HW 8 - terry (ect328) homework 08 Turner (59130) 1 This...

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Unformatted text preview: terry (ect328) homework 08 Turner (59130) 1 This print-out should have 10 questions. Multiple-choice 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 5 . 4 10 6 V. a) How much kinetic energy has the proton acquired? Correct answer: 8 . 64 10 13 J. Explanation: Let : V = 5 . 4 10 6 V and q = 1 . 60 10 19 C . K = U = q V = (1 . 60 10 19 C) (5 . 4 10 6 V) = 8 . 64 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: 3 . 21384 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 (8 . 64 10 13 J) 1 . 673 10 27 kg = 3 . 21384 10 7 m / s . 003 10.0 points Points A (3 m, 2 m) and B (3 m, 6 m) are in a region where the electric field is uniform and given by vector E = E x + E y , where E x = 6 N / C and E y = 1 N / C. What is the potential difference V A- V B ? Correct answer: 4 V. Explanation: Let : E x = 6 N / C , E y = 1 N / C , ( x A ,y A ) = (3 m , 2 m) , and ( x B ,y B ) = (3 m , 6 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 = (6 N / C) (3 m- 3 m) + (1 N / C) (6 m- 2 m) = 4 V . Note that the potential difference is inde- pendent of the path taken from A to B. 004 10.0 points Consider a circular arc of constant linear charge density as shown below. terry (ect328) homework 08 Turner (59130) 2 x y 2 5 + + + + + + + + + + O r What is the potential V O at the origin O due to this arc? 1. V O = 1 7 2. V O = 3 20 3. V O = 3 28 4. V O = 1 5 5. V O = 5 28 6. V O = 0 7. V O = 3 16 8. V O = 5 32 9. V O = 1 10 correct 10. V O = 5 24 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 2 5 d = 1 4 integraldisplay 2 5 d = 4 vextendsingle vextendsingle vextendsingle vextendsingle 2 5 = 4 parenleftbigg 2 5 - parenrightbigg = 1 10 ....
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This note was uploaded on 02/17/2010 for the course PHY 59130 taught by Professor Turner during the Fall '09 term at University of Texas at Austin.

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HW 8 - terry (ect328) homework 08 Turner (59130) 1 This...

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