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Unformatted text preview: nguyen (jmn727) – homework 08 – Turner – (59070) 1 This printout should have 10 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 2 . 9 × 10 6 V. a) How much kinetic energy has the proton acquired? Correct answer: 4 . 64 × 10 − 13 J. Explanation: Let : Δ V = 2 . 9 × 10 6 V and q = 1 . 60 × 10 − 19 C . Δ K = Δ U = q Δ V = (1 . 60 × 10 − 19 C) (2 . 9 × 10 6 V) = 4 . 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: 2 . 35519 × 10 7 m / s. Explanation: Let : m = 1 . 673 × 10 − 27 kg . Since K i = 0 J , Δ K = K f = 1 2 mv 2 f v f = radicalbigg 2 K f m = radicalBigg 2 (4 . 64 × 10 − 13 J) 1 . 673 × 10 − 27 kg = 2 . 35519 × 10 7 m / s . 003 10.0 points Points A (3 m, 4 m) and B (4 m, 5 m) are in a region where the electric field is uniform and given by vector E = E x ˆ ı + E y ˆ , where E x = 2 N / C and E y = 3 N / C. What is the potential difference V A V B ? Correct answer: 5 V. Explanation: Let : E x = 2 N / C , E y = 3 N / C , ( x A ,y A ) = (3 m , 4 m) , and ( x B ,y B ) = (4 m , 5 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 ˆ ı + E y ˆ . Now consider the term E x ˆ ı · dvectors in the inte grand. E x is just a constant and ˆ ı · dvectors may be interpreted as the projection of dvectors onto x , so that E x ˆ ı · dvectors = E x dx. Likewise E y ˆ · dvectors = E y dy . Or more simply, dvectors = dx ˆ ı + dy ˆ dotting it with E x ˆ ı + E y ˆ 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 = (2 N / C) (4 m 3 m) + (3 N / C) (5 m 4 m) = 5 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 λ as shown below. nguyen (jmn727) – homework 08 – Turner – (59070) 2 x y 3 5 π + + + + + + + + + + + + + + + O r What is the potential V O at the origin O due to this arc? 1. V O = 3 28 λ ǫ 2. V O = 5 36 λ ǫ 3. V O = 3 14 λ ǫ 4. V O = 1 7 λ ǫ 5. V O = 5 28 λ ǫ 6. V O = 5 24 λ ǫ 7. V O = 3 20 λ ǫ correct 8. V O = 0 9. V O = 1 5 λ ǫ 10. V O = 5 32 λ ǫ 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 λ, dq = λds = λr dθ , so V = k e integraldisplay 3 5 π λdθ = 1 4 π ǫ integraldisplay 3 5 π λdθ = λ 4 π ǫ θ vextendsingle vextendsingle vextendsingle vextendsingle 3 5 π = λ 4 π ǫ parenleftbigg 3 5 π parenrightbigg = 3 20 λ ǫ ....
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This note was uploaded on 01/21/2010 for the course PHY 303L taught by Professor Turner during the Spring '08 term at University of Texas.
 Spring '08
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