# 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

$\displaystyle \mathbf{p}=\alpha\mathbf{E} \ \ \ \ \ (1)$

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

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

The induced dipole moment is therefore

$\displaystyle p=\frac{\alpha}{4\pi\epsilon_{0}}\frac{q}{r^{2}} \ \ \ \ \ (3)$

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

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

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

$\displaystyle \mathbf{E}=\frac{\alpha q}{\left(4\pi\epsilon_{0}\right)^{2}r^{5}}\left[2\cos\theta\hat{\mathbf{r}}+\sin\theta\hat{\theta}\right] \ \ \ \ \ (5)$

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

$\displaystyle F=\frac{2\alpha q^{2}}{\left(4\pi\epsilon_{0}\right)^{2}r^{5}} \ \ \ \ \ (6)$

## One thought on “Point charge and neutral atom”

1. Jaime

As far as I am aware, the solutions mauanl is only available to instructors. I am, however, aware that David Pace (administrator of DavidPace.com) used to have solutions for a large number of problems on his website. Unfortunatly, he has recently taken these down without significant comment on the reason for doing so, so they are somewhat unavailable. You can,
however, access a cached version of them from google if you really wish to attempt
to locate them.I encourage you, however, to struggle your way through each and every problem in electrodynamics before ever checking to see if you arrived at the correct result. It really is the only way to learn.Alternatively, if you’re having significant difficulty, you might try posting to Yahoo! answers in the physics section and I’m sure that someone would be interested in helping you get over your difficulty.I wish you good luck in your study of electrodynamics using one of the better books out there. Good luck with your exam.References :