Getting direction of force of two point charges on a third

1. Feb 10, 2017

cmkluza

1. The problem statement, all variables and given/known data
What is the direction of the force F on the -10 nC charge in the figure? Give your answer as an angle measured cw from the +x-axis.

2. Relevant equations
$F = k\frac{Qq}{r^2}$

3. The attempt at a solution
I started by getting the components of the net force. The force between the two negative charges will be $F_1$ and the one between the -10 and +15 charges will be $F_2$.
$F_1 = <0, -y> \longrightarrow F_{1x} = k\frac{(5\times10^{-9})(10\times10^{-9})}{0.01^2}$
$F_1 = <0, -0.004495>$
$F_2 = <-x, y> \longrightarrow F_{2t} = k\frac{(15\times10^{-9})(10\times10^{-9})}{(0.03^2 + 0.01^2)}$
$\theta = \tan^{-1}(\frac{1}{3}) \longrightarrow F_2 = <-\cos(\theta)\bullet F_{2t}, \sin(\theta)\bullet F_{2t}>$
$F_t = <-0.0012793, -0.0040685>$
All of this math appears to be correct, because when I calculate the net force ($F_t = \sqrt{x^2 + y^2} \approx 4.3\times10^{-3}$) I get the right answer.
So, now I just need to find the angle, in degrees, clockwise from the positive x-axis, but whatever I'm doing appears to be wrong. Given these components, the vector should be in the third quadrant. So, I tried calculating the inverse tangent of the y over the x components of the total force ($\tan^{-1}(\frac{y}{x}) \approx 17.45$°), and I should be getting the right angle (I've even drawn it out and checked multiple times) if I subtract this answer from 180°, giving 162.55°, which should be 160° to 2 significant figures. But this is incorrect. What am I missing?

Last edited: Feb 10, 2017
2. Feb 10, 2017

TSny

I haven't checked your entire calculation. But, does F1 point in the negative x direction?

3. Feb 10, 2017

cmkluza

Yeah, I mixed up a lot of signs and axes in my OP, but it looks like I did so in my work as well - I had the x- and y-components for the final force switched up, hence why I got the right force magnitude but not direction. Thanks for pointing it out! That let me get the right answer (just had to switch the numbers I was putting in the arctan).