Far-field sphere-plane dispersion energy ($U(L)=-2A_H{R}^3(1+6R^2/(5L^2)+\cdots )/(9L^3)$)

For pairwise-additive nonretarded molecular attraction and equal solid number densities $n$, let $A_H={\pi }^{2}B{n}^2$ and let $L>R$ be the distance from a sphere centre to a planar half-space. Integrating the atom–half-space potential energy givesFor $L\gg R$, expansion gives the displayed expression. Equivalently expand $(L+z{)}^{-3}$ over the sphere: its odd moments vanish and $⟨z^2⟩=R^2/5$. This yields the first relative correction $6R^2/(5L^2)$. In the opposite near-contact limit, the gap $h=L-R\ll R$ gives $U\simeq -A_{H}R/(6h)$ and the sphere-plane dispersion force magnitude $A_{H}R/(6h^2)$. The two limits use different distances and must not be confused.