Electrostatic equilibrium between 3 styrofoam balls

[paul11]

Homework Statement

Three identical small Styrofoam balls (m = 1.95 g) are suspended from a fixed point by three nonconducting threads, each with a length of 52.1 cm and with negligible mass. At equilibrium the three balls form an equilateral triangle with sides of 28.3 cm. What is the common charge q carried by each ball?

Homework Equations

F = (k q1 q2) / r^2
F = mg
k = 9 * 10^9

The Attempt at a Solution

First I need to find the angle the string makes with the vertical, I start from one of the vertices and imagine a right triangle that connects with the midpoint of the equilateral triangle connecting the 3 balls. Given that I know from the corner of this right triangle will be 30 degrees, and that the adjacent side will be (28.3/2) cm, I calculate the hypotenuse to be 16.339 cm then I use this value to find the angle I want.

θ = sin^-1 ( 16.339 / 52.1 )
= 18.2768 degrees with vertical

Now I can find F tension in the same plane as the electrostatic forces.

Ft = mg tan 18.2768
= 6.3179 * 10 ^(-3)

This is equal in magnitude to the resultant vector of the two opposing electrostatic charges. I take the components of each electrostatic force with 30 degree angles to the resultant.

resultant = kq^2 cos 30 /r^2 + kq^2 cos 30 / r^2

6.3179 * 10^(-3) = kq^2 ( cos 30 + cos 30 ) / r^2
q = 3.3168 * 10^(-7)

CAPA says this is wrong, and I have no idea where I'm going wrong.EDIT: I solved it, it seems I was making a mistake somewhere with the vector part.

 

[kuruman]

I am glad you found your mistake on your own.