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Unformatted text preview: Version 041/AACCB midterm 01 turner (56725) 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 A charge moves a distance of 1 . 7 cm in the direction of a uniform electric field having a magnitude of 200 N / C. The electrical potential energy of the charge decreases by 9 . 69095 10 19 J as it moves. Find the magnitude of the charge on the moving particle. (Hint: The electrical poten tial energy depends on the distance moved in the direction of the field.) 1. 2.35388e18 2. 2.41793e18 3. 2.85028e19 4. 2.86629e18 5. 5.41233e19 6. 2.93034e18 7. 1.52602e18 8. 1.31625e18 9. 3.2506e19 10. 1.21857e18 Correct answer: 2 . 85028 10 19 C. Explanation: Let : d = 1 . 7 cm , E = 200 N / C and U electric = 9 . 69095 10 19 J . U electric = q E d q = U electric E d = 9 . 69095 10 19 J (200 N / C)(0 . 017 m) = 2 . 85028 10 19 C . 002 10.0 points Three identical point charges, each of mass 80 g and charge + q , hang from three strings, as in the figure. 9 . 8 m / s 2 1 c m 80 g + q 1 c m 80 g + q 42 80 g + q If the lengths of the left and right strings are each 10 cm, and each forms an an gle of 42 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. 0.53041 2. 0.300523 3. 0.461041 4. 0.344116 5. 0.912431 6. 0.727291 7. 0.684735 8. 0.517167 9. 1.00443 10. 0.87924 Correct answer: 0 . 53041 C. Explanation: Let : = 42 , m = 80 g = 0 . 08 kg , L = 10 cm = 0 . 1 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 Version 041/AACCB midterm 01 turner (56725) 2 F = mg tan (1) = (0 . 08 kg) (9 . 8 m / s 2 ) tan42 = 0 . 705917 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 radicalbigg F 5 k e = 2 (0 . 1 m) sin 42 radicalBigg . 705917 N 5 (8 . 98755 10 9 N m 2 / C 2 ) = 5 . 3041 10 7 C = . 53041 C ....
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 Spring '08
 Turner
 Physics, Charge

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