Net Force of Point Charges, Coulomb's law

kgigs6

1. The problem statement, all variables and given/known data
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?



2. Relevant equations

F= kq1q2/r^2
F=ma


3. 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
1. The problem statement, all variables and given/known data



2. Relevant equations



3. The attempt at a solution

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

Rethink the direction of the vertical components.

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!