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Charge q located a large distance from a neutral atom

  1. May 10, 2014 #1
    1. The problem statement, all variables and given/known data

    A point charge q is situated a large distance r from a neutral atom of polarisability α. Find the force of attraction between them.

    2. Relevant equations
    [itex]\vec{E}_{mono}(r)=\frac{q}{4\pi\epsilon_0r^2}\hat{r}[/itex]

    [itex]\vec{E}_{dip}(r,\theta)=\frac{p}{4\pi\epsilon_0r^3}(2\cos\theta\hat{r}+ \sin\theta\hat{\theta})[/itex]

    [itex]\vec{p}=\alpha\vec{E}[/itex]

    [itex]\vec{F}=q\vec{E}[/itex]


    3. The attempt at a solution

    [itex]F=\frac{-2\alpha q^{2}}{\left(4\pi\epsilon_{0}\right)^{2}r^{5}}[/itex] attractive force
    My questions are that I wonder about the [itex]\frac{1}{r^{5}}[/itex], is't acceptable? and what is physically meaning of (large distance r from a neutral atom)
     
    Last edited: May 10, 2014
  2. jcsd
  3. May 10, 2014 #2

    haruspex

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    I agree with that answer.
    To get it, you had to make an approximation with regard to distances, right? That is the reason you are told r is large. I.e. it is large compared with the effective distance (whatever that means) of the dipole moment.
     
  4. May 10, 2014 #3
    • Force between two monopoles ~ 1/r2
    • Force between a monopole and a dipole ~ 1/r3
    • Force between a monopole and an induced dipole (the one you calculated) ~ 1/r5
    • Force between two dipoles ~ 1/r4
    • Force between dipole and induced dipole ~ 1/r7
    • Force between two induced dipoles (each one induces the other) ~ I'm leaving that as an exercise. Can you figure it out?
     
  5. May 11, 2014 #4

    rude man

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    I agree with this also. Your E_r expression implicitly assumes r >> distance between dipole charges.

    I'm attaching a good set of notes for this topic.
     

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