## Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (4th Edition)

$F = 7.3\times 10^{-3}~N$ The angle ccw from the +x axis is $~~97.9^{\circ}$
We can find the upward force due to the $8.0~nC$ charge: $F_y = \frac{k~\vert q_1 \vert~\vert q_2 \vert}{r^2}$ $F_y = \frac{(9.0\times 10^9~N~m^2/C^2)(8.0\times 10^{-9}~C)(10\times 10^{-9}~C)}{(0.010~m)^2}$ $F_y = 7.2\times 10^{-3}~N$ We can find the leftward force due to the $10~nC$ charge: $F_x = \frac{k~\vert q_1 \vert~\vert q_2 \vert}{r^2}$ $F_x = \frac{(9.0\times 10^9~N~m^2/C^2)(10\times 10^{-9}~C)(10\times 10^{-9}~C)}{(0.030~m)^2}$ $F_x = 1.0\times 10^{-3}~N$ We can find the magnitude of the net force: $F = \sqrt{(1.0\times 10^{-3}~N)^2+(7.2\times 10^{-3}~N)^2} = 7.3\times 10^{-3}~N$ We can find the angle ccw from the +y axis: $tan~\theta = \frac{F_x}{F_y}$ $\theta = tan^{-1}~(\frac{1.0\times 10^{-3}~N}{7,2\times 10^{-3}~N})$ $\theta = 7.9^{\circ}$ The angle ccw from the +x axis is $~~97.9^{\circ}$