# 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 $\phi = 0$?

• $\phi = 0$ at infinity is arbitrary, it doesnt matter, all change in potential is the same and all yield the same E field. Apr 15, 2022 at 9:27
• Look at my comments below Apr 15, 2022 at 9:42

Maxwell's equations and Lorentz force law do indeed summarize all electrodynamics. However there are physical situations in which you do not know the charge distribution a priori, but instead you specify some surfaces on which the electric field is always normal, which happens when you have metals around.

In such situations you can in principle solve the three 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 Neumann BCs. So it's 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).

• Thanks for the response. How can coloumb's law be derived from Maxwell's law and Lorentz's law without resorting to ambiguities or assumptions? Is there a link? Most texbooks include a proof using the integral form of Maxwell's equations, but the spherical symmetry of the electric field is assumed, without justification.
– user70720
Mar 12, 2015 at 2:46
• Also, would it make sense for a different boundary condition to be chosen such that $F(x)$ is a nonconstant function?
– user70720
Mar 12, 2015 at 2:55
• My answer was misleading, i'll edit it. Mar 12, 2015 at 2:59
• The justification for spherical symmetry when deriving coulombs law using Gauss's law is inevitable, because if a spherically symmetric charge resulted in an asymmetric electric field this would be a very nontrivial statement about the electric permittivity of the vacuum. Mar 12, 2015 at 3:17
• 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, 2015 at 3:18

Causality requires any physical effect outside the light cone of a charge to vanish. So whether or not you believe that you can add an arbitrary constant, or a gauge field, $$\phi=0$$ at infinity makes sense.