direction T cos θ m g = 0 . These can be rewritten as F = T sin θ and m g = T cos θ . Dividing the former by the latter, we find F m g = tan θ , or F = m g tan θ (1) = (0 . 16 kg) (9 . 8 m / s 2 ) tan 42 = 1 . 41183 N . The distance between the right charge and the middle charge is L sin θ , and the distance to the left one is twice that. Since all charges are of the same sign, both forces on the right charge are repulsive (pointing to the right). We can add the magnitudes F = k e q q ( L sin θ ) 2 + k e q q (2 L sin θ ) 2 or F = 5 k e q 2 4 L 2 sin 2 θ . (2) We have already found F , and the other quan- tities are given, so we solve for the squared charge q 2 q 2 = 4 F L 2 sin 2 θ 5 k e (3) or, after taking the square root (we know q > 0) and substituting F from Eq. 1 into Eq. 3 and solving for q , we have q = 2 L sin θ radicalbigg m g tan θ 5 k e = 2 (0 . 109 m) sin 42 × radicalBigg (0 . 16 kg) (9 . 8 m / s 2 ) tan 42 5 (8 . 98755 × 10 9 N · m 2 / C 2 ) = 8 . 17624 × 10 7 C = 0 . 817624 μ C .
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solution_pdft1 - Version 096/ABCAA midterm 01 Turner(59070...

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