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solution957967_pdf - kinsey(cbk323 homework 08 Turner(56725...

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kinsey (cbk323) – homework 08 – Turner – (56725) 1 This print-out should have 12 questions. Multiple-choice questions may continue on the next column or page – fnd all choices beFore answering. 001 (part 1 oF 2) 10.0 points A proton is accelerated through a potential di±erence oF 3 . 0 × 10 6 V. a) How much kinetic energy has the proton acquired? Correct answer: 4 . 8 × 10 13 J. Explanation: Let : Δ V = 3 . 0 × 10 6 V and q = 1 . 60 × 10 19 C . Δ K = Δ U = q Δ V = (1 . 60 × 10 19 C) (3 . 0 × 10 6 V) = 4 . 8 × 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 . 39545 × 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 = r 2 K f m = R 2 (4 . 8 × 10 13 J) 1 . 673 × 10 27 kg = 2 . 39545 × 10 7 m / s . 003 10.0 points Points A (3 m, 3 m) and B (6 m, 7 m) are in a region where the electric feld is uniForm and given by ± E = E x ˆ ı + E y ˆ , where E x = 2 N / C and E y = 4 N / C. What is the potential di±erence V A - V B ? Correct answer: 22 V. Explanation: Let : E x = 2 N / C , E y = 4 N / C , ( x A ,y A ) = (3 m , 3 m) , and ( x B B ) = (6 m , 7 m) . We know V ( A ) - V ( B ) = - i A B ± E · d±s = i B A ± E · ²or a uniForm electric feld ± E = E x ˆ ı + E y ˆ . Now consider the term E x ˆ ı · in the inte- grand. E x is just a constant and ˆ ı · may be interpreted as the projection oF onto x , so that E x ˆ ı · = E x dx. Likewise E y ˆ · = E y dy . Or more simply, = 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 i x B x A dx + E y i y B y A dy = (2 N / C) (6 m - 3 m) + (4 N / C) (7 m - 3 m) = 22 V . Note that the potential di±erence is inde- pendent oF the path taken From A to B. 004 10.0 points
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kinsey (cbk323) – homework 08 – Turner – (56725) 2 Consider a circular arc of constant linear charge density λ as shown below. x y 4 5 π + O r What is the potential V O at the origin O due to this arc? 1. V O = 1 5 λ ǫ 0 correct 2. V O = 5 24 λ ǫ 0 3. V O = 5 28 λ ǫ 0 4. V O = 5 32 λ ǫ 0 5. V O = 3 14 λ ǫ 0 6. V O = 5 36 λ ǫ 0 7. V O = 1 7 λ ǫ 0 8. V O = 7 36 λ ǫ 0 9. V O = 0 10. V O = 3 22 λ ǫ 0 Explanation: The potential at a point due to a continuous charge distribution can be found using V = k e i dq r . In this case, with linear charge density λ, dq = λds = λr dθ , so V = k e i 4 5 π 0 λdθ = 1 4 π ǫ 0 i 4 5 π 0 = λ 4 π ǫ 0 θ v v v v 4 5 π 0 = λ 4 π ǫ 0 p 4 5 π - 0 P = 1 5 λ ǫ 0 . 005 10.0 points A dipole Feld pattern is shown in the Fgure. Consider various relationships between the electric potential at di±erent points given in the Fgure. G M C Y X + - Notice: ²ive potential relationships are given below. a) V M = V Y > V C b) V M = V Y = V C c) V M = V Y < V C d) V G < V C < V X e) V G > V C > V X Which relations shown above are correct? 1. ( d ) only 2. ( a ) and ( d ) only 3. ( c ) and ( e ) only 4. ( c ) only 5. ( b ) and ( d ) only correct 6. ( e ) only 7. ( a ) and ( e ) only
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kinsey (cbk323) – homework 08 – Turner – (56725) 3 8. ( b ) and ( e ) only 9. ( c ) and ( d ) only 10. ( a ) only Explanation: The electric potential due to one single point charge at a distance r from the charge is given by V = k q r .
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