Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (4th Edition)

Published by Pearson
ISBN 10: 0133942651
ISBN 13: 978-0-13394-265-1

Chapter 25 - The Electric Potential - Exercises and Problems - Page 709: 2


$v=1.4\times 10^{5} m/s$

Work Step by Step

By the conservation of energy; Loss in energy = Gain in energy Thus; $q\Delta V=\frac{1}{2}mv^{2}$ $v=[\frac{2q\Delta V}{m}]^{1/2}$ [ Since, $\Delta V= Ed$] $v=[\frac{2\times 1.6\times 10^{-19}\times 50, 000\times 2\times 10 ^{-3}}{1.67 \times 10^{-27} }]^{1/2}$ $v=1.4\times 10^{5} m/s$
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