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Unformatted text preview: Version 096/ABCAA midterm 01 Turner (59070) 1 This printout should have 18 questions. Multiplechoice 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 xdirection F T sin = 0 . In the ydirection 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.
 Spring '08
 Turner

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