Answer
$F = 1.25\times 10^{-4}~N$
Work Step by Step
We can find the distance $r$ between the $10~nC$ charge and the $-1.0~nC$ charge:
$\frac{r}{5.0~cm} = cos~60^{\circ}$
$r = (5.0~cm)~cos~60^{\circ}$
$r = 2.5~cm$
We can find the magnitude of $F$:
$F = \vert \frac{k~q_1~q_2}{r^2} \vert~sin~60^{\circ}$
$F = \frac{(9.0\times 10^9~N~m^2/kg^2)(1.0\times 10^{-9}~C)(10\times 10^{-9}~C)}{(0.025~m)^2}~sin~60^{\circ}$
$F = 1.25\times 10^{-4}~N$
