College Physics (4th Edition)

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

Chapter 17 - Problems - Page 652: 8


$U = -1.125\times 10^{-5}~J$

Work Step by Step

We can find the electric potential energy: $U = \frac{k~q_1~q_2}{r}$ $U = \frac{(9.0\times 10^9~N~m^2/C^2)(10.0\times 10^{-9}~C)(-10.0\times 10^{-9}~C)}{0.080~m}$ $U = -1.125\times 10^{-5}~J$
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