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Point of equilibrium between charges |
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| Jun25-12, 11:54 PM | #1 |
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Point of equilibrium between charges
1. The problem statement, all variables and given/known data
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.) 2. Relevant equations Coulomb's Law: Fe=keq1q2/r2 3. 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? |
| Jun26-12, 01:39 AM | #2 |
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Hi pilotguy! Welcome to PF!
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| Jun26-12, 11:05 AM | #3 |
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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? |
| Jun26-12, 11:14 AM | #4 |
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Point of equilibrium between chargesThink 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. |
| Jun26-12, 11:23 AM | #5 |
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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? |
| Jun26-12, 11:25 AM | #6 |
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| Jun26-12, 11:30 AM | #7 |
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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.
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| Jun26-12, 11:38 AM | #8 |
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| Jun26-12, 11:40 AM | #9 |
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| Jun26-12, 11:42 AM | #10 |
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I see. Any guess as to what was my professor trying to get at with his hint about the discriminant in the quadratic equation?
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| Jun26-12, 11:43 AM | #11 |
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| Jun26-12, 11:54 AM | #12 |
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Perfect! Thanks for the help!
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| Jun26-12, 04:25 PM | #13 |
Recognitions:
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ehild |
| Jun26-12, 04:34 PM | #14 |
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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?
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| Jun26-12, 04:51 PM | #15 |
Recognitions:
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ehild |
| Jun27-12, 04:48 AM | #16 |
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| Jun27-12, 05:00 AM | #17 |
Recognitions:
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ehild |
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