1. Jan 12, 2008

### BuBbLeS01

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

What is the net electric force on charge A? (s=1.41 cm, q1=1.1 nC, q2=0.72 nC, q3=5.21 nC. Force is positive if it points to +x direction.)

2. Relevant equations
F = (K * Q * Q) / r^2

3. The attempt at a solution
F2on1 = (K * Q1 * Q2) / r^2 = 3.59 x 10^-5 N
F3on1 = (K * Q2 * Q3) / 2r^2 = 6.486 x 10^-5 N
F3on1 - F2on1 = 2.9 x 10^-5 N

It's not right...what am I doing wrong?

Last edited: Jan 12, 2008
2. Jan 12, 2008

### hage567

3. Jan 12, 2008

### BuBbLeS01

Is it working now?

4. Jan 12, 2008

### hage567

Maybe this is a typing error, but this should be Q1 and Q3.

What are the directions of the forces on Q1? (Remember you are taking the force as positive if it is in the +ve x direction.)

5. Jan 12, 2008

### BuBbLeS01

Oh yea that was an error....
F2on1 = right, positive
F3on1 = left, negative

6. Jan 12, 2008

### hage567

OK, so what is the net force according to this?

7. Jan 12, 2008

### BuBbLeS01

to the right, positive because the negative charge is closest to A which exerts a positive force on A

8. Jan 12, 2008

### hage567

Yes, the negative will draw A towards +ve x, but the other positive charge C will push it away (-ve x direction). You must sum up the forces (they are vectors).

9. Jan 12, 2008

### BuBbLeS01

Oh ok so you add them because A is on the end. If it was in the middle you have to subtract them right?

F2on1 = (K * Q1 * Q2) / r^2 = 3.59 x 10^-5 N
F3on1 = (K * Q2 * Q3) / 2r^2 = 6.486 x 10^-5 N
F2on1 + F3on1 = 1.01 x 10^-4 N

10. Jan 12, 2008

### hage567

F2on1 is drawing A towards B. So the force is positive according to your convention. F3on1 is pushing A away since like charges repel. That force is in the negative direction. So Fnet = F1on2 - F1on3. It is the difference between the forces. The sign of the answer tells you which direction the of net force.

11. Jan 12, 2008

### BuBbLeS01

So I do...
(3.59e^-5) - (-6.486e^-5) = 1.01e^-4