# Three charged particles, finding the net force on each particle

1. Jun 28, 2014

### chococho

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

Three charged particles are placed at the corners of an equilateral triangle of side 1.20 m (see figure). The charges are Q1 = +3.5 µC, Q2 = -7.2 µC, and Q3 = -5.1 µC. Calculate the magnitude and direction of the net force on each due to the other two. (Assume the +x axis points to the right, that is, from Q2 toward Q3.)

2. Relevant equations

F=kq1q2 / r2

3. The attempt at a solution

It's a triangle with Q1 on top, Q2 on the bottom left and Q3 on the bottom right.

For Q3:

I have Q3 going up and left towards Q1, and going right due to Q2.

So I have F31x as a negative number, F31y as positive, and F32x as positive.
But my friend told me it's wrong, because it should be Q1 and Q2 moving towards/away from Q3, and Q3 is stationary. I thought it was the opposite; the other two forces are stationary and only Q3 moves.

So I tried it again with F31x as positive, F31y negative, and F32x as negative. It's still wrong.

2. Jun 29, 2014

### Simon Bridge

Remember opposite charges attract, same charges repel.
Since all the distances are the same, the bigger charge has the bigger effect.
Did you sketch the diagram?

"positive" and "negative" don't make sense as a way to describe the force directions.

3. Jun 29, 2014

### dauto

Last edited: Jun 29, 2014
4. Jun 29, 2014

### haruspex

It's impossible to answer your question about the signs of the forces because you have not defined whether F31 is the force Q3 exerts on Q1 or the force Q1 exerts on Q3,etc.

5. Jun 29, 2014

### chococho

When I said positive and negative, I was referring to the directions of the vector components of the forces.

So my reasoning is that Q3 will move towards Q1 because they are opposite. I said this was F31 (Q3 due to Q1, I think). And I broke F31 down into vector components with x going to the left (so, negative) and y going up (positive).

And F32 will have only x component because Q3 will repel Q2. So F32x will be positive, because it is along the positive x axis.

So, the net force will be:
Fx= -F31x*cos60 + F32x
Fy= F31y*sin60

sqrt(Fx^2 + Fy^2)

Does this look right?

Haruspex, that's what I'm not sure about. I'm not sure which charge acts on which, and which one stays stationary. But I don't know if this makes a difference.

6. Jun 29, 2014

### haruspex

No, you're the one defining the notation. When you write F31, I assume that represents a force. But do you mean it to be the force that Q1 exerts on Q3 or the other way around? They are equal and opposite.
I don't see why any of the charges should remain stationary. They all appear to be subject to an unbalanced combination of forces.

7. Jul 1, 2014

### Simon Bridge

It is common to teach beginning students to treat all charges but one as stationary and see which direction the last one would move to help visualize the force direction.
It can lead to some sloppy commentary though.

In these problems you are basically asked to work out the instantaneous forces on each one due to the others.

If you define $\vec r_n$ to be the position of charge n wrt the origin, and $\vec r_{mn}=\vec r_m-\vec r_n$ is the position of charge m≠n wrt charge n, then we could define a force $$\vec F_{nm} = \frac{kq_nq_m}{r_{nm}^3}\vec r_{nm} = -F_{mn}:n\neq m$$ ... this would be the force on charge m due to charge n.

But some text books define Fnm as the force on n due to m, so it is very important t be clear about the definitions you are using.