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Unformatted text preview: saldana (avs387) homework 08 Turner (56705) 1 This printout should have 12 questions. Multiplechoice questions may continue on the next column or page find all choices before answering. 001 (part 1 of 2) 10.0 points A proton is accelerated through a potential difference of 4 . 4 10 6 V. a) How much kinetic energy has the proton acquired? Correct answer: 7 . 04 10 13 J. Explanation: Let : V = 4 . 4 10 6 V and q = 1 . 60 10 19 C . K = U = q V = (1 . 60 10 19 C) (4 . 4 10 6 V) = 7 . 04 10 13 J . 002 (part 2 of 2) 10.0 points b) If the proton started at rest, how fast is it moving? Correct answer: 2 . 90104 10 7 m / s. Explanation: Let : m = 1 . 673 10 27 kg . Since K i = 0 J , K = K f = 1 2 mv 2 f v f = radicalbigg 2 K f m = radicalBigg 2 (7 . 04 10 13 J) 1 . 673 10 27 kg = 2 . 90104 10 7 m / s . 003 10.0 points Points A (1 m, 4 m) and B (3 m, 7 m) are in a region where the electric field is uniform and given by vector E = E x + E y , where E x = 4 N / C and E y = 3 N / C. What is the potential difference V A V B ? Correct answer: 17 V. Explanation: Let : E x = 4 N / C , E y = 3 N / C , ( x A ,y A ) = (1 m , 4 m) , and ( x B ,y B ) = (3 m , 7 m) . We know V ( A ) V ( B ) = integraldisplay A B vector E dvectors = integraldisplay B A vector E dvectors For a uniform electric field vector E = E x + E y . Now consider the term E x dvectors in the inte grand. E x is just a constant and dvectors may be interpreted as the projection of dvectors onto x , so that E x dvectors = E x dx. Likewise E y dvectors = E y dy . Or more simply, dvectors = dx + dy dotting it with E x + E y gives the same result as above. Therefore V A V B = E x integraldisplay x B x A dx + E y integraldisplay y B y A dy = (4 N / C) (3 m 1 m) + (3 N / C) (7 m 4 m) = 17 V . Note that the potential difference is inde pendent of the path taken from A to B. 004 10.0 points saldana (avs387) homework 08 Turner (56705) 2 Consider a circular arc of constant linear charge density as shown below. x y 3 7 + + + + + + + + + + + O r What is the potential V O at the origin O due to this arc? 1. V O = 5 36 2. V O = 3 28 correct 3. V O = 5 24 4. V O = 1 7 5. V O = 3 14 6. V O = 5 32 7. V O = 5 28 8. V O = 0 9. V O = 1 5 10. V O = 3 22 Explanation: The potential at a point due to a continuous charge distribution can be found using V = k e integraldisplay dq r . In this case, with linear charge density , dq = ds = r d , so V = k e integraldisplay 3 7 d = 1 4 integraldisplay 3 7 d = 4 vextendsingle vextendsingle vextendsingle vextendsingle 3 7 = 4 parenleftbigg 3 7  parenrightbigg = 3 28 ....
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This note was uploaded on 10/26/2010 for the course PHYS 56705 taught by Professor Turner during the Spring '10 term at University of Texas at Austin.
 Spring '10
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