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Electric charge

  1. Jun 4, 2005 #1
    I'm stuck on what I think should be a fairly easy question. The question is that there are three charged particles in a line, equidistant (distance L) from each other. From left to right, we'll call them 1, 2, and 3. 1 and 2 are fixed in place, but 3 is free to move. However, the net charge (q) of 1 and 2 on three is keeping 3 in equilibrium. I'm supposed to find the ratio of the charge on 1 to the charge on 2.

    I figured that basically, that means that q1q3/(2L)² = q2q3/L² . So I end up with the ratio being q1/q2 = 4/1 .

    But I know that is the wrong answer. So where am I going wrong? :confused:
    Any tips would be appreciated.
     
  2. jcsd
  3. Jun 4, 2005 #2

    Pyrrhus

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    It looks ok to me, can you post the complete statement of the problem?
     
  4. Jun 4, 2005 #3
    salthon,

    If q1/q2 = 4, what could you say about the signs of the two charges? And would that answer be consistent with q3 being in equlibrium?
     
  5. Jun 5, 2005 #4
    The entire question is this:
    Three charged particles lie on an x axis. Particles 1 and 2 are fixed in place. Particle 3, is free to move, but the net electrostatic force on it from particles 1 and 2 happens to be zero. If the distance between particles 1 and 2 is equal to the distance between 2 and 3, what is the ratio q1/q2?

    Since the signs are never mentioned, I just arbitrarily decided to make 1 positive, 2 negative, and 3 positive. Then I set up the equation so that the force of 1 on 3 equalled the force of 2 on 3. So I've got:

    (k*q1*q3) / (2L)² = (k*q2*q3) / L²

    where L is the distance between 2 adjacent particles.

    After crossing out everthing that's on both sides, I end up with:

    q1/q2 = 4/1

    When I submitted the answer though, that didn't work. I just can't figure out what's wrong with it though. Am I just way off base?

    Thanks.
     
  6. Jun 5, 2005 #5
    Are you sure you want the two forces on q3 to be equal?
     
  7. Jun 5, 2005 #6
    The question says particle 3 is free to move, but the net electrostatic force on it from particles 1 and 2 happens to be zero. Doesn't that mean that the forces have to be equal but with opposite charges?
     
  8. Jun 5, 2005 #7
    That doesn't make any sense. Forces don't have charges. Forces are just forces. If they're equal, then they have the same magnitude and the same direction. So if two equal forces are acting on q3, can it be at equilibrium?
     
  9. Jun 5, 2005 #8

    Pyrrhus

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    Ok so, because we don't know the charges

    [tex] \vec{F}_{1} = \pm k_{e} \frac{|q_{1}||q_{3}|}{(2L)^2} \vec{i} [/tex]

    [tex] \vec{F}_{2} = \pm k_{e} \frac{|q_{2}||q_{3}|}{L^2} \vec{i} [/tex]

    But if the net sum of coulomb forces at particle 3 is 0, one of the particles, either 1 or 2, must have the same charge as particle 3, so you are right when you set them equal. I still don't see a problem with that, the problem states the vectorial addition of coulombic forces equal 0, so they must have same magnitude and different direction Forces 1 and 2 in this problem.
     
    Last edited: Jun 5, 2005
  10. Jun 5, 2005 #9

    OlderDan

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    I think you are just not catching the subtlety of the reply you have been given, and not paying enough attention to your own scenario. You need change only one word in your question to make the answer yes, and you need only look at your original assumption of the charge configuration to know why

    q1/q2 = 4/1

    is not the correct answer. What did you assume to be true about those two charges to calculate the ratio?
     
  11. Jun 6, 2005 #10
    Sorry, I meant to say the forces have equal magnitude and opposite directions.

    What word? I'm really not getting this. I basically did the same thing that Cyclovenom did.

    I assumed that the sign of one charge would be the same as the 3rd one, and that the other would be opposite. Also, that the sum of the forces caused by the first two charges on the 3rd would be zero. Is one of those wrong, or is it something I'm leaving out, or both?
     
  12. Jun 6, 2005 #11

    Pyrrhus

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    Is the book asking for [itex] \frac{|q_{1}|}{|q_{2}|} [/itex] or [itex] \frac{q_{1}}{q_{2}} [/itex]?, remember [itex] \frac{|q_{1}|}{|q_{2}|} \ne \frac{q_{1}}{q_{2}} [/itex] for all values of [itex] q_{1} [/itex] and [itex] q_{2} [/itex] with the obvious restriction [itex] q_{2} \ne 0[/itex]
     
    Last edited: Jun 6, 2005
  13. Jun 6, 2005 #12

    OlderDan

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    You fixed the word I put in bold in your quote. Now look carefully at Cyclovenom's last post.
     
  14. Jun 6, 2005 #13
    Thanks everybody! All I had to do was put a negative on it.
     
  15. Jun 6, 2005 #14

    Pyrrhus

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    salthom, sorry for not just saying, put the negative sign on it! :tongue: , but it was better if you figured it out yourself :biggrin:

    Welcome to PF! :smile:
     
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