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# solution4 - fullwood(tf5349 Homework 2 Electric Field and...

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fullwood (tf5349) – Homework 2 Electric Field and Potential – Catala – (12054) 1 This print-out should have 20 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points A positron (a particle with a charge of + e and a mass equal to that of an electron) that is accelerated from rest between two points at a fixed potential difference acquires a speed of 9 . 3 × 10 7 m / s. What speed is achieved by a proton acceler- ated from rest between the same two points? (Disregard relativistic effects.) Correct answer: 2 . 17005 × 10 6 m / s. Explanation: Let : v f,positron = 9 . 3 × 10 7 m / s , m positron = 9 . 109 × 10 - 31 kg , m pr = 1 . 673 × 10 - 27 kg , and q pr = q positron = 1 . 60 × 10 - 19 C . Since K i = 0 J , K f = Δ U 1 2 m v 2 f = q Δ V . Δ V = m positron ( v f,positron ) 2 2 q positron = (9 . 109 × 10 - 31 kg) (9 . 3 × 10 7 m / s) 2 2 (1 . 60 × 10 - 19 C) = 24619 . 9 V . Thus v f,pr = radicalBigg 2 Δ U electric m pr = radicalBigg 2 ( q pr Δ V ) m pr = radicalBigg 2 (1 . 60 × 10 - 19 C) (24619 . 9 V) 1 . 673 × 10 - 27 kg = 2 . 17005 × 10 6 m / s . 002 (part 1 of 2) 10.0 points An electric field does 9 J of work on a 0 . 0009 C charge. What is the voltage change? Correct answer: 10000 V. Explanation: Let : W = 9 J and q = 0 . 0009 C . Work is W = qV V = W q = 9 J 0 . 0009 C = 10000 V . 003 (part 2 of 2) 10.0 points The same electric field does 18 J of work on a 0 . 0018 C charge. What is the voltage change? Correct answer: 10000 V. Explanation: Let : W = 18 J and q = 0 . 0018 C . The voltage change is V = W q = 18 J 0 . 0018 C = 10000 V . 004 10.0 points 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.

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fullwood (tf5349) – Homework 2 Electric Field and Potential – Catala – (12054) 2 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 . 005 10.0 points The Coulomb constant is 8 . 98755 × 10 9 N m 2 / C 2 and the acceleration of gravity is 9 . 81 m / s 2 . Find the potential difference between a point infinitely far away and a point 1 . 2 cm from a proton. Correct answer: 1 . 19867 × 10 - 7 V. Explanation: Let : r = 1 . 2 cm , k e = 8 . 99 × 10 9 N · m 2 / C 2 , and q p = 1 . 60 × 10 - 19 C .
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