Net Force of Point Charges, Coulomb's law

[kgigs6]

Homework Statement

The drawing shows three point charges fixed in place. The charge at the coordinate origin has a value of q1 = +8.02 C; the other two have identical magnitudes, but opposite signs: q2 = -4.73 C and q3 = +4.73 C. (a) Determine the net force exerted on q1 by the other two charges. (b) If q1 had a mass of 1.50 g and it were free to move, what would be its acceleration?

http://edugen.wiley.com/edugen/courses/crs1507/art/qb/qu/c18/ch18p_17.gif

Homework Equations

F= kq1q2/r^2
F=ma

The Attempt at a Solution

F = k(8.02E-6C)(4.73E-6C)/(1.3^2)
F= 0.20179N

The horizontal componenets of the Force vectors cancel out and the vertical components are both pointing straight up. I think I want to find the net vertical force so I did:

sin23 = x/0.2017
x = -0.1707N <--I thought this was the Force on q1 from one of the charges so in order to get the net force I doubled it = -0.3415N

This answer didn't look right and it wasn't but I'm really confused how to get the net force.
For part b I'm pretty sure I understand how to figure it out I just need the answer from part a to solve it.
F = ma --> a=F/m
m=1.5g -->0.0015kg

a= (?F?)/0.0015kg

 

[Kruum]

"sin23 = x/0.2017
x = -0.1707N"

If you are using a calculatro, check the settings, I get something different from this one. Also pay attention to the sign, now it's contradicting with the picture.

 

[Doc Al]

The horizontal componenets of the Force vectors cancel out and the vertical components are both pointing straight up.

Rethink the direction of the vertical components.

I think I want to find the net vertical force so I did:

sin23 = x/0.2017
x = -0.1707N

The angle is 23 degrees, not radians. (You have your calculator set to radian mode.)

 

[kgigs6]

Thanks! I got it right, and because I changed my calculator back to degrees I also was able to figure out why another problem wasn't working - Thanks for your help!