## College Physics (4th Edition)

Published by McGraw-Hill Education

# Chapter 17 - Problems - Page 652: 12

#### Answer

$U = 0.00285~J$

#### Work Step by Step

We can find the electric potential energy due to the $+4.0~\mu C$ charge and the $+3.0~\mu C$ charge: $U = \frac{k~q_1~q_2}{r}$ $U = \frac{(9.0\times 10^9~N~m^2/C^2)(4.0\times 10^{-6}~C)(3.0\times 10^{-6}~C)}{5.0~m}$ $U = 0.0216~J$ We can find the electric potential energy due to the $+4.0~\mu C$ charge and the $-1.0~\mu C$ charge: $U = \frac{k~q_1~q_3}{r}$ $U = \frac{(9.0\times 10^9~N~m^2/C^2)(4.0\times 10^{-6}~C)(-1.0\times 10^{-6}~C)}{3.0~m}$ $U = -0.012~J$ We can find the electric potential energy due to the $+3.0~\mu C$ charge and the $-1.0~\mu C$ charge: $U = \frac{k~q_2~q_3}{r}$ $U = \frac{(9.0\times 10^9~N~m^2/C^2)(3.0\times 10^{-6}~C)(-1.0\times 10^{-6}~C)}{4.0~m}$ $U = -0.00675~J$ We can find the total potential energy of the system of three charges: $U = 0.0216~J-0.012~J-0.00675~J$ $U = 0.00285~J$

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