homework 08-solutions - howard (cah3459) homework 08 Turner...

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howard (cah3459) – homework 08 – Turner – (56705) 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 . 7 × 10 6 V. a) How much kinetic energy has the proton acquired? Correct answer: 5 . 92 × 10 - 13 J. Explanation: Let : Δ V =3 . 7 × 10 6 V and q =1 . 60 × 10 - 19 C . Δ K = Δ U = q Δ V = (1 . 60 × 10 - 19 C) (3 . 7 × 10 6 V) = 5 . 92 × 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 . 66028 × 10 7 m / s. Explanation: Let : m . 673 × 10 - 27 kg . Since K i = 0 J , Δ K = K f = 1 2 mv 2 f v f = ± 2 K f m = ² 2 (5 . 92 × 10 - 13 J) 1 . 673 × 10 - 27 kg = 2 . 66028 × 10 7 m / s . 003 10.0 points Points A (3 m, 2 m) and B (5 m, 4 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 = 3 N / C. What is the potential di±erence V A - V B ? Correct answer: 10 V. Explanation: Let : E x = 2 N / C , E y = 3 N / C , ( x A ,y A ) = (3 m , 2 m) , and ( x B B ) = (5 m , 4 m) . We know V ( A ) - V ( B )= - ³ A B * E · d*s = ³ B A * E · d*s ²or a uniForm electric feld * E = E x ˆ ı + E y ˆ  . Now consider the term E x ˆ ı · d*s in the inte- grand. E x is just a constant and ˆ ı · d*s may be interpreted as the projection oF d*s onto x , so that E x ˆ ı · d*s = E x dx. Likewise E y ˆ · d*s = E y dy . Or more simply, d*s = 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 ³ x B x A dx + E y ³ y B y A dy = (2 N / C) (5 m - 3 m) + (3 N / C) (4 m - 2 m) = 10 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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howard (cah3459) – homework 08 – Turner – (56705) 2 Consider a circular arc of constant linear charge density λ as shown below. x y 5 7 π + O r What is the potential V O at the origin O due to this arc? 1. V O = 7 44 λ ( 0 2. V O = 7 52 λ ( 0 3. V O = 3 14 λ ( 0 4. V O = 5 32 λ ( 0 5. V O = 7 36 λ ( 0 6. V O = 5 36 λ ( 0 7. V O =0 8. V O = 5 28 λ ( 0 correct 9. V O = 3 22 λ ( 0 10. V O = 1 6 λ ( 0 Explanation: The potential at a point due to a continuous charge distribution can be found using V = k e ± dq r . In this case, with linear charge density λ , dq = λ ds = λ r dθ , so V = k e ± 5 7 π 0 λ dθ = 1 4 π( 0 ± 5 7 π 0 λ dθ = λ 4 0 θ ² ² ² ² 5 7 π 0 = λ 4 0 ³ 5 7 π - 0 ´ = 5 28 λ ( 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. K R G D L + - Notice: ²ive potential relationships are given below. a) V R = V D >V G b) V R = V D = V G c) V R = V D <V G d) V K G L e) V K G L Which relations shown above are correct? 1. ( a ) only 2. ( b ) and ( d ) only correct 3. ( c ) only 4. ( d ) only 5. ( a ) and ( d ) only 6. ( c ) and ( d ) only 7. ( c ) and ( e ) only
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howard (cah3459) – homework 08 – Turner – (56705) 3 8. ( a ) and ( e ) only 9. ( b ) and ( e ) only 10. ( e ) 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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homework 08-solutions - howard (cah3459) homework 08 Turner...

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