Electric force on a charge

1. Jan 11, 2012

kraigandrews

1. The problem statement, all variables and given/known data
Q1-----Q2-----Q3; Q1=1.78E-6C;Q2=Q3=-Q1, r1=r2=1.82m
Calculate the total force on Q2. Give your answer with a positive number for a force directed to the right.

2. Relevant equations

F= kq1q2/r^2

3. The attempt at a solution
F2 = F12+F23
F12 = k(1.78E-6)(-1.78E-6)/(1.82^2)
F23 = k(-1.78E-6)(-1.78E-6)/(1.82^2)
The problem is these forces cancel and I know that they shouldn't, so I am sure is some thing I am just not seeing here.

2. Jan 11, 2012

SammyS

Staff Emeritus
F12 is attractive, so it's in what direction?

F32 is repulsive, so it's in what direction?

3. Jan 11, 2012

kraigandrews

F12 is negative so to the right and F32 is positive so it is to the right because Q2 is negative

4. Jan 12, 2012

SammyS

Staff Emeritus
Therefore, they don't cancel !

Usually right is positive.

5. Jan 12, 2012

kraigandrews

okay so then I am not sure where I am going wrong since I know that it shouldn't cancel the charges and the distances are the same for F12 and F32 so it should be 2*F12 = F2, I am not sure what I am doing wrong here this shouuld be a fairly simple problem.

6. Jan 12, 2012

SammyS

Staff Emeritus
The sign given by these two equations really only is useful in showing whether you have attraction (negative sign) or you have repulsion (positive sign).
F1,2 = k(1.78E-6)(-1.78E-6)/(1.822)

F2,3 = k(-1.78E-6)(-1.78E-6)/(1.822)​
You can't blindly add the results together.

Since each force is in the positive x direction, they are both positive and should be added together.