## College Physics (4th Edition)

Published by McGraw-Hill Education

# Chapter 16 - Problems - Page 614: 20

#### Answer

The magnitude of the electric force exerted on the −0.60 μC point charge due to the other two point charges is $1.64~N$

#### Work Step by Step

We can find the magnitude of the electric force directed upward on the $-0.60~\mu C$ point charge due to the $+0.80~\mu C$ point charge: $F_y = \frac{k~\vert q_1 \vert~\vert q_2 \vert}{r^2}$ $F_y = \frac{(9.0\times 10^9~N~m^2/C^2)(0.80\times 10^{-6}~C)(0.60\times 10^{-6}~C)}{(0.080~m)^2}$ $F_y = 0.675~N$ We can find the magnitude of the electric force directed to the right on the $-0.60~\mu C$ point charge due to the $+1.0~\mu C$ point charge: $F_x = \frac{k~\vert q_1 \vert~\vert q_2 \vert}{r^2}$ $F_x = \frac{(9.0\times 10^9~N~m^2/C^2)(1.0\times 10^{-6}~C)(0.60\times 10^{-6}~C)}{(0.060~m)^2}$ $F_x = 1.50~N$ We can find the magnitude of the net force: $F = \sqrt{F_x^2+F_y^2}$ $F = \sqrt{(0.675~N)^2+(1.50~N)^2}$ $F = 1.64~N$ The magnitude of the electric force exerted on the −0.60 μC point charge due to the other two point charges is $1.64~N$

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