# HW6Sol - Platt, David Homework 6 Due: Sep 19 2005, 4:00 am...

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Unformatted text preview: Platt, David Homework 6 Due: Sep 19 2005, 4:00 am Inst: Ken Shih 1 This print-out should have 12 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. The due time is Central time. Note: Please give your answer for 003 in unit of eV (electron volt). 001 (part 1 of 1) 10 points Four charges are placed at the corners of a square, where q is positive ( q > 0). There is no charge in the center of the square. Q 4 =- q Q 1 =- q Q 3 = + q Q 2 = + q a The magnitude of the total electrostatic energy of the system is given by 1. U = 4 k e q 2 a . 2. U = 2 k e q 2 a 2 . 3. U = 4 k e q 2 a 2 . 4. U = 2 k e q 2 a . correct 5. U = 8 k e q 2 a 2 . 6. U = 4 2 k e q 2 a . 7. U = 2 k e q 2 a 2 . 8. U = 2 2 k e q 2 a 2 . 9. U = 2 k e q 2 a . 10. U = 4 2 k e q 2 a 2 . Explanation: The total electrostatic energy is U = U 12 + U 13 + U 14 + U 23 + U 24 + U 34 = k e q 2 a +1- 1 2- 1- 1- 1 2 + 1 =- 2 k e q 2 a | U | = 2 k e q 2 a . 002 (part 1 of 2) 10 points A voltage of 9 V is applied across the ends of a piece of copper wire 9 cm long. The mass of an electron is 9 . 11 10- 31 kg and its charge is 1 . 6 10- 19 C. What is the magnitude of the electrons acceleration? Correct answer: 1 . 75631 10 13 m / s 2 . Explanation: Let : V = 9 V , d = 9 cm = 0 . 09 m , m = 9 . 11 10- 31 kg , and q = 1 . 6 10- 19 C . Since the electric field is E = V d , and F = ma = q E , we have a = q V md = (1 . 6 10- 19 C) (9 V) (9 . 11 10- 31 kg) (0 . 09 m) = 1 . 75631 10 13 m / s 2 . 003 (part 2 of 2) 10 points After traveling 1 10- 8 m, if it does not col- lide with a copper ion over this distance, what is the kinetic energy of the electron?...
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## This note was uploaded on 04/10/2009 for the course PHY 303L taught by Professor Turner during the Fall '08 term at University of Texas at Austin.

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HW6Sol - Platt, David Homework 6 Due: Sep 19 2005, 4:00 am...

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