# Information content of the electrostatic Maxwell equations vs Coulomb's Law vs Poisson's Equation

In electrostatics, we have Maxwell's equations:

$\nabla \cdot E = \rho$

$\nabla \times E = 0$

These four equations (the second line standing for three equations) can also be written in terms of the electrostatic potential:

$-\nabla^2 V=\rho$

$E = -\nabla V$

Now if we know the positions of every charge in our system, we can find the electrostatic field (completely and entirely, with no additional information required) using Coulomb's law:

$E(x) = -\nabla \int \frac{\rho(x')}{4\pi|x-x'|} \mathrm{d}^3 x'$

My question is: to what extent can we do the same with Maxwell's Equations? For instance, whenever Coulomb's Law is derived from Maxwell's equations, an appeal needs to be made to spherical symmetry. Must we do this? Can we not use the vanishing curl somehow to reach the same conclusion? Similarly, if we derive Coulomb's Law from Poisson's equation, we must specify boundary conditions. We must specify that the potential is some constant, say zero, at infinity.

It appears that the information content is less.

I have read a little bit about the Helmholtz decomposition, and it appears that the curl and divergence of a vector field (E, in this case) do completely determine the vector field, providing certain restrictions are placed on the smoothness and decay of the field at infinity. In other words, it appears that in fact Maxwell's Equations (in the context of electrostatics) do have less information content than Coulomb's Law.

• The Maxwell's equations (together with the Lorentz force law) completely specify all of classical electrodynamics. – Danu Oct 1 '13 at 16:33
• If this is so, then why (for instance) does a derivation of Coulomb's Law from these equations seemingly require the assumption of spherical symmetry? – gj255 Oct 1 '13 at 18:17
• Possible duplicates: physics.stackexchange.com/q/44418/2451 and links therein. – Qmechanic Oct 1 '13 at 19:45

As for Poisson's equation versus Coulomb's Law, no requirement of spherical symmetry is necessary. Start with Poisson's equation, and set the charge density to be that of a point charge, namely \begin{align} \rho(\mathbf x) = q\delta(\mathbf x - \mathbf x_0). \end{align} Next, use the vanishing curl of the electric field (which works in the absence of explicitly time-varying magnetic fields) to write $\mathbf E = -\nabla \Phi$ and plug this into Poisson's equation to obtain \begin{align} \nabla^2\Phi = -\frac{q}{\epsilon_0}\delta(\mathbf x-\mathbf x_0) \end{align} In other words, we want to determine $\Phi$ that is a Green's function for the Laplace equation. The general solution is \begin{align} \Phi = \frac{1}{4\pi\epsilon_0}\frac{q}{|\mathbf x - \mathbf x_0|} + F(\mathbf x) \end{align} where $F$ is a harmonic function, namely one which satisfies Laplace's equation. If we then invoke the boundary condition that the potential vanishes at infinity, then this forces $F$ to be identically zero, and we obtain \begin{align} \Phi = \frac{1}{4\pi\epsilon_0}\frac{q}{|\mathbf x - \mathbf x_0|} \end{align} which, upon taking the gradient, yields he electric field of a point charge which is essentially the content of Coulomb's Law.