Charge q located a large distance from a neutral atom

[gfxroad]

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

\vec{E}_{mono}(r)=\frac{q}{4\pi\epsilon_0r^2}\hat{r}

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

\vec{p}=\alpha\vec{E}

\vec{F}=q\vec{E}

The Attempt at a Solution

F=\frac{-2\alpha q^{2}}{\left(4\pi\epsilon_{0}\right)^{2}r^{5}} attractive force
My questions are that I wonder about the \frac{1}{r^{5}}, is't acceptable? and what is physically meaning of (large distance r from a neutral atom)

 

[haruspex]

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.

 

[dauto]

• 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?

 

[rude man]

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.