PHY303L HW3 - thwing (jrt2329) Homework 03 Yao (59110) 1...

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Unformatted text preview: thwing (jrt2329) Homework 03 Yao (59110) 1 This print-out should have 21 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. 001 10.0 points It takes 146 J of work to move 3 C of charge from a positive plate to a negative plate. What voltage difference exists between the plates? Correct answer: 48 . 6667 V. Explanation: Let : W = 146 J and q = 3 C . The voltage difference is V = W q = 146 J 3 C = 48 . 6667 V . 002 (part 1 of 2) 10.0 points An electron moving parallel to the x axis has an initial speed of 5 10 6 m / s at the origin. Its speed is reduced to 1 10 5 m / s at the point x P , 2 cm away from the origin. The mass of the electron is 9 . 10939 10 31 kg and the charge of the electron is- 1 . 60218 10 19 C. Calculate the magnitude of the potential difference between this point and the origin. Correct answer: 71 . 042 V. Explanation: Let : m e = 9 . 10939 10 31 kg , q e =- 1 . 60218 10 19 C , v i = 5 10 6 m / s , v f = 1 10 5 m / s , and d = 2 cm . By conservation of energy ( K + U ) f = ( K + U ) i U f- U i U = K i- K f , Using the definition of kinetic energy and solv- ing for U yields U = 1 2 m e ( v 2 i- v 2 f ) = 1 2 (9 . 10939 10 31 kg) bracketleftBig ( 5 10 6 m / s ) 2- (1 10 5 m / s) 2 bracketrightBig = 1 . 13822 10 17 J , Finally, the potential difference is V = U- q e = 1 . 13822 10 17 J- (- 1 . 60218 10 19 C) = 71 . 042 V . 003 (part 2 of 2) 10.0 points Consider the setup from part one. Let U and V denote the potential energy and the electric potential at the origin. Let U P and V P denote the potential energy and the electric potential at the final point x P of the electron. Which of the following statements is true? 1. V P < V and U P = U 2. V P > V and U P < U 3. V P > V and U P = U 4. V P = V and U P = U 5. V P = V and U P > U 6. V P < V and U P > U correct 7. V P = V and U P < U 8. V P < V and U P < U 9. None of these. 10. V P > V and U P > U Explanation: As the electron moves from the origin to point P its kinetic energy decreases. Because thwing (jrt2329) Homework 03 Yao (59110) 2 energy is conserved, the potential energy of the electron increases U = K- K P > U P > U . Because the change in potential energy and the change in electric potential are related by U = q V , the electron moves downward in potential ( V P < V ) when its potential energy increases. The changes in potential and potential energy differ by a negative sign because the electron has negative charge. 004 (part 1 of 3) 10.0 points A proton is released from rest in a uniform electric field of magnitude 40000 V / m di- rected along the positive x axis. The proton undergoes a displacement of 0 . 5 m in the di- rection of the electric field as shown in the figure....
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This note was uploaded on 11/11/2009 for the course PHY 59110 taught by Professor Yao during the Spring '09 term at University of Texas at Austin.

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PHY303L HW3 - thwing (jrt2329) Homework 03 Yao (59110) 1...

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