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solution2_pdf - duong(ktd359 Homework 2 Spurlock(44203 This...

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duong (ktd359) – Homework 2 – Spurlock – (44203) 1 This print-out should have 81 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0points You have a potential difference of 11 V. How much work is done to transfer 0 . 22 C of charge through it? Correct answer: 2 . 42 J. Explanation: Let : V = 11 V and q = 0 . 22 C . The potential difference is V = W q W = V q = (11 V) (0 . 22 C) = 2 . 42 J . 002 10.0points The magnitude of a uniform electric field be- tween the two plates is about 2 × 10 5 N / C. If the distance between these plates is 0 . 1 cm, find the potential difference between the plates. Correct answer: 200 V. Explanation: Let : E = 2 × 10 5 N / C and Δ d = 0 . 1 cm = 0 . 001 m . The magnitude of the potential difference is Δ V = E Δ d = (2 × 10 5 N / C) (0 . 001 m) = 200 V . 003 10.0points An ion accelerated through a potential dif- ference of 137 V experiences an increase in kinetic energy of 5 . 92 × 10 17 J. Find the magnitude of the charge on the ion. Correct answer: 4 . 32117 × 10 19 C. Explanation: Let : Δ V = 137 V and K = 5 . 92 × 10 17 J . By conservation of energy Δ K = q Δ V q = Δ K Δ V = 5 . 92 × 10 17 J 137 V = 4 . 32117 × 10 19 C , which has a magnitude of 4 . 32117 × 10 19 C . 004 10.0points Points A (1 m, 2 m) and B (3 m, 4 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 = 4 N / C and E y = 4 N / C. What is the potential difference V A V B ? Correct answer: 16 V. Explanation: Let : E x = 4 N / C , E y = 4 N / C , ( x A , y A ) = (1 m , 2 m) , and ( x B , y B ) = (3 m , 4 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  .
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duong (ktd359) – Homework 2 – Spurlock – (44203) 2 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 = (4 N / C) (3 m 1 m) + (4 N / C) (4 m 2 m) = 16 V . Note that the potential difference is inde- pendent of the path taken from A to B. 005(part1of2)10.0points k e = 8 . 98755 × 10 9 N m 2 / C 2 , and the charge on a proton is 1 . 60218 × 10 19 C . Find the potential 1 . 05 cm from a proton. Correct answer: 1 . 37139 × 10 7 V. Explanation: Let : r 1 = 1 . 05 cm = 0 . 0105 m , k e = 8 . 98755 × 10 9 N m 2 / C 2 , and q p = 1 . 60218 × 10 19 C . The potential V 1 (1 . 05 cm away) is Δ V 1 = k e q p r 1 = (8 . 98755 × 10 9 N m 2 / C 2 ) × (1 . 60218 × 10 19 C) (0 . 0105 m) = 1 . 37139 × 10 7 V . 006(part2of2)10.0points What is the magnitude of the potential differ- ence between two points that are 1 . 05 cm and 2 . 31 cm from a proton?
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