= Solution
Write $C_4=\alpha Q^2/[2(4\pi\epsilon_0\epsilon_r)^2]$. Differentiating the stipulated potential gives
$$
\boxed{F_r=-\frac{du}{dr}=-\frac{4C_4}{r^5}
=-\frac{2\alpha Q^2}{(4\pi\epsilon_0\epsilon_r)^2r^5}}.
$$
For positive $\alpha$, the <force> points toward smaller separation for either sign of $Q$.
Its physical origin is \b[<charge-induced dipole attraction>], rather than <London dispersion forces>. A point-charge field induces $\mathbf p=\alpha\mathbf E$ and the <energy> cost of inducing that moment gives $u=-\alpha E^2/2$. The charge changes the <electric dipole moment>; it does not, within this linear model, change the intrinsic coefficient $\alpha$. For a conducting sphere in a uniform medium, $\alpha=4\pi\epsilon_0\epsilon_r a^3$; other dielectric contrasts give another <electric polarizability>.
There is a distance inconsistency in the stated model: the point-charge field uses center-to-center separation $R$, whereas the question calls $r$ a surface gap. For a point ion and sphere, \b[$R=a+r$]. A controlled far-field dipole model therefore uses $u\simeq-C_4/(a+r)^4$, with $R\gg a$. The boxed derivative is the literal result for the given mathematical $u(r)$; it is not a justification for extending that dipole formula to contact.
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