Question on boundary condition for Maxwell's Equations and Coulomb's law

When deriving Coulomb's law using the differential forms of Maxwell's equation, the boundary condition that $\phi = 0$ at infinity is also used.

From $\nabla × E = 0, E = \nabla \phi$ for some $V$, plugging this in to Gauss's law we get poisson's equation $\nabla^2 \phi = \rho/e_0$, or $\nabla^2 \phi = q\delta\lvert r-r_0\rvert/e_0$ for a point charge. The general solution is $\phi = \frac{1}{4\pi \epsilon_0} \frac{q}{\lvert r-r_0\rvert} + F$ where $F$ is a harmonic function which satisfies laplace's equation. We have to invoke the condition that at infinity $\phi = 0$ for $F(x)$ to disappear and thus for Coulomb's law to follow.

Maxwell's Equations and the Lorentz force law summarize all of electrodynamics, and that boundary conditions come from the constraints in the problem of consideration. The situation in consideration is a point charge in empty space, so what physical principle motivates this boundary condition, that $\phi = 0$?

In such situations you can in principle you can solve the of thee partial differential equations for $E_x, E_y$ and $E_z$ in three variables $x, y,$ and $z$, with the specified Dirichlet boundary conditions. All that introducing the potential did to you was convert this horrible system of PDEs with Dirichlet BCs to a single PDE with Neuman BCs. So its just a mathematical trick. All physical meaning is still in $E$ alone (classically).
Now once you agreed to introduce $\phi$ to make your life easier, you have to impose the same boundary conditions that you would have otherwise imposed on $E$ itself. For a point particle without introducing $\phi$, to solve the PDEs you had to impose $E\rightarrow 0$ as $r\rightarrow \infty$. This implies $\phi\rightarrow$ constant as $r\rightarrow\infty$. But then we can choose this constant at our disposal because it is physically meaningless (only difference in potential is meaningful).
• Also, would it make sense for a different boundary condition to be chosen such that $F(x)$ is a nonconstant function? – user70720 Mar 12 '15 at 2:55
• Thanks for the re-edit. Why does $E→0$ as $r→∞$? Do there exist situations where $E$ is not zero at $r = ∞$, or am I missing something obvious? – user70720 Mar 12 '15 at 3:18