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Unformatted text preview: Version 068/ABABA midterm 01 turner (56705) 1 This printout should have 19 questions. Multiplechoice questions may continue on the next column or page find all choices before answering. 001 10.0 points Three identical point charges, each of mass 120 g and charge + q , hang from three strings, as in the figure. 9 . 8 m / s 2 1 . 4 c m 120 g + q 1 . 4 c m 120 g + q 37 120 g + q If the lengths of the left and right strings are each 10 . 4 cm, and each forms an an gle of 37 with the vertical, determine the value of q . The acceleration of gravity is 9 . 8 m / s 2 , and the Coulomb constant is 8 . 98755 10 9 N m 2 / C 2 . 1. 0.618551 2. 0.567833 3. 0.488958 4. 0.651312 5. 0.775448 6. 1.014 7. 0.555881 8. 0.934425 9. 0.711318 10. 0.517167 Correct answer: 0 . 555881 C. Explanation: Let : = 37 , m = 120 g = 0 . 12 kg , L = 10 . 4 cm = 0 . 104 m , g = 9 . 8 m / s 2 , and k e = 8 . 98755 10 9 N m 2 / C 2 . Consider the forces acting on the charge on the right. There must be an electrostatic force F acting on this charge, keeping it balanced 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 . F sin F cos = F e mg F = mg tan (1) = (0 . 12 kg) (9 . 8 m / s 2 ) tan37 = 0 . 88618 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). We can add the magnitudes F = k e q q ( L sin ) 2 + k e q q (2 L sin ) 2 F = 5 k e q 2 4 L 2 sin 2 (2) q 2 = 4 F L 2 sin 2 5 k e (3) q = 2 L sin F 5 k e = 2 (0 . 104 m) sin 37 . 88618 N 5 (8 . 98755 10 9 N m 2 / C 2 ) = 5 . 55881 10 7 C = . 555881 C . 002 10.0 points Three small spheres carry equal amounts of electric charge. They are equally spaced and lie along the same line. + What is the direction of the net electric force on each charge due to the other charge? Version 068/ABABA midterm 01 turner (56705) 2 1. + 2. + 3. + 4. + 5. + correct 6. + 7. + 8. + 9. + 10. + Explanation: Since like charges repel and unlike charge attract, + 003 10.0 points A charge of 8 . 4 C is 20 cm above the center of a square of side length 40 cm. The permittivity of free space is 8 . 85 10 12 C 2 / N m 2 . Find the flux through the square. 1. 154426.0 2. 73446.3 3. 116761.0 4. 158192.0 5. 139360.0 6. 82862.5 7. 90395.5 8. 165725.0 9. 124294.0 10. 94162.0 Correct answer: 1 . 58192 10 5 N m 2 / C....
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This note was uploaded on 10/11/2010 for the course PHY 56715 taught by Professor Turner during the Spring '10 term at University of Texas at Austin.
 Spring '10
 Turner

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