## Point charge and neutral atom

Required math: calculus

Required physics: electrostatics

Reference: Griffiths, David J. (2007) Introduction to Electrodynamics, 3rd Edition; Prentice Hall – Problem 4.4.

In an earlier post, we saw that placing a neutral atom in an external field induces a dipole moment ${p}$ where for small fields the experimentally determined relation is

$\displaystyle \mathbf{p}=\alpha\mathbf{E}$
where ${\alpha}$ is the atomic polarizability.

If this external field is due to a point charge ${q}$ at a distance ${r}$ from the atom then the field at the atom due to the charge is

$\displaystyle E=\frac{1}{4\pi\epsilon_{0}}\frac{q}{r^{2}}$

The induced dipole moment is therefore

$\displaystyle p=\frac{\alpha}{4\pi\epsilon_{0}}\frac{q}{r^{2}}$

We’ve seen that the electric field due to a dipole is (in spherical coordinates):

$\displaystyle \mathbf{E}=\frac{p}{4\pi\epsilon_{0}r^{3}}\left[2\cos\theta\hat{\mathbf{r}}+\sin\theta\hat{\theta}\right]$

The field from the dipole induced by the point charge is therefore

$\displaystyle \mathbf{E}=\frac{\alpha q}{\left(4\pi\epsilon_{0}\right)^{2}r^{5}}\left[2\cos\theta\hat{\mathbf{r}}+\sin\theta\hat{\theta}\right]$
where ${\hat{\mathbf{r}}}$ points along the line from the atom to the point charge. The point charge is thus located at ${\theta=0}$ so the force on the charge due to the dipole is

$\displaystyle F=\frac{2\alpha q^{2}}{\left(4\pi\epsilon_{0}\right)^{2}r^{5}}$