Two conducting spheres

For point charges the answer is that they always repel each other ("like charges repel..."). For our case this is not true. Let us assume that $q$ is a negative charge The point charge will push the electrons on the conducting sphere in such a way that the side of the sphere that faces the point charge will develop a net positive charge, and the other side, that is farther away from the point charge, will have a larger negative charge. The net force is attractive if the point charge is close to the metallic sphere.

Let us look at the two extremes, first looking at the situation when the separation between the point charge and the surface of the sphere, $s$ , is small; $s<<R$ . The net charge over the sphere is distributed over a large area, and most of that is far away from the point charge. That creates a small repulsion that becomes smaller and smaller as s is getting smaller. From the perspective of the point charge the charged sphere will look like an uncharged conducting plane. So now we need to find out the nature of the force between a point charge and a large metallic plate. This is a well know problem, solved by the method of image charges. The answer is that the force is always attractive, because the image charge has the opposite sign relative to the point charge.

In the other case when $s>>R$ the size of the sphere does not matter and we can replace the sphere with a point charge. Clearly, the force will be repulsive.

Therefore we conclude that the force is attractive for small $s$ and repulsive for large $s$ .

This problem can be also solved exactly with the method of image charges, although the solution is a bit more complicated than the treatment of a point charge and conducting plane. The force is

$F=kq^2\left(\frac{1}{R+s}-\frac{R}{s(s+2R)}\right)$

The first term is the repulsive force between the two charges of $q$ , placed a a distance of $s+R$ . The second term is the attractive force between the point charge and the image charge. For small $s$ the second term dominates, because of the $1/s$ factor. For large $s$ the second term is small (it drops off as $1/s^2$ ) and the first term is larger. Also, it is easy to see that the net force is negative (attractive) for $s=R/2$ and repulsive for $s=R$ .

For two metallic spheres the situation remains similar, except for some very special combination of radii and charges, see "Electrostatics of two charged conducting spheres" by John Lekner