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Chapter 24 Solutions

# Chapter 24 Solutions - wright(daw2557 Chapter 24 de(14443...

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wright (daw2557) – Chapter 24 – de – (14443) 1 This print-out should have 7 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0points At distance r from a point charge q , the elec- tric potential is 414 V and the magnitude of the electric field is 221 N / C. Determine the value of q . Correct answer: 8 . 62913 × 10 - 8 C. Explanation: Let : k e = 8 . 98755 × 10 9 N · m 2 / C 2 , V = 414 V , and e = 221 N / C . E = k e q r 2 and V = k e q r , so that V E = r . The potential is V = k e q r = k e q V E = k e qE V q = V 2 k e E = (414 V) 2 (8 . 98755 × 10 9 N · m 2 / C 2 ) (221 N / C) = 8 . 62913 × 10 - 8 C . 002(part1of2)10.0points Calculate the speed of a proton that is acceler- ated from rest through a potential difference of 188 V. Correct answer: 1 . 8978 × 10 5 m / s. Explanation: Let : m p = 1 . 67262 × 10 - 27 kg , q = 1 . 60218 × 10 - 19 C , and Δ V = 188 V . By conservation of energy Δ K + q Δ V = 0 , 1 2 mv 2 = q Δ V v p = radicalBigg 2 q Δ V m p = radicalBigg 2 (1 . 60218 × 10 - 19 C) (188 V) 1 . 67262 × 10 - 27 kg = 1 . 8978 × 10 5 m / s .

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Chapter 24 Solutions - wright(daw2557 Chapter 24 de(14443...

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