# Force between pt charge and polarized atom

1. Nov 30, 2008

### musemonkey

1. A point charge q is distance r from a neutral atom of polarizability $$\alpha$$. What's the power of r of the force of attraction between them?

2. Relevant equations

Dipole moment induced by an external field is

$$p = \alpha E_{ext}$$.

The dipole moment (or its magnitude to be precise) is

$$p = dq$$

where d is the distance between the charges +q and -q.

Field from a dipole goes as

$$E_{dip} \propto \frac{p}{r^3}$$.

Field from a point charge goes as

$$E_{ext} \propto \frac{q}{r^2}$$.

3. The attempt at a solution

Force of attraction is

$$q E_{dip} \propto \frac{q p}{r^3} = \frac{q \alpha E_{ext}}{r^3} \propto q^2\alpha~\frac{1}{r^5}$$.

Sanity check: force of attraction between a dipole $$p = dq$$ and a point charge q distance r away (on the axis of the dipole) using Coulomb's law.

$$F = \frac{q^2}{4\pi\epsilon_0} \left ( \frac{1}{(r-d/2)^2} - \frac{1}{(r+d/2)^2} \right ) = \frac{q^2}{4\pi\epsilon_0}\frac{1}{r^2} \left ( \frac{1}{(1-d/2r)^2} - \frac{1}{(1+d/2r)^2} \right )$$

For $$r >> d$$

$$F \approx \frac{q^2}{4\pi\epsilon_0}\frac{1}{r^2} \left ( 1 + 2\left(\frac{d}{2r}\right)^2 - 1 + 2\left(\frac{d}{2r}\right)^2 \right ) = \frac{1}{4\pi\epsilon_0}\frac{(dq)^2}{r^4} = \frac{1}{4\pi\epsilon_0}\frac{p^2}{r^4}$$.

The dipole moment has r dependence

$$p = \alpha E_{ext} \propto \frac{\alpha}{r^2}$$,

which implies

$$F \propto \frac{p^2}{r^4} \propto \frac{\alpha^2}{r^8}$$.

Both calculations cannot be correct. So which one is right, and why is the other one wrong? Thanks!

Last edited: Nov 30, 2008