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Unformatted text preview: pasha (sep635) HW 3-3 Antoniewicz (56445) 1 This print-out should have 17 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. 001 (part 1 of 2) 4.0 points A moving electron passes near the nucleus of a gold atom, which contains 79 protons and 118 neutrons. At a particular moment, the electron is a distance of 4 . 7 10- 9 m from the gold nucleus. What is the magnitude of the force exerted by the gold nucleus on the electron? Correct answer: 8 . 22835 10- 10 N. Explanation: The magnitude of the electric force is given by F = 1 4 q 1 q 2 r 2 . Plugging in the appropriate values, we obtain F = 1 4 (79 e )( e ) (4 . 7 10- 9 m) 2 = 8 . 22835 10- 10 N . where e is the fundmental unit of charge 1 . 6 10- 19 C. 002 (part 2 of 2) 4.0 points What is the magnitude of the force exerted by the electron on the gold nucleus? Correct answer: 8 . 22835 10- 10 N. Explanation: By Newtons third law, interacting objects must exert equal and opposite forces on each other. Therefore the force on the gold nucleus due to the electron is the same as the force on the electron due to the gold nucleus. 003 8.0 points The figure below shows two positively charged objects and one negatively charged object.- + + What is the direction of the net elec- tric force on the negatively charged object? Choose your answer from the diagram labeled I-IX. I II III IV V VI VII VIII IX zero net force 1. I 2. V 3. VIII 4. IX 5. II 6. IV 7. VII correct 8. III 9. VI Explanation: We can take advantage of symmetries in this problem. With respect to the vertical axis, the two positive charges on the left are equidistant from the negative charge on the right, and since they are on opposite sides pasha (sep635) HW 3-3 Antoniewicz (56445) 2 of the negative charge, their vertical (or y ) components will cancel out. That leaves their horizontal (or x ) components. So the force will either point to the right or to the left (along the + x or- x axis, respectively). Given that the charges on the left are positive, the force must be attractive, so the net electric force points toward the left....
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This note was uploaded on 11/17/2011 for the course PHY 303K taught by Professor Turner during the Fall '08 term at University of Texas at Austin.
- Fall '08