1. Coulomb's law does not completely describe the behaviour of charges. Maxwell's equations do.
2. Changes in frame do not affect Maxwell's laws. A change in frame from an electrostatic situation that is described by Coulomb's law results in you getting the magnetic field and everything else in Maxwell's laws anyway (thanks to SR, Ziggurat's posted something on this before I believe). All this talk of motion being relative gains you nothing over what we already know.
3. Maxwell's equations do not explain the atomic nucleus.
4. Maxwell's equations do not explain the atomic nucleus.
(I thought that point 3 was important enough to mention twice)
Gains me nothing over what you already know? Really? You should consider it possible that the things that you believe that you know are, besides being incomplete with regard to the real aspects of the universe, possibly a mixture of true knowledge and pseudo-knowledge... (read false knowledge).
Let's first discuss two deuterons that are at rest with respect to one another. You may believe that they each possess an 'electro-static' field that will prevent them from obtaining an 'at rest' condition with each other. However, there isn't an experiment in the world where it has been demonstrated that elementary charged particles that are at rest with respect to one another actually behave like the charged pith balls in Coulomb's laboratory. Please don't cite accelerator beams requiring magnets to prevent their dispersion because of Coulomb repulsion. That is a nonsensical argument easily disposed of by the facts. Monoenergetic beams, so far, have proven to be impossible to generate and maintain not because of 'electrostatic repulsion' of the beam components as is commonly believed but rather because of the methods used to accelerate beams produce beams with a wide spectrum of energies so that internal beam temperatures are not even close to absolute zero.
If Maxwell's equations suggest that the motion of a charged particle generates or causes to emerge from the particle's apparent location a vector field that propagates outward at the speed of light on a plane that is normal to the translational axis of that motion, does it matter if the motion of the particle is not perceived in its rest frame? In fact, in its rest frame that vector field will not exist. 'Transformed away' is the usual verbiage. Suppose some remote particle, P, is accelerated to a point where it has a constant velocity with respect to a pair of deuterons, D1 and D2, that are overlapping in the same momentum space (defined as having a common de Broglie wavelength that is equal to or greater than their inter-particle distance as calculated from a center of momentum frame). If that remote particle, P, has some component of its motion that is normal to a plane that contains D1 and D2, then in that particle frame, there is generated by the relative motion of D1 and D2 a pair of vector fields that will intersect on a plane that contains D1 and D2 such that at the intersection point they will be exactly anti-parallel (pointing in exact opposite directions) and will produce a null point or negentropic point such that the particles D1 and D2 will move towards that they might occupy that point (based on the quantum axiom that 'all quanta obtain to the lowest energy state available'). Parallel current carrying wires only differ from parallel wires not carrying a current because of the presence of the vector fields that are produced by the motion of the charges. They produce vector fields that at their intersection point are exactly anti-parallel (if the current is in the same directional sense) and the wires are now appearing to be attractively interactive. They show no interaction if only one wire or neither wire is carrying a current. If the currents are in opposite directions the wires seem to repel one another.
For our pair of deuterons D1 and D2 we must consider that there are very many remote particles in the universe that have some component of their motion that is normal to a plane that contains D1 and D2. For each and every remote particle that has some component of its motion that is normal to a plane that contains D1 and D2 there is generated a unique pair of vector fields that intersect and produce a null or negentropic energy point at the point of intersection. When one sums all those negentropic energy points, it creates a very 'deep' negentropic energy point or 'well' to which the particles, D1 and D2, will strongly fall towards. This would also be true of two protons but because there is no neutron that would ordinarily be the source of a time rate gradient structure (a quantum gravity 'field') the two protons could be easily perturbed by the collision between them and some local third particle and they would easily separate. It is the presence of neutrons as quantum gravity source particles that hold the nuclei of atoms together because they maintain the near particles in the same momentum space and the motion of the rest of the particles in the universe do the rest by generating the vector fields that produce the strong negentropic point towards which the particles must move to abide in the lowest energy state available. Particles near a strong gravitational source will nearly always overlap in the same momentum space and a strong gravitational source that contains particles of positive charge will always repel electrons. Why? Because a positive charged particle and a negatively charges particle at rest with respect each other will by virtue of the motion of very many remote particles that have some component of their motion normal to a plane that contains a proton and an electron generate pairs of vector fields that at their intersection point will be parallel, implying not a negentropic point but rather just the opposite, perhaps best described as a high energy hill (as opposed to a 'well') away from which the particles will move to obtain to a lower energy state. Because of the low mass of an electron with respect to a proton and the neutrons in the nucleus the electron receives the lion's share of the energy for the conservation of momentum's sake when a neutron decays in a nucleus into a proton, electron and anti-neutrino.
So, in fact:
1. Maxwell's equations do explain the atomic nucleus.
2. Maxwell's equations do explain the atomic nucleus.
(I thought that point 1 was important enough to mention twice)
CC