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Dipole interaction

  1. Feb 25, 2013 #1
    1. The problem statement, all variables and given/known data

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    2. Relevant equations



    3. The attempt at a solution

    Thus far I believe I am supposed to use calculate the interaction with kQq/Δr? I have tried summing over interactions so that 1/4∏ε*[(-Q)(-q)/(r-d/2+D/2) + (-Q)(q)/(r+d/2+D/2) + (Q)(-q)/(r-d/2-D/2) + (Q)(q)/(r+d/2-D/2)] , but this doesn't work, could someone point me in the right direction.
     

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  3. Feb 25, 2013 #2

    mfb

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    With taylor series for those factors of 1/(r+d/2) and similar (and assuming d/2 << r), you should get the correct result.
     
  4. Feb 25, 2013 #3
    Ok, thanks, is kQq/Δr the correct expression for the interaction energy between 2 charges, and more importantly is that what i'm supposed to be using? Because if so I don't see how I get any 1/(r-d/2) terms, I just get stuff like 1/(r-d/2+D/2) unless I neglect the D/2 because r >> D?
     
  5. Feb 26, 2013 #4

    mfb

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    This is correct.

    And right, you do not get "exactly" 1/(r+d/2), but you get similar expressions.
     
  6. Feb 26, 2013 #5
    Ok thanks for the help, I can't seem to figure out what point i'm supposed to be taylor expanding about? For instance for the interaction between -Q and -q, 1/4∏ε*[(-Q)(-q)/(r-d/2+D/2)] take r = 1/(r-d/2+D/2) and that r >> D & d, d/2 vanish & D/2 vanish? Then I am just left with 1/r, bit confused, even if i try to expand that without making the approximation mentioned I still am not quite sure about what point I am expanding about.
     
    Last edited: Feb 26, 2013
  7. Feb 26, 2013 #6

    mfb

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    The difference between 1/r and your terms is the point of the expansion.

    $$\frac{1}{r+x} = \frac{1}{r} \cdot \frac{1}{1+\frac{x}{r}} \approx \frac{1}{r} (1 \pm \dots)$$
     
  8. Feb 26, 2013 #7
    Yeh I got that bit but I mean if f(x) = 1/r * 1/(1 + x/r) and this expands to 1/r [f(a) + f'(a)(x-a) . . .] what is "a" supposed to be?
     
    Last edited: Feb 26, 2013
  9. Feb 26, 2013 #8

    mfb

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    That is the point where you develop your taylor expansion. It depends on the definition of the f you choose, but I would expect 1 there.
     
  10. Feb 26, 2013 #9
    Sorry I think I'm confused because I'm not entirely sure what the point is i'm developing the taylor expansion from, because if r = (r-d/2+D/2) that's over a length not a point :S? so kQq/(r-d/2+D/2) where i take f(x) = 1/r*(1/(1-x/r)) where x = d/2-D/2, still can't see what the point i'm expanding about is? Is it a=d/2 or a=D/2 or even a=r? so f(x) ≈ 1/r*[1/(1-a/r) + 1/r*(1/(1-a/r)2)*(x-a)] but so confused about the a :(.
     
    Last edited: Feb 26, 2013
  11. Feb 26, 2013 #10

    vela

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    You're expanding about x/r = 0. To be a bit more explicit, the function you're expanding is ##f(z) = \frac{1}{1-z}## about ##z=0## and then plugging in ##z=x/r##.
     
  12. Feb 26, 2013 #11
    Yey, thanks very much to both of you for the help. :!!)
     
  13. Feb 26, 2013 #12
    Actually I just tried that and my expansion ≈ 1/r + (d/2 + D/2)/r^2, then i tried for the remainder terms and the final result came to (2rQq - 2rQq)/r^2 = 0, i imagine my expansion came out wrong, is it wrong?
     
    Last edited: Feb 26, 2013
  14. Feb 26, 2013 #13

    vela

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    Nope, that's fine. If you look at the result you're trying to arrive at, you should notice it has r3 on the bottom, so you want the 1/r and 1/r2 terms to cancel out.
     
  15. Feb 26, 2013 #14
    Ok brilliant, thanks a bunch for the help, I'll take it from here just got to get the manipulation right :D.
     
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