Suppose I have a sphere of radius $R$ with potential $V_o$. Since the volume inside the sphere is bounded, then the lack of curvature of the potential (i.e. $\nabla^2\phi = 0$) gives a potential $\phi(r\le R)=V_0$. Outside the sphere we have a choice of making $\phi(r)=V_0$ or $\phi(r)=V_0R/r$. Both are solutions of $\nabla^2\phi=0$. The constant potential does not produce an electric field, but the last one does.
Why do we choose the last form of the potential? (I know that a sphere of potential $V_0$ represents a point charge in the middle of value $Q=4\pi\epsilon_0RV_0$, and that this charge should produce an electric field, but I never mentioned any charge at all...)
I appreciate if you can add something to this discussion.