[theThinker (OP)]

I have a mile long parallel plate capacitor AB.  It may be taken to be two parallel conducting wire with rectangular cross section and the wires are close to each other forming the capacitor. The capacitor is fully charged to a voltage V. It is then discharged at the end B through shorting it with a resistance R.
Question: What is the time taken for the capacitor to fully discharge?
   
What I have in mind is that when fully charged, the plates have uniform inner surface charge density +d/-d along the wires. While discharging, current flows along the wire. As it is well know that the drift velocity of current is extremely small - order of mm/sec - compared to one mile, I expect it would take ages for the excess electrons from the end A to finally travel to reach end B. So the time to fully discharge such a capacitor should be a rather large figure.     

[syhprum]

1/186000 Secs the energy moves in the form of an electromagnetic wave that moves at c in an air spaced device or more typically .7c in a coaxial cable 

[theThinker (OP)]

1/186000 Secs the energy moves in the form of an electromagnetic wave that moves at c in an air spaced device or more typically .7c in a coaxial cable 

I know you are talking about signal speed in transmission line. But I am actually talking about the movement of the excess electrons. After the capacitor is charged, the battery source is removed. The excess electrons at end A may be individually tagged; they must finally move and cross end B and electron drift speed is extremely slow.

[wolfekeeper]

Your mental model of this is wrong.

Think of a water tap connected to the mains.

When you open the tap, the water starts flowing, instantly.

Why is that? If you think about it, there's a huge column of water going all the way to the nearest water tower, and there's only a few tonnes of pressure behind it as head, but there's a mass- thousands of tonnes of water per square metre per kilometer. It should barely dribble out to start with and then gradually get faster and faster. But it doesn't! Why not?

The reason is- there's a bit of give in the pipe, and as you open the tap, the water gets squeezed out like toothpaste until finally the whole column is flowing.

And the electricity is fairly similar; the electrons repel each other really, really, REALLY strongly, so the electric 'fluid' is largely incompressible. The slight 'give' in a water pipe is exactly like capacitance, it can store a little extra electricity, which is plenty enough to get the electricity flowing; but as with the give in the pipe it doesn't take many electrons to flow to completely discharge the entire capacitor in your example. So even though the electrons flow very slowly, because the electric forces are so very strong, electricity can carry a lot of power.

Feynman gave an example where he imagined if there was suddenly an electrical imbalance of 1% in the electrons in your finger, you would explode at pretty much the speed of light. That's how incredibly strong the electric forces are.

[theThinker (OP)]

Your mental model of this is wrong.

Think of a water tap connected to the mains.

When you open the tap, the water starts flowing, instantly.

Why is that? If you think about it, there's a huge column of water going all the way to the nearest water tower, and there's only a few tonnes of pressure behind it as head, but there's a mass- thousands of tonnes of water per square metre per kilometer. It should barely dribble out to start with and then gradually get faster and faster. But it doesn't! Why not?

The reason is- there's a bit of give in the pipe, and as you open the tap, the water gets squeezed out like toothpaste until finally the whole column is flowing.

And the electricity is fairly similar; the electrons repel each other really, really, REALLY strongly, so the electric 'fluid' is largely incompressible. The slight 'give' in a water pipe is exactly like capacitance, it can store a little extra electricity, which is plenty enough to get the electricity flowing; but as with the give in the pipe it doesn't take many electrons to flow to completely discharge the entire capacitor in your example. So even though the electrons flow very slowly, because the electric forces are so very strong, electricity can carry a lot of power.

Feynman gave an example where he imagined if there was suddenly an electrical imbalance of 1% in the electrons in your finger, you would explode at pretty much the speed of light. That's how incredibly strong the electric forces are.

I still don't get it. I don't find relevance with the analogy of flow of water.
 
 In my example, the battery source has been removed from the capacitor; the upper plate has a mile long uniform +d charge density, the lower -d. We don't need to consider how the charges could maintain their equilibrium distribution despite the "enormous" electrical repulsive forces.
 
When the right end B is short by resistance R, a electron current flow from lower end left A  to B and from B to A right along the top +ve conductor. So how fast will a single identified electron at lower end A flow right to B? It is all about electron flow.

I don't see how this "mental" model is not acceptable. The classical electron is an identifiable particle. We can trace its history from A to B.       

[wolfekeeper]

No electron will EVER flow from A to B. A few electrons will flow from near B and then the capacitance will exhausted.

Going back to the water pipe; imagine there's no reservoir at the far end, the water is in a pipe at significant pressure, for exampe, somebody has closed the valve at the far end, but it's still under pressure. If you open one end the water will spurt out until the 'give' in the pipe has contracted back; squeezed out the water. When that has happened, only a small amount of water near the exit will have left the pipe, and then the flow will stop. It's exactly the same with electricity.

[Bored chemist]

What is the time taken for the capacitor to fully discharge?

Forever.
It's exponential decay.

[theThinker (OP)]

What is the time taken for the capacitor to fully discharge?

Forever.
It's exponential decay.

It is only my guess that you could be right. Is it possible to give some details. Is the decay found in some standard textbook derivation.   

[wolfekeeper]

It can actually have different behaviors depending on R and other parameters. If R is reasonably large then it's just an exponential decay and it will act as a capacitor and take a long while to decay. If R is smaller and the inductance is signficant then you can get wave behaviors because the wire can act as a transmission line. But the wave amplitude will exponentially decay as well.

[Bored chemist]

For what it's worth you can model this big capacitor as a transmission line.
https://en.wikipedia.org/wiki/Transmission_line

And if you discharge it into a resistor equal to the characteristic impedance then you will get a pretty near square wave pulse with a duration equal to the time taken for em radiation to travel a mile in whatever dielectric you used in the capacitor.

[theThinker (OP)]

For what it's worth you can model this big capacitor as a transmission line.
https://en.wikipedia.org/wiki/Transmission_line

And if you discharge it into a resistor equal to the characteristic impedance then you will get a pretty near square wave pulse with a duration equal to the time taken for em radiation to travel a mile in whatever dielectric you used in the capacitor.

Is there a simple way not to talk about em wave propagation along transmission line. I want to trace the actual history of the movement of the lower -ve line electrons. Finally, the plates must be electrically "neutral"; the electrons  must move and go to fill the +ve ions electrons  holes in the upper plates.

[wolfekeeper]

That's the thing, the electrons don't move much at all, they all just shuffle along a bit.

The primary reason is that a mole of electrons going past a point over a whole second is an amp. But in this case the discharge will happen in a tiny fraction of a second and the wire will have many, many moles of electrons in it that will barely move.

So charge on a capacitor is:

Q = cV

I don't know the equation for square wires, but the capacitance between two round parallel wires is here:

https://en.wikipedia.org/wiki/Capacitance#Capacitance_of_conductors_with_simple_shapes

What diameter wire and how far apart and what voltage?

[theThinker (OP)]

I might have found the answer.

From "capacitor discharge":
https://web.northeastern.edu/afeiguin/p1220-Fall2011/slides/chapter26-RC.pdf

The current i or charge Q decays exponentially - in other words "forever".
i(t) = i₀ exp(-t/RC);
where i₀ = large initial current in R. 

It take ages for the electrons at the end A of the lower conductor to cross the end B. There is a true electron movement from A to B - very slowly.

As the decay formula is not dependent on the geometry of the capacitor, the "mile-long" factor is only relative. In our world a mile-long capacitor and a 1cm long are vastly different, but compared to the size of an electron, both distances are still infinitely large. So it seems adding the "mile-long" geometry is only a distraction to our ability to recognize the problem.   

I am not sure how such a solution compares to signal speed in transmission lines.

[Colin2B]

The current i or charge Q decays exponentially - in other words "forever".
i(t) = i₀ exp(-t/RC);
where i₀ = large initial current in R. 

Although theoretically forever, in practice the current drops to a negligible level way, way before that. As you can see, it all depends on R. If you connect the plates with a short cct (almost 0 resistance)  the charge will dissipate way before any electrons have moved from one end of the wire to the other.

It take ages for the electrons at the end A of the lower conductor to cross the end B. There is a true electron movement from A to B - very slowly.

It is important to separate the extremely slow electron movement (electron drift velocity) from movement of charge.

I am not sure how such a solution compares to signal speed in transmission lines.

When you flick a switch the current will start to come out of the other end of the wire at between 50 and 99% of the speed of light depending on the characteristics of the wire. Electrons move excruciatingly slowly, but the wire is full of them, push at one end and those at the other end will move.
The water analogy given earlier is a good one, if a hose is empty then it will take a while to fill and water come out the end; but a wire is like a hose full of water, turn the tap on no delay.
Reread the answers from @wolfekeeper and @Bored chemist , it’s all there

[evan_au]

Oops - overlap with Colin...

The current i or charge Q decays exponentially - in other words "forever".

Engineers talk about the time constant RC, rather than the time for the charge to completely dissipate.
- The Capacitance is C, in Farads
- The Resistance is R, in Ohms
- The initial voltage is V, in Volts (and doesn't affect the answer!)
The time to discharge the capacitance to V/e = 37% of the initial voltage is RC seconds
Where e = 2.718281828..., the base of the natural logarithms
...assuming that inductance is negligible
...and the time for electromagnetic waves to propagate through the circuit is fast compared to RC.

If you plug in resistances around ohms, and capacitances of microfarads, RC is around microseconds.
And the speed of light effects are also measured in microseconds

it is well know that the drift velocity of current is extremely small - order of mm/sec - compared to one mile, I expect it would take ages for the excess electrons from the end A to finally travel to reach end B.

It is true that drift velocity is around mm/second, when the current is near the maximum continuous current the wire can carry.

The mistake here is in assuming that an electron must travel 2km to reach the other end, at a speed of mm/sec.

1 coulomb of charge represents 6x10¹⁸ electrons.

Since Charge Q = CV, a 1uF capacitor charged to 1000V carries 0.001 coulombs, spread out along your 1km capacitor.

If the wire is made of copper, there is 1 electron per atom in the conduction band, ie there is Avogadro's number of conduction electrons (6x10²³) in every 64g of copper. This occupies 7cc, or 7m of wire with 1mm square cross-section.
Each mm length of this wire contains 6x10²³/7000 = 9x10¹⁹ electrons

To discharge 0.001 coulombs of charge, the electrons in the wire at the connected end have to move only:
0.001 x 6x10¹⁸ / 9x10¹⁹ mm = 0.00007mm

However, when you discharge a capacitor in microseconds, the peak current is far higher than the wire's maximum continuous current capacity - it often produces a spark if you have a capacitor charged to 100V or more. But it won't melt the wire, because the duration is so short (microseconds).

So the electrons near the join would need to move micrometers in microseconds - not a difficult feat.
And the electrons at A and B would hardly move at all.

If you make the wire very thin, the electrons have to move farther, but then the wire's resistance is much higher, and must be taken into account in the RC time constant.

[theThinker (OP)]

It is important to separate the extremely slow electron movement (electron drift velocity) from movement of charge.
...
When you flick a switch the current will start to come out of the other end of the wire at between 50 and 99% of the speed of light depending on the characteristics of the wire. Electrons move excruciatingly slowly, but the wire is full of them, push at one end and those at the other end will move.
The water analogy given earlier is a good one, if a hose is empty then it will take a while to fill and water come out the end; but a wire is like a hose full of water, turn the tap on no delay.

I have great issue with the above. I would say my understanding is totally at odds with the above.

drift velocity - it is exactly the average electron speed movement in a current - actual movement of charges. It is said in a neutral conductor, the free electrons have random Fermi speed, but there is zero net current. Only when there is an emf, e.g. a battery that a steady current flows in the wire; the drift velocity here is very small, but proportional to current I (for this instance). So I don't understand why you say "separate the extremely slow electron movement (electron drift velocity) from movement of charge".
 
My capacitor discharge problem cannot be compared using the full water pipe analogy. My capacitor is no more connected to the battery. The discharge of a capacitor is a real flow of charge Q of electrons (Q = n x electron charge) from the lower plate to fill the +ve ion holes in the upper plate.

We can conceptually forget about the free electrons of a neutral conductor. Our lower conductor plate is -ve charged with Q; we assume they are not substituted by any of the valence electrons (outer orbital) of the atoms in the lattice. So we even may identify one such electron at the left end A and tagged it as "Happy". Happy has to move all the way from A to cross R and to fill one of the +ve ion holes. Only in this manner could the upper plates finally be electrically neutral when we say the capacitor has been fully discharged. So we may conceptually say there is a real movement of Q charges from the lower plate to the upper plates - all the Q electrons move at drift-velocity speed crossing R to the upper plates.     

Note that the movement of current here is all about drift velocity (within the transmission wire) but nothing related to signal speed in transmission line theory.

[theThinker (OP)]

...The mistake here is in assuming that an electron must travel 2km to reach the other end, at a speed of mm/sec.

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.

[chiralSPO]

As can be seen in the picture below, the same final state can be produced by the movement of one electron across the entire distance, or by the coordinated motion of multiple electrons across fractions of the entire distance (in the picture shown, three electrons each must travel 1/3 of the distance--in reality, it is zillions of electrons each barely moving):

simple picture.png (85.11 kB . 1068x445 - viewed 6074 times)

[Colin2B]

...The mistake here is in assuming that an electron must travel 2km to reach the other end, at a speed of mm/sec.

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.

If you reject that statement you are making a mistake. No electron makes it’s way along the 1 mile, or even along the short piece of wire connecting A & B. The discharge rate is not due to electron speed. Evan & I are saying the same thing and so is @chiralSPO - his diagram sums it up nicely.

[evan_au]

So we even may identify one such electron at the left end A and tagged it as "Happy".

I am afraid that we have found no distinguishing marks on electrons that allow us to distinguish one electron in the conduction band  of a metal from an adjacent electron in the conduction band.

We can conceptually forget about the free electrons of a neutral conductor.

More usefully, you could forget about that concept.

The neutral conductor is filled with a mobile sea of electrons.
The slightest voltage gradient across this conductor causes this sea of electrons to slosh around.
And there are waves on this sea, representing the propagation of electromagnetic waves whenever there is a change in current or voltage in the conductor.

Our lower conductor plate is -ve charged with Q; we assume they are not substituted by any of the valence electrons (outer orbital) of the atoms in the lattice.

The excess electrons on the negative plate will appear as a small number of extra electrons in the conduction band. In the absence of an electric field, these excess electrons will space themselves out so they are as far apart as possible. But the sea of electrons can still slosh around.

The fact that the negative plate of the capacitor is close to the positive plate means that most of the excess electrons in the negative plate will be spaced out along the side nearest the positive plate.

Similarly, the positive plate will have a small number of electrons missing from the conduction band.  But the sea of electrons can still slosh around.

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

An analogy: The Moon's gravitational field affects water molecules in the ocean in a way similar to the way an electric field affects the sea of electrons in a conductor.

Water molecules are pretty interchangeable (unless you do something tricky like tag them with tritium).
The tide can go out in Europe at the same time the tide is coming in on a USA beach.
This does not mean that the water molecules from the beach in Europe must travel the width of the Atlantic ocean to be deposited on a beach in the USA.

Instead, the water molecules on the beach in Europe move a little farther offshore (only to come in again on the next tide), while water molecules just offshore in the USA move up the beach a little (only to go out again on the next tide).

Similarly, conduction electrons at A do not travel 2 miles to reach B. They just move a tiny bit away from A, while conduction electrons already adjacent to B just move a little closer to B. The whole process occurs at close to the speed of light, and can be modeled as a wave propagating down a transmission line.

rectangular cross section

When you have sudden changes in currents (microseconds), the current mostly flows in a thin skin around the outside of the conductor.

The effective resistance is far higher than you would expect from DC resistance measurements, and the current will take a little longer to discharge.

This is why the earth conductors attached to lightning rods are usually a woven wire mesh, rather than a single solid conductor; the individual strands have a far greater surface area than a solid rectangular conductor, and much lower resistance to rapidly-changing currents.

See: https://en.wikipedia.org/wiki/Skin_effect