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A particle with a charge of +4.20 nC is in a uniform

University Physics | 13th Edition | ISBN: 9780321675460 | Authors: Hugh D. Young, Roger A. Freedman ISBN: 9780321675460 31

Solution for problem 14E Chapter 23

University Physics | 13th Edition

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University Physics | 13th Edition | ISBN: 9780321675460 | Authors: Hugh D. Young, Roger A. Freedman

University Physics | 13th Edition

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Problem 14E

A particle with a charge of +4.20 nC is in a uniform electric field directed to the left. It is released from rest and moves to the left; after it has moved 6.00 cm, its kinetic energy is found to be +1.50 × 10?6 J. (a) What work was done by the electric force? (b) What is the potential of the starting point with respect to the end point? (c) What is the magnitude of .

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Solution 14E The work done in moving a charge q across a potential difference V is given by W = qV . From the data given, we can calculate the potential difference and then the electric field. Given, charge q = + 4.20 nC = + 4.20 × 10 9C Kinetic energy = 1.50 × 10 6 J (a) Here, the kinetic energy of the particle is manifested as the work done on it. The particle gets its kinetic energy because of the work done on it by the electric force. Hence 6 the magnitude of the work done is 1.50 × 10 J . (b) The potential difference can be calculate from the equation W = qV . Therefore, the 6 9 potential difference V = W/q = 1.50 × 10 /4.20 × 10 V = 357 V (c) Distance moved by the particle d = 6.00 cm = 6.00 × 10 2 m Therefore, the magnitude of electric field = 357 2V /m = 5950 V /m 6.00×10

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Chapter 23, Problem 14E is Solved
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Textbook: University Physics
Edition: 13
Author: Hugh D. Young, Roger A. Freedman
ISBN: 9780321675460

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A particle with a charge of +4.20 nC is in a uniform