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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.

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Both solutions are valid, and the correspond to two very different spheres.

Choosing $\phi(r)=V_0$ corresponds to a sphere with no charge. Choosing $\phi(r)=V_0R/r$ corresponds to a sphere with total charge $4\pi\epsilon_0RV_0$.

Remember, when solving Laplace's equation, we don't just need $\nabla^2\phi=0$. We also need to match boundary conditions. On the surface of a conductor, $\phi$ must satisfy the boundary conditions $\frac{\partial}{\partial r}(\phi_{out}(r)-\phi_{in}(r))=-\frac{\sigma}{\epsilon_0}$. So you're not free to chose between the different solutions! You have to pick the solution that obeys your boundary conditions. $\phi(r)=V_0$ everywhere satisfies Laplace's equation, but not the boundary conditions.

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