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Force of a dipole

  1. Feb 8, 2005 #1
    mk here's the Q

    Calculate the magnitude of the force, due to an electric dipole moment 3.76e-29 C.m, on an electron 2.23e-8 m from the center of the dipole, along the dipole axis. Assume that this distance is large relative to the dipole's charge separation.

    Ok, i have a feeling this is a really easy problem, but i've like gone retarded and i can't do any physics it seems right now. So, first problem, i didn't understand what the center of the dipole was or the dipole axis. as in, halfway between the two charges, or when they're lined up the line that connects them, extended. I tried both to no avail... probably cause i wasn't sure what to do from there either. I tried just Kq/r^2, but i didn't know q, so i tried solving for q, but i know i did that wrong, because my equations weren't right. which i decided didn't matter cause i figured i was doing it wrong and that i should binomial expansion. But i realized i didn't actually understand how i was supposed to use it. So, i'm just stuck, i have no idea how to even start right. So some guidance in the right direction would be great, thanks....

    ~gale~
     
  2. jcsd
  3. Feb 8, 2005 #2

    Doc Al

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    Staff: Mentor

    The dipole axis is the line joining the two charges. You need to find the field of a dipole along that axis. You can either look it up or derive it. To derive it, just use the field from each charge (call them q and -q) separated by distance d (thus the dipole moment is p = qd). So the net field would be:
    [tex]E = kq(\frac{1}{(r-d/2)^2} - \frac{1}{(r+d/2)^2})[/tex]
    I'll leave it to you to simplify this. (It's not hard.)
     
    Last edited: Feb 8, 2005
  4. Feb 8, 2005 #3
    i actually did get that far. and i use binomial expansion, um, reduce and i guess what i get is 2kqd/r^3 kd=p=3.76e-29
    so i get 2kp/r^3 plug in p=3.76e-29 and r=2.23e-8
    and it doesn't work, so i did something wrong. help!
     
  5. Feb 8, 2005 #4

    Doc Al

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    Staff: Mentor

    That's the correct formula for the on-axis field from a dipole. Did you then calculate the force on the electron? What makes you think you did something wrong?
     
  6. Feb 8, 2005 #5
    because i submitted the answer and it was wrong.
    ok i realize this sounds stupid, but i'm getting tired, how am i supposed to be calculating the force on the electron?
     
  7. Feb 8, 2005 #6

    Doc Al

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    Staff: Mentor

    So far, all we've done is calculate the field ([itex]E = 2kp/r^3[/itex]). Now you must use the field to calculate the force: [itex]F = Eq_e[/itex].
     
  8. Feb 8, 2005 #7
    right right, i knew that sorry. i got there, but i don't know q... so then what? i'm really tired now, if i'm asking too much, you can stop answering me, i feel like my brains turned off anyways.
     
  9. Feb 8, 2005 #8
    did you submit it through some electronic system? Some of these systems can be very picky about the terms you use.

    for example

    they may want the equation in terms of [tex] \epsilon_0 [/tex] instead of k, so your equation becomes

    [tex] \frac{qd}{2 \pi \epsilon_0 r^3} [/tex]

    ...just a thought

    .
     
    Last edited: Feb 8, 2005
  10. Feb 8, 2005 #9

    Doc Al

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    Of course you know q.... it's an electron. What's the charge on an electron? (Look it up!) :smile:
     
  11. Feb 8, 2005 #10


    UUUUGH, Jeeze i'm tired, and ya its electronic, and its super picky... ugh
    thank you!
     
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