Charge q located a large distance from a neutral atom

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Homework Statement



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

Homework 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]

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)
 
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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.
 
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  • 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?
 
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