## Net Force of Point Charges, Coulomb's law

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
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 Blog Entries: 1 "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.

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Blog Entries: 1
 Quote by kgigs6 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