[Bored chemist]

a mole of electrons going past a point over a whole second is an amp.

Out by a factor of about 100,000, not that it matters much.
https://en.wikipedia.org/wiki/Faraday_constant

The problem is that the "thinker" is misattributing  the delay (which is infinite in principle- even for a physically small capacitor), to the time it would take electrons to drift through the circuit.

As I said, you can reduce that delay to a fairly well defined, finite, time if you account for the inductance.
 In that case the capacitor is "empty" after 1/186000 seconds whether he likes it or not.

It is also a fundamental observation in physics that you can not "paint an electron purple". The wave function for the electrons in a wire can not meaningfully be ascribed to each of them individually, it's a property of the whole bunch.


 

Is there a simple way not to talk about em wave propagation along transmission line.

Yes, you just don't ask questions like "What happens when you discharge a mile long capacitor?"

I have to totally reject the above statement.

Why do you choose to completely reject reality?

[theThinker (OP)]

I have committed an embarrassing mistake - a big one!   What everyone can see as obvious, I make a mess out of it.

Some clarifications. I don't have a B.Sc physics, dropped out of first year Engineering. Did some further reading all on my own and know some rudiments of EM theory. I did not do the basics thoroughly. I vaguely understand something about the relation between current, drift velocity, carrier charge density, current density, etc. I don't know what happened to my thinking going haywire and off the track.

There is no excess electrons traveling the full 2 miles from the lower A end to the upper A end. It is all about current flow - flow of charge carriers electrons and nothing else. It must be the diagram  of chiralSPO - the three dark circles - that make me remember that current flow is just a movement of a "train of electrons". The analogy of water flow in a pipe is correct.

I just made a mental slip-up concerning current flow. There are two ways to view current flow, as coulomb/sec or as actual motion of the electrons at a drift speed v. In water flow, it would be volume/sec or speed v of water flow. Either way is ok.     

Wiki :"Copper has one free electron per atom, so n is equal to 8.5×1028 electrons per cubic metre". A typical drift speed of electron of 1 Amp current may be  2 x 10¯⁵ m/s. With such drift speed, in 1 sec, the  electrons move a distance of 0.02mm, but the charges passing a section of the wire (due to high N/m³)may be of the order of Q, the initial charge of the capacitor. So a slight distance shift in the "train of charge carrier" electrons would have enough electrons to pass R and to neutralize the upper +Q charge fully discharging the capacitor. None of the electrons do a full 1 or 2 mile trip.

Some other considerations. At any moment where the discharge current is i(t) in R, the current in the wires is higher nearer the end B and dropping gradually where it is near zero nearer the end A. This is because near end A, only a small amount of charge flow is enough to neutralize the excess charge imbalances.

Electric field considerations (may be wrong). Before discharge, only the gap between the plates have a uniform electric field E perpendicular to the plates. At the very moment of shorting the capacitor, the field lines in the air gap disappears altogether (?).  The only field lines would now be within the conductors giving the electrons motion due to : E = ρJ where J = current density, ρ = resistivity. At end A, J is small and E weak; at end B, J is big and so E is strong. I assume a sudden change in the electric fields based on instantaneous-action-at-a-distance for electrical interactions.       

[theThinker (OP)]

To evan-eu:
I actually realized my  mistake right after reading chiralSPO's post and the diagram. I posted my reply above without reading your recent post at about the same time.  Thanks for your humorous description of water molecules crossing tha Atlantic ocean.

[evan_au]

the current in the wires is higher nearer the end B and dropping gradually where it is near zero nearer the end A.

I'm afraid not. The current at ends A and B will be small and equal in magnitude (but opposite in sign).
The current at the other end will be far higher.

The reason is that initially, the charge Q is equally spread out along the mile.
Average Current I is the amount of charge Q passing a given point in 1 second.
Where the wires join (1 mile from A and B), all of charge Q must pass in order to equalise the voltage.
Near A and B, only Q/1760 will be present in the last yard of the conductor (1760 yards =1 mile).
So only Q/1760 will flow out of that last yard when the capacitor is discharged.

[Colin2B]

I have committed an embarrassing mistake - a big one!   What everyone can see as obvious, I make a mess out of it.

I wouldn’t feel too badly about it, this is an error made by many physics teachers, and some school textbooks.
One of the interesting things about physics is realising some yrs later that your teacher got a few things wrong, or rather, not quite correct. Makes for interesting discussions at parent’s evening when you have children.

[chiralSPO]

I have committed an embarrassing mistake - a big one!   What everyone can see as obvious, I make a mess out of it.

Some clarifications. I don't have a B.Sc physics, dropped out of first year Engineering. Did some further reading all on my own and know some rudiments of EM theory. I did not do the basics thoroughly. I vaguely understand something about the relation between current, drift velocity, carrier charge density, current density, etc. I don't know what happened to my thinking going haywire and off the track.

There is no excess electrons traveling the full 2 miles from the lower A end to the upper A end. It is all about current flow - flow of charge carriers electrons and nothing else. It must be the diagram  of chiralSPO - the three dark circles - that make me remember that current flow is just a movement of a "train of electrons". The analogy of water flow in a pipe is correct.

I just made a mental slip-up concerning current flow. There are two ways to view current flow, as coulomb/sec or as actual motion of the electrons at a drift speed v. In water flow, it would be volume/sec or speed v of water flow. Either way is ok. 

Glad we could help

Don't worry about the slip-up, we are all here to learn. It's only when members refuse to change their thinking when presented with counterarguments that other members begin to take offense.

[theThinker (OP)]

the current in the wires is higher nearer the end B and dropping gradually where it is near zero nearer the end A.

I'm afraid not. The current at ends A and B will be small and equal in magnitude (but opposite in sign).
The current at the other end will be far higher.

The reason is that initially, the charge Q is equally spread out along the mile.
Average Current I is the amount of charge Q passing a given point in 1 second.
Where the wires join (1 mile from A and B), all of charge Q must pass in order to equalise the voltage.
Near A and B, only Q/1760 will be present in the last yard of the conductor (1760 yards =1 mile).
So only Q/1760 will flow out of that last yard when the capacitor is discharged.

I think we have a misunderstanding about A and B. In my original post, A is the left end of the capacitor - a mile away from the shorting R. B is the other end of the capacitor at R.

I think our analysis about the current flow along the 1 mile lines is the same. Nearer ends A, the current gets smaller and nearing zero; at the R ends B, the current are strongest. The signs of the upper and lower wire currents are of course reversed. The current within R is uniform.   

[Bored chemist]

I make a mess out of it.

It's not as if the rest of us were born knowing it; we all screw up.
However, it's not generally sensible to say  stuff like "

I have to totally reject the above statement. See my answer to Colin2B.

An electron at the lower plate end A must travel all the way crossing R covering 1 mile at drift-velocity speed.

After you have been told by a whole bunch of people that  you are mistaken.