= Charge-induced dipole attraction
{title2=$U=-\alpha Q^2/[2(4\pi\epsilon_0)^2r^4]$}
A neutral isotropic atom with static <electric polarizability> $\alpha>0$ in the field of a distant point charge $Q$ has induced dipole $\mathbf p=\alpha\mathbf E$ and energy $U=-\alpha E^2/2$. Since $E=|Q|/(4\pi\epsilon_0r^2)$,
$$
U(r)=-\frac{\alpha Q^2}{2(4\pi\epsilon_0)^2r^4},\qquad F_r=-\frac{dU}{dr}=-\frac{2\alpha Q^2}{(4\pi\epsilon_0)^2r^5}.
$$
The negative radial force is attractive for either sign of $Q$. Reversing $Q$ reverses the induced dipole but not the energy or leading force. The factor $1/2$ accounts for the work of inducing the dipole; using the permanent-dipole energy $-\mathbf p\cdot\mathbf E$ without that work would double the answer. The approximation assumes weak fields, a slowly varying field across the atom and negligible electronic overlap.
Back to article page
