Answer
$F = 7.3\times 10^{-3}~N$
The angle ccw from the +x axis is $~~97.9^{\circ}$
Work Step by Step
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 $~~97.9^{\circ}$
