I also suspect you're correct about the magnetic fields in the scenario that the rod is perpendicular to the direction of motion. Doing the calculation for point charges would be a pain. However, if you consider that parallel currents attract via their magnetic interaction, and that two point charges moving in the same direction are "like" two parallel currents, the attraction from the magnetic field most likely will exactly cancel the added repulsion due to the increased electrical field.
Actually, it's not as bad as I thought, and it demonstrates that the total energy works out to be the same. We've already solved for the electric fields, and found
[latex]\[ E(\theta=0) = \frac{q}{R^2} (1 - v^2/c^2) \][/latex]
[latex]\[ E(\theta=90) = \frac{q}{R^2} \frac{1}{[1 - (v^2/c^2)]^{1/2}} \][/latex]
The magnetic field has a similar (though not quite the same) angular dependence. Again dropping the force constants and the unit vector indicating field direction, we have
[latex]\[ B = \frac{qv}{R^2} \frac{(1 - v^2/c^2)sin\theta}{[1 - (v^2/c^2)sin^2\theta]^{3/2}} \][/latex]
And for the two cases of interest:
[latex]\[ B(\theta=0) = 0[/latex]
[latex]\[ B(\theta=90) = \frac{qv}{R^2} \frac{1}{[1 - (v^2/c^2)]^{1/2}} \][/latex]
You are correct that this is an attractive force for the theta=90 case (for like charges). So what is the force? Well, it's q vxB for magnetic case, and qE for the electric case. So the total force ends up being (ignoring force constants)
[latex]\[ F(\theta=0) = \frac{q^2}{R^2} (1 - v^2/c^2) \][/latex]
[latex]\[ F(\theta=90) = \frac{q^2}{R^2} \frac{1 - v^2/c^2}{[1 - (v^2/c^2)]^{1/2}} = \frac{q^2}{R^2} (1 - v^2/c^2)^{1/2}\][/latex]
where the -v
2 on top in the second equation comes from the magnetic field contribution, and the 1/c
2 comes from converting the force constants (which I left out otherwise). Now these forces aren't equal, and they aren't even equal if we consider the length-contracted R for the theta=0 case. But we're not interested just in the force, we're interested in the potential. And for that, we need to integrate over R. The integral turns 1/R
2 into 1/R, and when you evaluate this at the length contracted position, you add in the missing (1-v
2/c
2)
1/2, which will make the
potential for co-moving charges at both locations exactly equal. As expected from conservation of energy arguments. So there's nothing funky going on with quantum mechanics and relativity: the conservation of energy arguments I made works perfectly, the actual electrodynamics calculations back them up, and the potentials at the Lorentz-contracted positions all end up exactly the same, modified by the same (1-v
2/c
2)
1/2 factor you'd expect from just looking at relativistic equation for total energy of a system.