solution_pdft1 - Version 096/ABCAA midterm 01 Turner...

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Unformatted text preview: 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 Earths 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 = 9 . 7 . Flux is = E A cos = E ab cos = (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 1 . 9 c m 160 g + q 1 . 9 c m 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 C. Explanation: Let : = 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 . Newtons 2nd law: summationdisplay F = ma. 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 Version 096/ABCAA midterm 01 Turner (59070) 2 against the force of gravity mg . The electro- static force is due to the other two charges and is therefore horizontal. In the x-direction F T sin = 0 . In the y-direction T cos mg = 0 . These can be rewritten as F = T sin and mg = T cos . Dividing the former by the latter, we find F mg = tan , or F = mg tan (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 , 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)....
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This note was uploaded on 12/11/2009 for the course PHY 303L taught by Professor Turner during the Spring '08 term at University of Texas at Austin.

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solution_pdft1 - Version 096/ABCAA midterm 01 Turner...

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