Calculating Electric Force on a Charge

[kraigandrews]

Homework Statement

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.

Homework Equations

F= kq1q2/r^2

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.

 

[SammyS]

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

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

 

[kraigandrews]

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

 

[SammyS]

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

Therefore, they don't cancel !

Usually right is positive.

 

[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.

 

[SammyS]

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).

F_1,2 = k(1.78E-6)(-1.78E-6)/(1.82²)

F_2,3 = 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.