This preview shows pages 1–3. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: sanne (as42476) – Homework 03 – Yao – (59110) 1 This printout should have 21 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. 001 10.0 points It takes 127 J of work to move 2 . 3 C of charge from a positive plate to a negative plate. What voltage difference exists between the plates? Correct answer: 55 . 2174 V. Explanation: Let : W = 127 J and q = 2 . 3 C . The voltage difference is V = W q = 127 J 2 . 3 C = 55 . 2174 V . 002 (part 1 of 2) 10.0 points An electron moving parallel to the x axis has an initial speed of 3 × 10 6 m / s at the origin. Its speed is reduced to 6 × 10 5 m / s at the point x P , 4 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: 24 . 5619 V. Explanation: Let : m e = 9 . 10939 × 10 − 31 kg , q e = 1 . 60218 × 10 − 19 C , v i = 3 × 10 6 m / s , v f = 6 × 10 5 m / s , and d = 4 cm . By conservation of energy ( K + U ) f = ( K + U ) i U f U i ≡ Δ U = K i K f , Using the definition of kinetic energy and solv ing for Δ U yields Δ U = 1 2 m e ( v 2 i v 2 f ) = 1 2 (9 . 10939 × 10 − 31 kg) × bracketleftBig ( 3 × 10 6 m / s ) 2 (6 × 10 5 m / s) 2 bracketrightBig = 3 . 93526 × 10 − 18 J , Finally, the potential difference is Δ V = Δ U q e = 3 . 93526 × 10 − 18 J ( 1 . 60218 × 10 − 19 C) = 24 . 5619 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 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 correct Explanation: As the electron moves from the origin to point P its kinetic energy decreases. Because sanne (as42476) – Homework 03 – Yao – (59110) 2 energy is conserved, the potential energy of the electron increases Δ U = K K P > U P > U . Because the change in potential energy and the change in electric potential are related by Δ U = q Δ 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 1 . 3 × 10 5 V / m di rected along the positive x axis. The proton undergoes a displacement of 0 . 4 m in the di rection of the electric field as shown in the figure....
View
Full
Document
This note was uploaded on 11/03/2009 for the course PHY 303L taught by Professor Turner during the Spring '08 term at University of Texas.
 Spring '08
 Turner
 Physics, Charge, Work

Click to edit the document details