Your answer would have been correct if, for example, the spheres were non-conducting and if the charges were distributed uniformly over their surfaces.
However, since the spheres are conducting, the surface charge distribution on each sphere will be altered because of the repulsion from the charges on the other sphere. In particular, the charges on each sphere will be pushed away by the charges on the other sphere. This will cause the charges on opposite spheres to be further away from each other, and the force of repulsion to be less than in the case of a uniform surface charge distribution.
Addendum 1. Why would the force be the same if the spheres were non-conducting with uniform surface charge densities? Well, without going into mathematical detail, note that using Gauss's law on each sphere would show that the electric field outside of that sphere would be the field of a point charge with the same total charge. Therefore, each sphere would simply "see" the other sphere as a point charge, and the result follows.
Addendum 2. Let's be a bit more precise in showing that when the spheres are conducting, the force will be lower than in the point charge case. By addendum 1 above, we know that the force would be the same of the spheres had uniform surface charge distribution with total charge equal to that of each point charge. Therefore, it suffices to show that the the surface charge distributions in the conducting case would lead to a smaller force.
To show this, divide each sphere into a large number of charge elements, and suppose that the charge distributions on the surfaces of the conducting spheres begin uniform. How exactly do the charge elements rearrange themselves on each sphere because of the influence of the other sphere? Well, note that if the other sphere were not there, then the charges would stay put and the distributions would remain uniform. Since the other sphere is there, the charges feel a force that pushes them away from the other sphere. In particular, picking any pair of charge elements from opposite spheres, we see that the distance between such a pair will be larger in the new surface charge distribution than in the uniform case. In particular, if we perform a sum over all pairs of surface charge elements on opposite spheres of the coulomb force between them, then every term in the sum will be smaller than in the uniform case, so the total force will also be smaller.
The potential at the surface of a charged sphere or radius $r$ is:
$$ V = k\frac{Q}{r} \tag{1} $$
Since the area of the sphere is $4\pi r^2$, the charge density is:
$$ \rho = \frac{Q}{4\pi r^2} $$
and rearranging gives:
$$ Q = \rho 4\pi r^2 $$
and substituting for $Q$ in equation (1) we get:$$ V = 4\pi k\rho r $$
Can you take it from here?
Best Answer
The basic condition that has to hold inside a conductor is that the electric field vanishes everywhere, i.e. $\vec{E}=0$. If you join the two halves of the sphere, the electrons redistribute themselves in such a way that the configuration reaches equilibrium. The latter state can only be reached with vanishing electric field, otherwise there would be a force between the electrons causing acceleration.
As a consequence, the electrons have to distribute evenly, which results in half of the charge being carried by each half of the sphere.
Vanishing electric field implies constant potential $V$, as seen from the following picture: