# Equilibrium position of a charge between other 2 charges

1. Oct 7, 2014

### Bedeirnur

1. The problem statement, all variables and given/known data
Place a third charge q between two charges Q1 = -2Q and Q2= 3Q so they stay in electrostatic equilibrium.
Knowing that the distance between Q1 and Q2 is equal L

2. Relevant equations
(KQ1Q)/(QQ1)2 = (KQ2Q)/(QQ2)2

3. The attempt at a solution

Let's start by saying that i know how to solve that problem, or at least how to formally solve it, there is just something i can't understand...

So, let's try to solve...

To have equilibrium |F1|=|F2|

We can say that the distance between Q1 and Q, Q1Q=x and the distance between Q2 and Q, Q2Q=(L-x)

We put the equation

(KQ1Q)/(x)2 = (KQ2Q)/(L-x)2

At the end we get x = sqrt(2)*L/(sqrt(3)-sqrt(2) ----> A positive x

And till here i get it...

But in case we put x = Q2Q and (L-x) = Q1Q , we get at the end a Negative x

I can't actually understand the meaning of a negative/positive distance...

My question is, what is the real meaning of the 2 x's, why is one positive and why is one negative?

And how can i understand if the charge q that we have placed is at the left of the left charge or at the right of the right charge?

We know for example here, that the charge q MUST be in either the right purple circle or the left circle (non between the Q1 and Q2.

2. Oct 7, 2014

### haruspex

I note the question reads "so they stay in equilibrium". They, plural. I agree it is not possible to place a charge between two opposite charges so that it is in equlibrium. However, I also do not see how you could place a single charge between them so that neither of the two original charges experiences a force.

3. Oct 7, 2014

### Bedeirnur

Well, say that the first 2 are blocked at their position, i have to place a third one, so it doesn't move...

4. Oct 7, 2014

### BvU

Your answer in the first case can't be right: for x > L the charge would be pushed to the right.

You don't have a correct relevant equation. (You don't have a relevant equation at all, in fact :( ).

Your first step is also not sufficient: You want F1 + F2 = 0, or rather $|\vec F_1 + \vec F_2| = \vec 0$, not $|\vec F_1| = |\vec F_2|$ !

Last edited: Oct 7, 2014
5. Oct 8, 2014

### haruspex

I already agreed with you that there is no solution with that interpretation. Or the other interpretation.
But the equation $|\vec F_1| = |\vec F_2|$ must hold, so it is a reasonable way of finding possible solutions. Some must then be discarded because the forces add instead of cancelling. The problem here is that there is no point between the charges where the forces cancel. Do you see a solution?

6. Oct 8, 2014

### Bedeirnur

I actually already said that there is No point in between where the charge may stay in equilibrium.

The text itself here is not correct...

I just can't understand the meaning of the negative positions that you get, for example, if you put X as the distance between the higher charge and the one we have to place, we get a negative distance.

7. Oct 8, 2014

### BvU

Look at x as a coordinate, not as a distance. If you have x as the coordinate in your picture wrt the postion of the negative charge, and with the postitive x-direction to the right, then a negative x means the point you found is to the left of the negatve charge.

I think we can agree that there is no F=0 point between Q1 and Q2

Choosing the positive charge as the point where x' = 0 should bring you to the exact same point, so there must be something wrong with the signs in your equations if you then get a positive x'. You should get x'= x - L

I do get the same value for x as you got.

Last edited: Oct 8, 2014
8. Feb 18, 2015

### Arjun J

Dear Friend, the equilibrium position only depends on the magnitude of the two charges not their polarity thus the equation will change to:
x = L sqrt (|Q1|) / (sqrt (|Q2|) - sqrt (|Q1|))

9. Feb 18, 2015

### Staff: Mentor

That's not true.

If the two fixed charges have opposite polarities then there can be no equilibrium position for a charge set between them: their electric fields reinforce each other everywhere on the line between them. If they have the same polarity then the field directions are opposed and an equilibrium position can be found between them.

Clearly then the location of the equilibrium position depends upon the polarities.

10. Feb 18, 2015

### haruspex

Your problem may be that the force one charge exerts on another is not given by kq1q2/x2. That gives the magnitude, but if you want it to give the direction then it becomes kq1q2x/|x|3.

11. Feb 18, 2015

### Arjun J

Then the negative sign could indicate that the distance hence obtained is in the opposite direction.

12. Feb 18, 2015

### Staff: Mentor

How does your formula distinguish between the cases if you eliminate all polarity information by taking absolute values?

13. Feb 18, 2015

### Arjun J

The negative sigh is because you chose Q1 as first charge and x as Q2Q