How to Calculate Net Electric Force Correctly?

[BuBbLeS01]

Net Electric Charge...Please help!

Homework Statement

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.)

Homework Equations

F = (K * Q * Q) / r^2

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?

 

[hage567]

Your picture isn't working.

 

[BuBbLeS01]

Is it working now?

 

[hage567]

F3on1 = (K * Q2 * Q3) / 2r^2 = 6.486 x 10^-5 N

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.)

 

[BuBbLeS01]

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

 

[hage567]

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

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

 

[BuBbLeS01]

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

 

[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).

 

[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

 

[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.

 

[BuBbLeS01]

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