## College Physics (4th Edition)

The magnitude of the electric force exerted on the $+1.0~\mu C$ point charge due to the other two point charges is $1.2~N$
We can find the magnitude of the electric force directed to the left on the $+1.0~\mu C$ point charge due to the $-0.60~\mu C$ point charge: $F_1 = \frac{k~\vert q_1 \vert~\vert q_2 \vert}{r^2}$ $F_1 = \frac{(9.0\times 10^9~N~m^2/C^2)(1.0\times 10^{-6}~C)(0.60\times 10^{-6}~C)}{(0.060~m)^2}$ $F_1 = 1.50~N$ We can find the magnitude of the electric force exerted on the $+1.0~\mu C$ point charge due to the $+0.80~\mu C$ point charge: $F_2 = \frac{k~\vert q_1 \vert~\vert q_2 \vert}{r^2}$ $F_2 = \frac{(9.0\times 10^9~N~m^2/C^2)(0.80\times 10^{-6}~C)(1.0\times 10^{-6}~C)}{(0.10~m)^2}$ $F_2 = 0.720~N$ Let $\theta$ be the angle above the horizontal of the line between the $+0.80~\mu C$ charge and the $+1.0~\mu C$ charge. We can find the x-component of the vector sum of the electric forces exerted on the $+1.0~\mu C$ point charge due to the other point charges: $F_x = -F_1+F_{2,x}$ $F_x = -F_1+F_2~cos~\theta$ $F_x = (-1.50~N)+(0.720~N)(0.6)$ $F_x = -1.068~N$ We can find the y-component of the net force exerted on the $+1.0~\mu C$ point charge due to the other point charges: $F_y = -F_{2,y}$ $F_y = -F_2~sin~\theta$ $F_y = -(0.720~N)~(0.8)$ $F_y = -0.576~N$ We can can find the magnitude of the net force: $F = \sqrt{F_x^2+F_y^2}$ $F = \sqrt{(-1.068~N)^2+(-0.576~N)^2}$ $F = 1.2~N$ The magnitude of the electric force exerted on the $+1.0~\mu C$ point charge due to the other two point charges is $1.2~N$