Osric, I'm having a bit of a hard time latching solidly enough onto your question and request for confirmation in
your post on the last page, and while I've been thinking about it, so many more posts have been added and it's hard to keep up; so this might be out of order, and certainly incomplete, as new material gets posted faster than I can address it.
Sorry for not being clear! I'll try to distill my question(s) into true/false questions so that they're easier to answer both in terms of time required and in terms of figuring out what I'm trying to ask. My goal in asking these questions is to confirm my understanding or shine a light on my misunderstandings, not to put people on the forum to the trouble of writing tomes in an attempt to educate me (heck, in your case you've already written and published your tome, it's totally unreasonable to expect more!). I do appreciate people's attempts to explain, too, of course, but I know it's very time consuming: please feel free to provide one-word true/false answers (or no answer at all).
In any event the questions from that post in true/false form are:
1. True or false: The electric field represents the force applied to charges in space, but it takes time to set up the electric field surrounding any conductor, and the time required depends on the “inductance”.
2. True or false: This setup time exists whether or not the circuit is open or closed
3. True or false: The electric field charges the wire to the potential that we’ve applied if the circuit is open
4. True or false: The electric field causes current to flow if the circuit is closed
If the answer to any of the above is "false", I need to go back and re-review the content that makes me believe these things to identify whether I have misunderstood it or if the content is wrong.
The electric field comes from the voltage. There doesn't have to be any current to form this field, although if it's across a capacitor, you'll need some current to charge up the capacitor to some voltage, just as it takes some movement to store energy in a spring.
Great, this would appear to correspond with my question 3 and imply that it is true, validating my understanding.
The magnetic field comes from the current, regardless of voltage, although to get the current going in an inductor, you'll need some voltage across it, just as it takes some torque to get a flywheel going. (It also takes torque in the opposite direction to slow it down and to stop it.)
This appears to indicate I have a misunderstanding. I thought that when we connect a voltage source to an open wire, the electric
and magnetic fields would get established over time. Then if we remove the voltage, the magnetic field would collapse. However based on what you're saying now I think that the magnetic field collapses once the fields settle down: that is, we connect voltage to the wire, the electric and magnetic fields get set up down the length of the wire at for the sake of argument 0.6c, bounce around a little and then settle to a 0 current situation where the electric potential is now available at all points on the wire but there is
no longer a magnetic field. When the circuit is closed and current flows, the magnetic field will set up along the length of the wire. It seems to me that this setup must occur "backwards", that is the voltage source on a long wire sets up an electric field that propagates from the voltage source along the length of the wire to its open end, but when we close the circuit the magnetic field will propagate from the open end back to the voltage source as some of the free electrons shuffle toward the end of the wire.
[True or false?]: As a separate matter, it is clear to me that a capacitor acts like a short circuit turning into an ever stronger resistor causing current to flow while its electric field gets set up, while an inductor (any wire or coil) acts as a open circuit turning into an ever less resistant resistor while its electric field gets set up. That'd better be true or I
really don't understand
The water-pipe analogy only goes so far, for various reasons, and I can't think of any way to make it extend to mutual inductance.
Fill the water pipe with bucky balls?

... indeed, I have completely abandoned the water pipe way of thinking about it as misleading and wrong.
Many CMOS inputs and outputs are more or less symmetrical, unlike TTL. WDC's output drivers can pull up all the way to Vcc, and pull up just as hard as down, and many 5V CMOS parts' inputs have their thresholds at 2.5V. Jeff Laughton has an excellent post (actually topic) about thresholds at
viewtopic.php?f=4&t=6594 .
I'll have to review my data sheets and maybe be more careful about not using a mishmash of whatever parts I have in my parts drawers.
A nice thing about a ground plane is that because of mutual inductance, the return path will go through the ground plane, taking on the shape of the trace carrying the signal, because that's easier for it than taking the shortest path, and it eliminates some problems.
This seems to correspond to what I was trying to say/ask of Paganini, that it's the path of least inductance not the past of least resistance that guides the current on the return path. Or maybe another way to say it is that as the fields in the ground plane need time to get set up, the field will set up closest to the signal trace first and so current will flow right underneath the signal trace initially, perhaps spreading into a wider and wider flow over time, but if the current flow is short lived fundamentally this should result in the current flowing the same path as the trace because there is no time for the field to spread out over the ground plane, and this is
not the path of least resistance, which could be directly across some shorter straight line in the ground plane, but a path determined by the mutual inductance ... which is why I questioned Paganini's point.