Point of equilibrium between charges

AI Thread Summary
Equilibrium between a positive charge (q+) and a negative charge (q-) does not exist between them due to the nature of the forces acting on a test charge placed in that region. When a test charge is placed between q+ and q-, it experiences a repulsive force from q+ and an attractive force from q-, resulting in a net force that does not balance out. The discussion suggests that equilibrium can only occur at infinity, where the forces effectively reduce to zero. The hint about the discriminant in the quadratic equation indicates that solving for equilibrium leads to imaginary roots, confirming the absence of a real equilibrium point between the two charges. The ambiguity regarding the magnitudes of the charges is noted, as it affects the potential for equilibrium outside the segment defined by the two charges.
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Homework Statement



Take two charges, one positive with charge q+, and another with charge q- are at a distance d away from each other. Under what conditions is there a point of equilibrium and, if it exists, where would it be located? Show mathematically. (Hint: use the quadratic equation and comment about the discriminant.)

Homework Equations



Coulomb's Law: Fe=keq1q2/r2

The Attempt at a Solution



For equilibrium to exist (with the positive charge a distance x from the origin and the negative charge therefore a distance (d-x) from the origin), F+=F-, so ke*(q1*q2)/x2=ke*(q1*q2)/(d-x)2.

I simplify this to (d-x)2=x2

Simplifying more, d-x=x, and x=d/2, meaning that there's equilibrium halfway between the charges. Though I'm VERY unsure of this. Could someone tell me if I'm headed the right direction?
 
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Hi pilotguy! Welcome to PF! :smile:

F+=F-, so ke*(q1*q2)/x2=ke*(q1*q2)/(d-x)2.

q1 and q2 are not the same for both the equalities. Let the test charge be q' . On one side is the repulsive force, and on the other side is an attractive force.
 
So does that mean that it's Frep=ke*(q'*q+)/x^2 and Fattr=ke*(q'*q-)/(d-x)^2?

Then equate them and solve for x?
 
pilotguy said:
So does that mean that it's Frep=ke*(q'*q+)/x^2 and Fattr=ke*(q'*q-)/(d-x)^2?

Then equate them and solve for x?

Do you think the equilibrium point will lie halfway?
Think about it, if you place a charge between the charges q+ and q-, there will never be a situation where Frep = Fattr. Can you see why is this so?
Make a diagram and place a charge midway and draw the direction of forces.
 
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Oh, I think I see what you're saying. If the + charge is on the left, and the - charge is on the right, with a positive test charge between them, the test charge will be repelled from the positive charge and attracted toward the negative charge, meaning that both forces would act to the right, so there wouldn't be equilibrium.

Am I on the right track?
 
pilotguy said:
Oh, I think I see what you're saying. If the + charge is on the left, and the - charge is on the right, with a positive test charge between them, the test charge will be repelled from the positive charge and attracted toward the negative charge, meaning that both forces would act to the right, so there wouldn't be equilibrium.

Am I on the right track?

Yes! There won't be an equilibrium position in between the charges. Think outside the box(charges)? :wink:
 
So there's no equilibrium between them; does that mean that there's only equilibrium at +/- infinity? That would reduce the force to essentially zero. I'm a little lost.
 
pilotguy said:
So there's no equilibrium between them; does that mean that there's only equilibrium at +/- infinity? That would reduce the force to essentially zero. I'm a little lost.

Seems so. This is basically an electric dipole. Field due to dipole is zero at infinity.
 
pilotguy said:
So there's no equilibrium between them; does that mean that there's only equilibrium at +/- infinity? That would reduce the force to essentially zero. I'm a little lost.

Yep! :approve:
 
  • #10
I see. Any guess as to what was my professor trying to get at with his hint about the discriminant in the quadratic equation?
 
  • #11
pilotguy said:
I see. Any guess as to what was my professor trying to get at with his hint about the discriminant in the quadratic equation?

Yes. You would get an negative sign under the square root, while solving the quadratic(imaginary roots)
 
  • #12
Perfect! Thanks for the help!
 
  • #13
pilotguy said:

Homework Statement



Take two charges, one positive with charge q+, and another with charge q- are at a distance d away from each other. Under what conditions is there a point of equilibrium and, if it exists, where would it be located? Show mathematically. (Hint: use the quadratic equation and comment about the discriminant.)

Are those charges q+ and q- equal in magnitude?

ehild
 
  • #14
I'm not sure; the question is a little ambiguous on that. Does it make a difference to the solution from earlier in the thread?
 
  • #15
pilotguy said:
I'm not sure; the question is a little ambiguous on that. Does it make a difference to the solution from earlier in the thread?

Yes, it makes a difference. There can be a neutral point outside the segment d. Try q+=12 mC and q-=-3 mC, and d=1 m, for example.

ehild
 
  • #16
ehild said:
Are those charges q+ and q- equal in magnitude?

ehild

I think it is to be interpreted as a '+q' charge and a '-q' charge, making them equal in magnitude.
 
  • #17
Infinitum said:
I think it is to be interpreted as a '+q' charge and a '-q' charge, making them equal in magnitude.

Then it would have been written as +q and -q, instead of writing "one positive with charge q+, and another with charge q-" and the problem were not be a real problem with discriminant and everything. Try to solve it with charges of different magnitudes. It is a good practice. :smile:

ehild
 
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