Finding Net Coulombic Force

[deezy]

Homework Statement

Three charges with magnitude $$8 × 10^{-4} C$$ are located at (1, 0), (0 , 0), and (-1, 0) meters. The middle charge is negative, and the other two positive. What is the net Coulombic force exerted by them on a negative $$8 × 10^{-5} C$$ charge at (0, 2)?

$$k = 9.0 × 10^9 N∙m^2/C^2$$

Homework Equations

$$F_{12} = \frac{k |q_1| |q_2|}{r^2}$$
$$F_{net on x} = F_{1 on x} + F_{2 on x} + F_{3 on x} + ...$$
$$F_{net} = \sqrt {F_x^2 + F_y^2}$$

The Attempt at a Solution

$$F_1 = \frac {(9 × 10^9)(8 × 10^{-4})(-8 × 10^{-5})}{(\sqrt{5})^2} = -115.2 (N)$$

$$F_2 = \frac {(9 × 10^9)(-8 × 10^{-4})(-8 × 10^{-5})}{(2)^2} = 144 (N)$$

$$F_3 = \frac {(9 × 10^9)(8 × 10^{-4})(-8 × 10^{-5})}{(\sqrt{5})^2} = -115.2 (N)$$

$$F_x = 115.2 * sin \theta + (-115.2 * sin \theta) = 115.2 * \frac{1}{\sqrt 5} - 115.2 * \frac{1}{\sqrt 5} = 0$$

$$F_y = -115.2 * cos \theta + (-115.2 * cos \theta) + 144 = -115.2 * \frac{1}{\sqrt 5} - -115.2 * \frac{1}{\sqrt 5} + 144 = -62.08 (N)$$

$$F_{net} = \sqrt {F_x^2 + F_y^2} = \sqrt {0^2 + (-62.08)^2} = 62.08 (N)$$

I'm pretty sure I need the magnitudes of the forces, but not sure what to do after that. I'm also not sure what should be positive or negative. The answer choices are in the 600s and 800s, and no matter what combination of positives or negatives I use, I can't get anywhere close to 600 or 800.

I attached the diagram I drew below.

 

[gneill]

Your calculations for the individual forces look fine, and your diagram indicates appropriate directions for each of the forces. By symmetry the net force should act along the y-axis, and by the actual forces calculated the net force should be directed downward (so it's negative).

I don't see how the result could have a magnitude even as high as 600N, given the magnitudes of the individual forces. There must be something wrong with either the problem statement or the answer key.