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Aproximation question

  1. Jun 16, 2009 #1
    why the "a" disappears
    i cant understand the process of the approximation .

    if r1 goes to be equal r2 =r1=r
    that "a " should be 0 .

    http://i42.tinypic.com/119ndqt.gif
     
  2. jcsd
  3. Jun 16, 2009 #2

    Cyosis

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    Yes if r2=r1 then a=0, exactly. But we're making an approximation. If r is way bigger than a then r2 is almost equal to r1, because r2-r1 is a very small part of r2. Therefore you can approximate r1r2 as r^2.
     
  4. Jun 16, 2009 #3
    but why "a" disappears
    ?

    i agree that if r>>a that r1r2=r^2
    because the are close to each other
    but there a multiplication by 'a'
    i cant understand mathematically
    why "a" disappears
    ?
     
  5. Jun 16, 2009 #4

    Cyosis

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    What is the definition of a dipole moment?
     
  6. Jun 16, 2009 #5
    electric dipole moment is p=qd

    where p is the moment created by the charges separated by a distance d. (on each end of the first body). It's an induced dipole moment.
     
  7. Jun 16, 2009 #6

    Cyosis

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    Yes and what is the electrical dipole moment of this particular configuration?
     
  8. Jun 16, 2009 #7
    "electric field created by two point charges +q and and -q, separated by a (small) distance d.

    If you reduce the distance and increase the charge so that qd remains constant, the field stays approximately the same, except for the region close to the charges.

    A dipole is the limiting case of this, as d goes to zero and q goes to infinity but qd remains finite."

    in my case how qd is a charge of one ball q
    d is 'a'
    and if there multiplication stays the same
    then the field has the value
    still it doesnt explain how 'a' dissapears
     
  9. Jun 16, 2009 #8

    Cyosis

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    You didn't answer my question, you just cited a book. I want you to calculate the electrical dipole moment for the configuration in your problem. Can you do that?
     
  10. Jun 16, 2009 #9
  11. Jun 16, 2009 #10

    Cyosis

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    Almost, but not quite. Can you describe the configuration quantitatively to me using terms such as positive charge = ? ,negative charge =? separation distance =?
     
  12. Jun 16, 2009 #11
    positive charge = +q
    negative charge =-q
    separation distance =a
     
  13. Jun 16, 2009 #12

    Cyosis

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    Nope the separation distance is not a, I suggest you take a good look at the picture. After that calculate the correct electrical dipole moment.
     
  14. Jun 16, 2009 #13
    separation distance =2a
     
  15. Jun 16, 2009 #14

    Cyosis

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    I would appreciate if you would reply to my entire post. Calculate the correct electrical dipole moment now that you have found the correct separation distance.
     
  16. Jun 16, 2009 #15
  17. Jun 16, 2009 #16

    Cyosis

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    Yes can you see now why the a "disappears"?
     
  18. Jun 16, 2009 #17
    lol its being substituted by dipole moment sign
    thanks
     
  19. Jun 16, 2009 #18

    Cyosis

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    Indeed, I am glad you finally see it!
     
  20. Jun 16, 2009 #19
    i cant understand why they change the drawing in that way.
    i know that if we go far far away than our dipole will look like a single charge.
    and there will not be any r1 and r2 only r.
    so i cant see why they change the drawing in such way
    ?
     
  21. Jun 16, 2009 #20

    Cyosis

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    Because the drawing represents a real physical dipole. They then show you how to calculate it when r>>a by using an approximation. Their intend most likely is that you realise the dipole potential is an approximation for a physical dipole. The dipole you're talking about is a mathematical dipole in the limit a->0, they don't exist in reality.
     
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