College Physics (4th Edition)

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

Chapter 17 - Problems - Page 652: 11

Answer

$U = -1.1\times 10^{-5}~J$

Work Step by Step

We can find the electric potential energy due to the $+10.0~nC$ charge and the $-10.0~nC$ charge: $U = \frac{k~q_1~q_2}{r}$ $U = \frac{(9.0\times 10^9~N~m^2/C^2)(10.0\times 10^{-9}~C)(-10.0\times 10^{-9}~C)}{0.080~m}$ $U = -1.125\times 10^{-5}~J$ We can find the electric potential energy due to the $+10.0~nC$ charge and the $-4.2~nC$ charge: $U = \frac{k~q_1~q_2}{r}$ $U = \frac{(9.0\times 10^9~N~m^2/C^2)(10.0\times 10^{-9}~C)(-4.2\times 10^{-9}~C)}{0.080~m}$ $U = -4.725\times 10^{-6}~J$ We can find the electric potential energy due to the $-10.0~nC$ charge and the $-4.2~nC$ charge: $U = \frac{k~q_1~q_2}{r}$ $U = \frac{(9.0\times 10^9~N~m^2/C^2)(-10.0\times 10^{-9}~C)(-4.2\times 10^{-9}~C)}{0.080~m}$ $U = 4.725\times 10^{-6}~J$ We can find the total potential energy of the system of three charges: $U = -1.125\times 10^{-5}~J-4.725\times 10^{-6}~J+4.725\times 10^{-6}~J$ $U = -1.1\times 10^{-5}~J$
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