Chapter 24 Solutions

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

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Unformatted text preview: 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 10.0 points 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 q E 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 (part 1 of 2) 10.0 points 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 V = 188 V ....
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This note was uploaded on 03/23/2011 for the course PHYS 1444 taught by Professor Sureshsharma during the Spring '09 term at UT Arlington.

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

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