College Physics (4th Edition)

Published by McGraw-Hill Education
ISBN 10: 0073512141
ISBN 13: 978-0-07351-214-3

Chapter 16 - Problems - Page 615: 53


(a) $F = 8.0\times 10^{-17}~N$ (b) The increase in kinetic energy is $~2.4\times 10^{-19}~J$

Work Step by Step

(a) We can find the magnitude of the force exerted on the electron: $F = E~\vert q \vert = (500.0~N/C)(1.6\times 10^{-19}~C) = 8.0\times 10^{-17}~N$ (b) We can find the work done by the electric field on the electron: $W = F~d = (8.0\times 10^{-17}~N)(3.00\times 10^{-3}~m) = 2.4\times 10^{-19}~J$ The increase in kinetic energy is equal to the work done on the electron which is $2.4\times 10^{-19}~J$
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