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Homework Help: Moving point charges

  1. Sep 10, 2010 #1

    somasimple

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    Hi all,

    Here is my problem:

    Two positive points charges are situated as the initial figure:
    1/ How to compute the voltage between A and B (or C)?

    2/ If the final condition is like described in the second figure, what is now the voltage between A and B.
    3/ if the transition between the two states takes a time t1, does the speed of this transition changes something about the voltage that exists between A and B?

    Thanks.
     

    Attached Files:

  2. jcsd
  3. Sep 10, 2010 #2

    Redbelly98

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    What is the potential or voltage due to a single point charge? That would be useful here.
     
  4. Sep 10, 2010 #3

    somasimple

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  5. Sep 10, 2010 #4

    somasimple

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    BTW, I received a warning/infraction but I'm not an undergrad student.
    The question is just outside my level of expertise (health).
     
  6. Sep 10, 2010 #5

    Redbelly98

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    Yes, exactly, that is the way to go.
    No, it's really the same computation. Use the distance from the point of interest (either A or B) to each charge.

    Okay, but that is not relevant. Any textbook-style problem, even if it's for your own independent study, is subject to our "homework help" rules. If you haven't already, you can check out our policy by clicking the "Rules" link at the top of this page, and then scroll down to the section titled Homework Help.
     
  7. Sep 11, 2010 #6

    somasimple

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    I made the subject. I made the images. I put the subject in the right forum (Classics Physics) but the subject was moved by a moderator.
     
  8. Sep 11, 2010 #7

    Redbelly98

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    If you'd like to continue with working on the problem, I am willing to help.

    As a start, I can suggest computing the potential at point B in your first figure.
     
  9. Sep 11, 2010 #8

    somasimple

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    My shortcut (?)
    k = coulomb's constant
    d2 = d-r2
    VAB=(k*q1/(r1+d2))+(k*q2/(r2+d2))
    it takes account of the space behind the charges (?)
     
  10. Sep 11, 2010 #9

    Redbelly98

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    Ah, no, it doesn't really work that way. For example, the potential at point B due to charge q1 alone would be

    k q1 / (distance from q1 to B)

    However, I am looking at the figures more closely and some things are confusing me:

    1. You have used r2 to refer to the position of q1. Did you mean to mix the subscripts like that? It is confusing to do so, but we can go with that definition if that is what you intend.

    2. Do "A" and "B" refer to points along the line joining the charges? It would make sense if they do, but then point "C" would be the same as point "B" in your first figure.
     
  11. Sep 11, 2010 #10

    somasimple

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    there is an error in the figure. r1 was meant to belong to q1.
    and r2 to q2.
     
  12. Sep 11, 2010 #11

    somasimple

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    Yes.
     
  13. Sep 11, 2010 #12

    Redbelly98

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    Okay, thanks.

    So, if you just had charge q1, the potential at B would be
    k q1 / r1
    But you also need to add the potential due to charge q2, so you'd have
    VB = (k q1 / r1) + ____​

    Then you'd do the same thing for VA, the potential at point A.
     
  14. Sep 11, 2010 #13

    somasimple

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    VB = (k q1 / r1) + (k q2 / r2)
    VA = (k q1 /(d-r1)) + (k q2 /(d- r2)) ?
     
  15. Sep 11, 2010 #14

    Redbelly98

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    Yes, that's right.

    And the voltage between A & B would be VA-VB ... or VB-VA, depending on whether A or B is being used as the reference point.

    EDIT:
    Your question #2 works the same way.
    #3: As long as we are ignoring relativistic effects, then no, the speed of the transition does not affect the potential.
     
  16. Sep 12, 2010 #15

    somasimple

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    Does it affect the electric current?
     
  17. Sep 12, 2010 #16

    Redbelly98

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    What current?
     
  18. Sep 13, 2010 #17

    somasimple

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    I thought there was a current since there is a voltage but and I'm certainly wrong.
     
  19. Sep 13, 2010 #18

    somasimple

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    I made new pictures:
    Hope they are more accurate.
     

    Attached Files:

  20. Sep 15, 2010 #19

    Redbelly98

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    There would have to be some type of conductor joining A and B for there to be a current.
    Looks good, that more accurately depicts the situation you are describing.
     
  21. Sep 16, 2010 #20

    somasimple

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    That is what I learned some time ago. A current may exist only if there is an electric circuit.

    Thanks.
     
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