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

• chococho
In summary, the problem involves three charged particles, Q1 = +3.5 µC, Q2 = -7.2 µC, and Q3 = -5.1 µC, placed at the corners of an equilateral triangle of side 1.20 m. The task is to calculate the magnitude and direction of the net force on each particle due to the other two, with the assumption that the +x axis points to the right. The forces are given by the equation F = kq1q2 / r2, where q1 and q2 are the charges of the particles and r is the distance between them. The directions of the forces are determined by the principle that opposite charges attract and like charges rep
chococho

## Homework Statement

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.)

F=kq1q2 / r2

## 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.

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.

Last edited:
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.

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.

chococho said:
Haruspex, that's what I'm not sure about. I'm not sure which charge acts on which, and which one stays stationary.
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.

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

## 1. What is the net force on a charged particle?

The net force on a charged particle is the overall force acting on the particle, taking into account all of the individual forces acting on it.

## 2. How do you calculate the net force on a charged particle?

To calculate the net force on a charged particle, you must add up all of the individual forces acting on the particle. This includes both attractive and repulsive forces from other charged particles.

## 3. What is the relationship between the magnitude and direction of the net force on a charged particle?

The magnitude of the net force on a charged particle is directly proportional to the amount of charge on the particle and the strength of the electric field it is in. The direction of the net force is determined by the direction of the electric field and the direction of the individual forces acting on the particle.

## 4. How does the distance between charged particles affect the net force?

The net force between two charged particles is inversely proportional to the square of the distance between them. This means that as the distance between them increases, the net force decreases.

## 5. Can the net force on a charged particle ever be zero?

Yes, it is possible for the net force on a charged particle to be zero if the individual forces acting on it cancel each other out. This can happen when the particle is in a neutral electric field or when the individual forces are equal in magnitude but opposite in direction.

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