#### Answer

The net force on the -10 nC charge is $7.3\times 10^{-3}~N$ and it is directed at an angle of $97.9^{\circ}$ ccw from the +x-axis.

#### Work Step by Step

We can find the magnitude of the force on the -10 nC charge which is directed upward by the +8.0 nC charge.
$F_y = \frac{(k)(8.0~nC)(10~nC)}{r^2}$
$F_y = \frac{(9.0\times 10^9~N~m^2/C^2)(8.0\times 10^{-9}~C)(10\times 10^{-9}~C)}{(0.010~m)^2}$
$F_y = 7.2\times 10^{-3}~N$
We can find the magnitude of the force on the -10 nC charge which is directed to the left by the +10 nC charge.
$F_x = \frac{(k)(10~nC)(10~nC)}{r^2}$
$F_x = \frac{(9.0\times 10^9~N~m^2/C^2)(10\times 10^{-9}~C)(10\times 10^{-9}~C)}{(0.030~m)^2}$
$F_x = 1.0\times 10^{-3}~N$
We can find the magnitude of the net force on the -10 nC charge.
$F = \sqrt{F_x^2+F_y^2}$
$F = \sqrt{(1.0\times 10^{-3}~N)^2+(7.2\times 10^{-3}~N)^2}$
$F = 7.3\times 10^{-3}~N$
We can find the angle $\theta$ ccw from the positive y-axis.
$tan(\theta) = \frac{1.0\times 10^{-3}~N}{7.2\times 10^{-3}~N}$
$\theta = arctan(\frac{1.0\times 10^{-3}~N}{7.2\times 10^{-3}~N})$
$\theta = 7.9^{\circ}$
The net force on the -10 nC charge is $7.3\times 10^{-3}~N$ and it is directed at an angle of $90^{\circ}+7.9^{\circ}$, which is $97.9^{\circ}$, ccw from the +x-axis.