College Physics (4th Edition)

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

Chapter 17 - Problems - Page 652: 18


$\Delta U = 1.92\times 10^{-17}~J$

Work Step by Step

We can find the change in electric potential energy: $\Delta U = q~\Delta V$ $\Delta U = q~(V_B-V_A)$ $\Delta U = (-1.6\times 10^{-19}~C)~[-360~V-(-240~V)]$ $\Delta U = 1.92\times 10^{-17}~J$
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