How Does Tension Affect Electric Charge Calculation?

[verd]

Hey,

I thought I did this correctly and my answer turned out incorrect. Here's the problem:

http://synthdriven.com/images/deletable/help.jpg"

This is what I did:

In order to use Coloumb's Law, I need r, the distance between the two charges.

$$\arcsin{4}=\frac{x}{0.24m}$$
$$x=0.24\arcsin{4}=0.098m$$
$$r=2x=0.196m$$

So that's r.

Then I need to pick one side of this thing apart and break it down into x and y-components.

With the weight, I have three forces acting on this thing. The tension in the rope, the weight, and the electrical force repelling each sphere.

My goal is to find the x-component of the electrical force so I can then plug it into Coulomb's law and find the charge.

For the weight,
Wx=0
Wy=-mg=-0.98

For the tension,
Tx=-Tsin4
Ty=Tcos4

For the electrical force,
Fx=Tsin4
Fy=0

To find T, all I did was:

FyNET=Wy+Ty+Fy=Tcos4-mg
(net force of y-component)

T=mg/cos4
Right??

So for T, I get 0.098239

Plugging that into Tx, I get 0.006853And then it goes into Coulomb's law. Which I have reformatted this way:

$$F=\frac{kq^2}{r^2}$$

...Because both charges are supposed to be equal... Moving that around, I get:
$$q=\sqrt{\frac{Fr^2}{k}}$$

Plugging in values, I got 1.711e-7What did I do wrong?

 

[Doc Al]

This is what I did:

In order to use Coloumb's Law, I need r, the distance between the two charges.

$$\arcsin{4}=\frac{x}{0.24m}$$
$$x=0.24\arcsin{4}=0.098m$$
$$r=2x=0.196m$$

So that's r.

That first equation should have sin(4), not arcsin(4).