College Physics (4th Edition)

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

Chapter 17 - Problems - Page 654: 49


The final kinetic energy is $2.8\times 10^{-16}~J$

Work Step by Step

We can use conservation of energy to find the final kinetic energy $K_2$: $K_2+U_2 = K_1+U_1$ $K_2 = U_1-U_2 + K_1$ $K_2 = -\Delta V~q +K_1$ $K_2 = -(-500~J/C)(2)(1.6\times 10^{-19}~C)+1.20\times 10^{-16}~J$ $K_2 = 2.8\times 10^{-16}~J$ The final kinetic energy is $2.8\times 10^{-16}~J$.
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