 Solution

At a large separation, the two like net charges give the familiar leading repulsion $k{Q}_1{Q}_2/R^2$, where $k=(4\pi {ϵ}_0{)}^{-1}$ and $R$ is centre separation. Each conductor also changes the other's surface charge distribution through electrostatic induction. The facing and remote surface charge densities need not remain equal, and an external insulating suspension prevents net charge transfer without preventing redistribution within either sphere.

Like net charges do not guarantee repulsion for arbitrary conducting spheres at every separation. Sufficiently unequal charges can give an attractive resultant after polarization. This does not contradict Coulomb's law: the law still applies to the distributed charges, while the point-charge approximation has lost the relevant induced contribution.

A useful controlled example takes a small conducting sphere of radius $a$ and fixed positive charge $Q_2$ in the nearly uniform field of a much more strongly charged distant sphere. For $a\ll R$, the small sphere has electric polarizability $\alpha =4\pi {ϵ}_0{a}^3$ and induced dipole moment $p=\alpha E$. Its leading interaction energy isDifferentiation gives the outward radial forceThe second term is attractive. With $\eta =Q_2/Q_1>0$, this leading model gives $F_R<0$ when $\eta <2(a/R{)}^3$. One can choose a sufficiently small positive $\eta$ while retaining $a/R\ll 1$, so the example does not require a neutral or earthed sphere. It is an asymptotic model, not an exact finite-sphere threshold.

An exact analysis confirms that close-range attraction is possible for like-charged conducting spheres and identifies a special contact-charge ratio for which the force remains repulsive.^[ref] This later analysis supplies a precise qualification to the elementary expectation; the source question gives neither a charge ratio nor a separation regime that would justify asserting universal repulsion.