College Physics (4th Edition)

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

Chapter 17 - Problems - Page 654: 48


The kinetic energy increased $2.4\times 10^{-19}~J$

Work Step by Step

We can use conservation of energy to find the change in kinetic energy $\Delta K$: $K_2+U_2 = K_1+U_1$ $K_2-K_1 = U_1-U_2$ $\Delta K = -\Delta V~q$ $\Delta K = -E~d~q$ $\Delta K = -(500.0~N/C)(3.0\times 10^{-3}~m)(-1.6\times 10^{-19}~C)$ $\Delta K = 2.4\times 10^{-19}~J$ The kinetic energy increased $2.4\times 10^{-19}~J$.
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