# HW 8 - benavides(jjb2356 homework 08 Turner(59130 This...

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benavides (jjb2356) – 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 . 3 × 10 6 V. a) How much kinetic energy has the proton acquired? Correct answer: 8 . 48 × 10 13 J. Explanation: Let : Δ V = 5 . 3 × 10 6 V and q = 1 . 60 × 10 19 C . Δ K = Δ U = q Δ V = (1 . 60 × 10 19 C) (5 . 3 × 10 6 V) = 8 . 48 × 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 . 18394 × 10 7 m / s. Explanation: Let : m = 1 . 673 × 10 27 kg . Since K i = 0 J , Δ K = K f = 1 2 m v 2 f v f = radicalbigg 2 K f m = radicalBigg 2 (8 . 48 × 10 13 J) 1 . 673 × 10 27 kg = 3 . 18394 × 10 7 m / s . 003 10.0 points Points A (1 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 = 6 N / C and E y = 4 N / C. What is the potential difference V A - V B ? Correct answer: 22 V. Explanation: Let : E x = 6 N / C , E y = 4 N / C , ( x A , y A ) = (1 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 = (6 N / C) (4 m - 1 m) + (4 N / C) (5 m - 4 m) = 22 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.

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benavides (jjb2356) – homework 08 – Turner – (59130) 2 x y 3 7 π + + + + + + + + + + + O r What is the potential V O at the origin O due to this arc? 1. V O = 3 14 λ ǫ 0 2. V O = 1 7 λ ǫ 0 3. V O = 1 5 λ ǫ 0 4. V O = 0 5. V O = 5 28 λ ǫ 0 6. V O = 5 32 λ ǫ 0 7. V O = 3 28 λ ǫ 0 correct 8. V O = 3 22 λ ǫ 0 9. V O = 5 24 λ ǫ 0 10. V O = 5 36 λ ǫ 0 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 = λ ds = λ r dθ , so V = k e integraldisplay 3 7 π 0 λ dθ = 1 4 π ǫ 0 integraldisplay 3 7 π 0 λ dθ = λ 4 π ǫ 0 θ vextendsingle vextendsingle vextendsingle vextendsingle 3 7 π 0 = λ 4 π ǫ 0 parenleftbigg 3 7 π - 0 parenrightbigg = 3 28 λ ǫ 0 . 005 10.0 points A dipole field pattern is shown in the figure. Consider various relationships between the electric potential at different points given in the figure.
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