Veritasium "How Electricity Actually Works"

[electrodacus]

Yes it will.   It's called the displacement current.   

Displacement current is "flowing" anywhere ,  even in an insulator ,  where the electric field is varying with time.

https://en.wikipedia.org/wiki/Displacement_current

This is the part you are referring to

"In physical materials (as opposed to vacuum), there is also a contribution from the slight motion of charges bound in atoms, called dielectric polarization."

We do not consider the limitations of dielectrics here as there is no such thing as a perfect dielectric but I'm fairly certain that Tim or you do not have the equipment to measure such small losses due to dielectric polarization.
And all examples where 3 to 20V with distance between the plates of 1m in air.  Plus that energy is not flowing from one plate to the other of the capacitor. That is lost energy in the form of heat in the dielectric (air in case of a transmission line). This is insignificant compared to IR radiation from the wires and that IR also will not transfer any energy to the load unless you say that an IR photon will help a bit the filament of a lamp but will sure not charge a discharged capacitor.

When you charge a capacitor the energy remains in the capacitor (except for the one ending up as heat due to capacitor plate resistance) so energy did not flow trough the capacitor so trough dielectric on the other side but remained stored so that if you disconnect the capacitor and apply a load (say a resistor or lamp) you can get that energy out.

[bsfeechannel]

Yes, this is how the hydraulic analogy is inaccurate. Which is why the word 'analogy' is used: it's very similar, not identical.

But it is brain-damaging engineers. It's time to get rid of it.

The energy arrives 1m/c where c is the speed of sound in air.

But not in vacuum.

For the same reason: opening the valve emits a sound wave, because you're violently compressing/accelerating water.
Just as in the experiment proposed, it's a tiny amount of energy.

The amount of energy is irrelevant. What you are not paying attention is that the energy that arrives first through space is not a spike of energy. It is a sustained continuous step and it does not disappear. When the rest of the energy flowing through space guided by the wires finally arrives, it is just added to the initial step.

Comparing Derek's and Alpha Phoenix's setup, you'll see that the step "duration" is proportional to the length of the line. If it were 300km, it would "last" 1s. If it were infinite, it would last as much as you'd want. So, it is not a "transient", it is "standient", that doesn't go away.

Another problem is that some people who do not accept that the energy flows through space, are prepared to understand that the Poynting vector, perpendicular to the surface of the load, really conveys the energy that it will dissipate. But they don't understand that the wires are also resistors, and that the Poynting vector, perpendicular to whatever surface of the wire, be it radial or axial, will cause that energy to be dissipated by the wire.

In other words, you cannot hand energy to a wire in one side and expect that this energy will reappear at the other side. The wire will dissipate it entirely.

You can prove that by connecting any piece of wire to the poles of a battery. The energy will not be transferred from the negative pole to the positive pole. It'll simply be dissipated. Completely.

Sounds ridiculous, but people do not connect the dots.

[snarkysparky]

dielectric polarization is not about losses.  It is about the atoms in the dielectric having a polar electrostatic potential and in the presence of an electric field they mostly rotate to line up with
that field thereby strengthening the field.

Energy absolutely does flow through a capacitor. 

https://en.wikipedia.org/wiki/Capacitive_power_supply

[TimFox]

Just to annoy those who dislike complex values, the dielectric "constant" or permittivity of material due to internal polarization with applied electric field is represented by a complex frequency-dependent value to include dielectric loss.
See:  https://en.wikipedia.org/wiki/Permittivity  for a detailed explanation.
The relationship with dielectric loss is discussed in  https://en.wikipedia.org/wiki/Dielectric_loss
As is well known, the dielectric loss of typical insulators at reasonable frequencies is measurable:  good plastics such as polypropylene have dielectric loss around 0.1%, but polyester (Mylar) is roughly 1%.  PVC is much worse.
My anecdote (already posted here several times) about the late Professor U Fano (at the University of Chicago) lecturing on dielectric phenomena was about a quantum-mechanical calculation of the polarization in a medium with harmonically-bound electrons.

[electrodacus]

dielectric polarization is not about losses.  It is about the atoms in the dielectric having a polar electrostatic potential and in the presence of an electric field they mostly rotate to line up with
that field thereby strengthening the field.

Energy absolutely does flow through a capacitor. 

https://en.wikipedia.org/wiki/Capacitive_power_supply

If energy does not flow trough a capacitor then it can only flow trough wire.
So what are we disagreeing on ?

[electrodacus]

Another problem is that some people who do not accept that the energy flows through space, are prepared to understand that the Poynting vector, perpendicular to the surface of the load, really conveys the energy that it will dissipate. But they don't understand that the wires are also resistors, and that the Poynting vector, perpendicular to whatever surface of the wire, be it radial or axial, will cause that energy to be dissipated by the wire.

In other words, you cannot hand energy to a wire in one side and expect that this energy will reappear at the other side. The wire will dissipate it entirely.

You can prove that by connecting any piece of wire to the poles of a battery. The energy will not be transferred from the negative pole to the positive pole. It'll simply be dissipated. Completely.

Sounds ridiculous, but people do not connect the dots.

How will you charge a discharged capacitor from a charged one if you can not transfer energy trough a wire ?
Do you just keep the capacitors close and wait for energy to be transferred or do you connect them with wires ?
And yes part of the energy (not all) will end up as heat due to wire resistance. In the case of two identical capacitors one charged and one discharged exactly half of the energy will end up as heat with the other half of the energy remaining in the two capacitors in equal quantity so a quarter of the initial energy in each capacitor.

[bsfeechannel]

How will you charge a discharged capacitor from a charged one if you can not transfer energy trough a wire ?

If you are charging or discharging, a wire will come in handy for the transfer of the charges.

Do you just keep the capacitors close and wait for energy to be transferred or do you connect them with wires ?

Have in mind that wires, capacitors, inductors, antennas, etc. are all conductors and that nature does not distinguish between them.

If you manage to put the plates of both capacitors in contact without the use of wires, the charges will redistribute and you'll have a loss of energy due to the current through the plates. And part of the rest, that was in the air between the plates of the first capacitor, will be transferred in between the plates of the second capacitor through the air.

Nice try, though.

[aetherist]

Capacitance/induction & induction/induction & radio/induction are all inductions.
Veritasium's  3.3 ns is due to induction (ie through the air), no matter whether u say that it is due to capacitance/induction or some other kind of induction.

However, Veritasium is of course wrong re his silly Poynting Vector having or transferring energy.

U & i agree that the main source of electric energy comes later via the wire (not via the silly Poynting field). But i don’t see how capacitance plays a part in that.

In any case the descriptions that i have seen re what happens with 2 capacitors in a circuit is all wrong.

All energy transfer from source to load/lamp is done trough wires and none of it radiated unless you want to take in consideration that some IR radiated photon from the wires heated the lamp filament.

It seems to me that u have defined electricity as being energy carried by wires. And then u say that electric energy can only be carried by wires.
It seems to me that u are ignoring wireless energy transfer.

[hamster_nz]

The very fact that in a given wire the electrons drift the same speed for the same current should be enough to conclusively prove that electrons are not carrying the energy, and the energy is in the fields.

1A @ 1V (1 J/s)through a 2mm diameter wire? anywhere along the wire 6.24 x 10^18 electrons will drift past, drifting at an average of 23 μm/s.

1A @ 1000V, (1000 J/s), the same 2mm diameter wire? anywhere along the wire 6.24 x 10^18 electrons will drift past, drifting at an average of 23 μm/s.

The motion of electrons is not transferring the energy - it does not change with energy being transferred.

Also, how much does that number of electrons weigh? (you can do the math if you like) - so little weight moving at such a slow speed transfers very close to zero kinetic energy. There is very little 'electron hammer' effect (analogous to water hammer).
 

[aetherist]

The very fact that in a given wire the electrons drift the same speed for the same current should be enough to conclusively prove that electrons are not carrying the energy, and the energy is in the fields.

1A @ 1V (1 J/s)through a 2mm diameter wire? anywhere along the wire 6.24 x 10^18 electrons will drift past, drifting at an average of 23 μm/s.

1A @ 1000V, (1000 J/s), the same 2mm diameter wire? anywhere along the wire 6.24 x 10^18 electrons will drift past, drifting at an average of 23 μm/s.

The motion of electrons is not transferring the energy - it does not change with energy being transferred.

Also, how much does that number of electrons weigh? (you can do the math if you like) - so little weight moving at such a slow speed transfers very close to zero kinetic energy. There is very little 'electron hammer' effect (analogous to water hammer).

Drifting electrons always weigh 1.7 kg per cubic m of Cu.
If drifting exists.
If electrons exist.
And i have my suspicions about wire.

[electrodacus]

It will make no sense for me to answer to you individually (last 4 posts).
You all lack significant understanding about what energy is and what energy storage is.
Today is late and I do not feel like but tomorrow I will take that oscilloscope screenshot from Derek's last video and I will explain exactly what each part means and what it represents. Maybe that will be helpful but I doubt that.

Derek I think mentioned the same thing is his last video, that he can not see how if same number of electrons exit the battery as they enter then how any energy was transferred by them.
The answer is very simple. The electrons could only exit one side and enter the other because there was an imbalance between the two sides.

Analogies are never good but imagine a compressed air cylinder with two chambers divided by an elastic membrane (dielectric equivalent in a capacitor).
When this cylinder has the same amount of gas particle (any gas say Nitrogen) on each side of the membrane it contains no energy but if you use say mechanical energy to move with a pump molecules from one side to the other side then you have a device that stored energy equivalent to a charged capacitor or battery.
Now if you connect a pipe connecting the two sides/chambers of this cylinder you have the equivalent to connecting a wire between the terminals of a charged capacitor. In both cases the stored energy will end up as heat.
You can use that pressure differential to convert that stored energy in to something more useful than heat say mechanical energy by spinning a turbine and still the same number of molecules will exit one side of the cylinder and enter the other side.   
     
 

[hamster_nz]

The very fact that in a given wire the electrons drift the same speed for the same current should be enough to conclusively prove that electrons are not carrying the energy, and the energy is in the fields.

1A @ 1V (1 J/s)through a 2mm diameter wire? anywhere along the wire 6.24 x 10^18 electrons will drift past, drifting at an average of 23 μm/s.

1A @ 1000V, (1000 J/s), the same 2mm diameter wire? anywhere along the wire 6.24 x 10^18 electrons will drift past, drifting at an average of 23 μm/s.

The motion of electrons is not transferring the energy - it does not change with energy being transferred.

Also, how much does that number of electrons weigh? (you can do the math if you like) - so little weight moving at such a slow speed transfers very close to zero kinetic energy. There is very little 'electron hammer' effect (analogous to water hammer).

Drifting electrons always weigh 1.7 kg per cubic m of Cu.
If drifting exists.
If electrons exist.
And i have my suspicions about wire.

So for 1m length of 2mm diameter copper wire (~30g of copper) that's about 5mg of drifting electrons.

At 1A those electrons are drifting at the average speed of about 23 μm/s, regardless of the if that wire is transferring 0.1W or 1000W...

The kinetic energy of those electrons is half of stuff all (or more formally 0.5 * 0.005g * 0.000023 m/s * 0.000023 m/s = 0.00000000000000132 J)

[aetherist]

It will make no sense for me to answer to you individually (last 4 posts).
You all lack significant understanding about what energy is and what energy storage is.
Today is late and I do not feel like but tomorrow I will take that oscilloscope screenshot from Derek's last video and I will explain exactly what each part means and what it represents. Maybe that will be helpful but I doubt that.

Derek I think mentioned the same thing is his last video, that he can not see how if same number of electrons exit the battery as they enter then how any energy was transferred by them.
The answer is very simple. The electrons could only exit one side and enter the other because there was an imbalance between the two sides.

Analogies are never good but imagine a compressed air cylinder with two chambers divided by an elastic membrane (dielectric equivalent in a capacitor).
When this cylinder has the same amount of gas particle (any gas say Nitrogen) on each side of the membrane it contains no energy but if you use say mechanical energy to move with a pump molecules from one side to the other side then you have a device that stored energy equivalent to a charged capacitor or battery.
Now if you connect a pipe connecting the two sides/chambers of this cylinder you have the equivalent to connecting a wire between the terminals of a charged capacitor. In both cases the stored energy will end up as heat.
You can use that pressure differential to convert that stored energy in to something more useful than heat say mechanical energy by spinning a turbine and still the same number of molecules will exit one side of the cylinder and enter the other side.     

Ok, i am keen to see what u have to say about the scope screenshot. In the meantime allow me to soften u up re energy.

(1) Energy is a source of force. The creation of a force needs mass (2). And it (3) needs a medium & a process to transmit the energy or force & (4) it needs a medium & process to tap into that source.
There are only 4 kinds of force, (5) gravitational force, & (6) inertial force, & (7) electric (charge) force, & (8 ) magnetic force.
There is only one medium, it is (9) the aether.
(10) There are at least 8 kinds of aether process, we need 2 ovem here today, (10a) the bulk macro flow of aether (giving 5 & 6), & (10b) the micro excitation of aether (giving 7 & 8 ).
(11) Everything we feel & see is made of photons, the fundamental (quasi) particle. (12) Photons have mass. (13) Photons emit photaenos (14). (15) Photaenos have mass. (16) Photaenos give us the em field (giving us 7 & 8 ).
(17) Energy is stored in photons & (18 ) photaenos. 
(19 ) Energy is stored in the kinetic movement of mass.
(20) Energy is stored in the position of mass (potential energy).
(21) Electric energy is stored in electons, ie photons that hug the surface of the wire. (22) Plus it is stored in the photaenos that are emitted as a part of every photon &  electron (ie the em field). (23) Plus it is stored in free-ish surface electrons on a wire. (24) Plus it is stored in drifting electrons.
(25) Re the other 6 processes not mentioned in (10), 5 of these 6 are (26) the creation of aether, (27) the annihilation of aether, (28) the creation of photons, (29) the annihilation of photons. (30) the annihilation of photaenos. The remaining process is my secret.
(31) Energy duznt exist, what we have is (32) force. (30) Force duznt exist, what we have is mass & position.
(31) Mass (gravitational mass) duznt exist, what we have is the desire to change position.
(32) Mass (inertial mass) duznt exist, what we have is the resistance to any change of position.
(33) Everything is a process (of the aether).

[aetherist]

The very fact that in a given wire the electrons drift the same speed for the same current should be enough to conclusively prove that electrons are not carrying the energy, and the energy is in the fields.

1A @ 1V (1 J/s)through a 2mm diameter wire? anywhere along the wire 6.24 x 10^18 electrons will drift past, drifting at an average of 23 μm/s.

1A @ 1000V, (1000 J/s), the same 2mm diameter wire? anywhere along the wire 6.24 x 10^18 electrons will drift past, drifting at an average of 23 μm/s.

The motion of electrons is not transferring the energy - it does not change with energy being transferred.

Also, how much does that number of electrons weigh? (you can do the math if you like) - so little weight moving at such a slow speed transfers very close to zero kinetic energy. There is very little 'electron hammer' effect (analogous to water hammer).

Drifting electrons always weigh 1.7 kg per cubic m of Cu.
If drifting exists.
If electrons exist.
And i have my suspicions about wire.

So for 1m length of 2mm diameter copper wire (~30g of copper) that's about 5mg of drifting electrons.

At 1A those electrons are drifting at the average speed of about 23 μm/s, regardless of the if that wire is transferring 0.1W or 1000W...

The kinetic energy of those electrons is half of stuff all (or more formally 0.5 * 0.005g * 0.000023 m/s * 0.000023 m/s = 0.00000000000000132 J)

They might not carry much energy, but they can transmit lots of energy (from the source).
If drifting electrons exist.

[electrodacus]

Ok, i am keen to see what u have to say about the scope screenshot. In the meantime allow me to soften u up re energy.

(1) Energy is a source of force. The creation of a force needs mass (2). And it (3) needs a medium & a process to transmit the energy or force & (4) it needs a medium & process to tap into that source.
There are only 4 kinds of force, (5) gravitational force, & (6) inertial force, & (7) electric (charge) force, & (8 ) magnetic force.
There is only one medium, it is (9) the aether.

I need to ask what is your qualification.
I will not bother to read the rest of what you wrote.

There is absolutely zero evidence that energy in this particular case can travel outside the wire.
I guess my only way to explain is some sort of mechanical analogies but you did not mentioned anything about my last analogy with compressed gas in a cylinder with two chambers separated by an elastic membrane.
If there are more molecules in one chamber then if they have a path they will like to move in the other chamber in order to get to the lowest energy state so equal amount of molecules in both chambers.
While as many molecules will enter the low pressure chamber as they will exit form the high pressure one there will be energy delivered by them.
Why will you think there will be any difference for electrons stored in a capacitor ?


Also why when transferring energy between two identical capacitors half of the energy is lost and using a thermal camera you can see where all that energy went and it is in conductors making it quite obvious that energy traveled trough wires and due to resistance half of transferred energy was lost.

[aetherist]

Ok, i am keen to see what u have to say about the scope screenshot. In the meantime allow me to soften u up re energy.

(1) Energy is a source of force. The creation of a force needs mass (2). And it (3) needs a medium & a process to transmit the energy or force & (4) it needs a medium & process to tap into that source.
There are only 4 kinds of force, (5) gravitational force, & (6) inertial force, & (7) electric (charge) force, & (8 ) magnetic force.
There is only one medium, it is (9) the aether.

I need to ask what is your qualification.
I will not bother to read the rest of what you wrote.

There is absolutely zero evidence that energy in this particular case can travel outside the wire.
I guess my only way to explain is some sort of mechanical analogies but you did not mentioned anything about my last analogy with compressed gas in a cylinder with two chambers separated by an elastic membrane.
If there are more molecules in one chamber then if they have a path they will like to move in the other chamber in order to get to the lowest energy state so equal amount of molecules in both chambers.
While as many molecules will enter the low pressure chamber as they will exit form the high pressure one there will be energy delivered by them.
Why will you think there will be any difference for electrons stored in a capacitor ?

Also why when transferring energy between two identical capacitors half of the energy is lost and using a thermal camera you can see where all that energy went and it is in conductors making it quite obvious that energy traveled trough wires and due to resistance half of transferred energy was lost.

I am a retired civil engineer.
That a half of the energy is lost during transfer is interesting, i don’t have a view re that, but it smells fishy.
I have kept out of capacitor to capacitor stuff here, except that i did say that almost everything said/written here has looked wrong to me. Remember i suggested u needed a 2nd switch.
I have agreed with u that the Poynting Field carries zero energy – u must be confusing me with someone else.

The gas cylinder with a membrane is a standard analogy for capacitors – something to do with waving away the need for displacement current or something.

Electrons can't cross from one plate to the other, they would have to jump through the air or whatever.
[G] If we shorted across the gap from one plate to the other then electrons would flow through the short, & would deliver energy (whatever that means). But this is never done.

[C]  If we shorted around the circuit from one plate to the other then electrons would flow through the wire, & would deliver energy (whatever that means). This is always done.

In both cases no energy is transferred via the air.

But the amount of energy transfer will usually not be the same in [G] & [C].

[SandyCox]

Let's return to Veritasium's original topic.

Consider the steady-state (DC) case where all the voltages and currents are constant. Also assume that the wires are perfect conductors. The electronics only carry a small amount of kinetic energy as they travel through the wires. Once they reach the resistor (lightbulb) they are accelerated by the electric field associated with the voltage across the resistor. In the process, electric potential energy of the electrons is converted to kinetic energy. This kinetic energy is, in turn, converted to thermal energy through collisions with atoms in the resistor.

Now let's consider the analogy of a spaceship returning to earth. It has both kinetic energy and gravitational potential energy. Both are converted to thermal energy as it reenters the atmosphere. So what carries the gravitational potential energy? Is it the spaceship or the Earth's gravitational field?

[aetherist]

Let's return to Veritasium's original topic.

Consider the steady-state (DC) case where all the voltages and currents are constant. Also assume that the wires are perfect conductors. The electronics only carry a small amount of kinetic energy as they travel through the wires. Once they reach the resistor (lightbulb) they are accelerated by the electric field associated with the voltage across the resistor. In the process, electric potential energy of the electrons is converted to kinetic energy. This kinetic energy is, in turn, converted to thermal energy through collisions with atoms in the resistor.

Why should electrons drift faster when they reach a resistor?
Why can't electrons go slower when they reach a resistor, but have more collisions or resistance than when going faster in the wire?
If electrons have zero resistance when in a perfect conductor then they must drift at almost c/1.  If so then how could they accelerate when they get to the resistor?

Now let's consider the analogy of a spaceship returning to earth. It has both kinetic energy and gravitational potential energy. Both are converted to thermal energy as it reenters the atmosphere. So what carries the gravitational potential energy? Is it the spaceship or the Earth's gravitational field?

The gravitational potential energy is not carried by anything.
The existence of the spaceship has no effect on any gravitational field.
The spaceship's gravitational potential energy is available, & it has a certain value, but it is due to the spaceship's relative position to Earth [caveat] in the surrounding gravitational field including Earth's gravitational field.
In any case, if Earth's escape velocity is  11.8 km/s, & if the Sun's escape velocity at Earth's radius is  42 km/s, then the spaceship's direction of approach to Earth, & the spaceship's velocity would affect whether the Earth or the Sun were the major factor.

[SandyCox]

I will reply to your post after you received the Noble prize.

[TimFox]

Remember that the drift of electrons (or holes in a semiconductor) can be measured using the Hall effect.
In solid-state physics, the solid structure imposes a mobility factor (in m²V s) on the charge carriers.
The conductivity of the medium is proportional to the product of mobility and carrier density.
In the article cited below, quite a few properties of the material contribute to the mobility:  drifting electrons encounter an obstacle course as they try to follow the E field.
Note that the Hall effect can discern the polarity of the charge carriers:  if both polarities be present, then there is a net effect.
For a summary, see  https://en.wikipedia.org/wiki/Electron_mobility

[IanB]

I posted this comment on Derek's video, slighly expanded:

Riddle me this, a new thought experiment: Let's go extreme and say you have a 100mm diameter copper conductor at pure steady state DC delivering a small amount of power to a pure resitive load, say 1W, but go as low as you want. No transients, no skin effect, no nothing, just pure steady state DC into a resistive load.
Is there NO energy WITHIN this comicly large wire? None? It's all on the OUTSIDE of the wire in the fields at DC? Really? REALLY?

The classicial field theory math might work at DC, but I just can't get over the feeling that it doesn't pass the sniff test at DC. I don't get The Vibe I get with AC and transients. Quantum Electrodynamics and probability theory in the electron fields within the wire better passes the sniff test at DC.

Can someone please convince me that there is no energy flow within this 100mm diameter wire at all, and that all the energy flows outside the wire at DC.

I was thinking about this in terms of a water analogy.

Let's suppose we have a large lake, with a weir on the far side, and the lake is full to the top of the weir. Now let's suppose we turn on a hose, and start adding a small stream of water on the opposite side of the lake. By and by, water will start flowing over the weir at the same rate it is being added from the hose.

Clearly water is flowing across the lake from the hose to the weir. However, if we try to measure any gradient in the surface level of the lake, we will likely fail--it will be quite level from one side to the other to the precision of our instruments. Similarly, if we try to measure any change in pressure throughout the lake we will also fail, it will once again be the same everywhere to the precision of our instruments.

Yet for the water to flow across the lake, there must be a driving force. There must be a pressure/level gradient, no matter how infinitesimal, for the water to flow.

Perhaps the analogy here with the surface electric charge would be the gravitational field. The level of the lake surface must be slightly higher on the hose side than on the weir side, and this change in level in the presence of a gravitational potential field leads to a driving force that moves the water across the lake. The slightly higher level will simultaneously lead to a slightly higher hydrodynamic pressure below the surface, which in this analogy would correspond to a small electric field inside the wire.

[electrodacus]

I am a retired civil engineer.
That a half of the energy is lost during transfer is interesting, i don’t have a view re that, but it smells fishy.
I have kept out of capacitor to capacitor stuff here, except that i did say that almost everything said/written here has looked wrong to me. Remember i suggested u needed a 2nd switch.
I have agreed with u that the Poynting Field carries zero energy – u must be confusing me with someone else.

The gas cylinder with a membrane is a standard analogy for capacitors – something to do with waving away the need for displacement current or something.

Electrons can't cross from one plate to the other, they would have to jump through the air or whatever.
[G] If we shorted across the gap from one plate to the other then electrons would flow through the short, & would deliver energy (whatever that means). But this is never done.

[C]  If we shorted around the circuit from one plate to the other then electrons would flow through the wire, & would deliver energy (whatever that means). This is always done.

In both cases no energy is transferred via the air.

But the amount of energy transfer will usually not be the same in [G] & [C].

There is no strange smell This is a particular case where both capacitors are identical and you charge one form another. You can imagine that a real resistor has a DC series resistance and for transient/AC impedance.
Since this are equal (identical capacitors) you will build a divider when you parallel them so if you measure symmetrical you will always measure 1/2 of the charged capacitor voltage from the time you parallel them to when they are in steady state so no more current flow.

If capacitors are not identical then more or less than half of the energy will be lost as heat the exact half is for the identical capacitor case.

Both G and C will convert the same amount of energy to heat.

[TimFox]

"That a half of the energy is lost during transfer is interesting, i don’t have a view re that, but it smells fishy."

You can do a very simple experiment about energy loss when connecting two capacitors.
Obtain (and measure) two 10 uF polypropylene capacitors.  Make sure that each is discharged before proceeding.
Wire any available switch between the two capacitors.  It's your choice:  DPST or SPST and a wire.
Connect a 10 megohm voltmeter across one of them (time constant = 100 seconds), called C₁.
Then, connect a reasonable and safe (say, 9 V battery) DC supply to that capacitor C₁ and wait until the voltage stabilizes at voltage V₁.
Disconnect the DC supply, note the voltage, close the switch, and note the voltage V₂ before it decays due to the 10 megohm resistance.
The only reason to measure the capacitors before the test is to improve the accuracy.
Before closing the switch, the total energy in the system is (1/2) x C₁ x V₁².
After closing the switch, the total energy is (1/2) x (C₁ + C₂) x V₂².
I don't need to repeat this test, since I already know the answer.

[electrodacus]

I promised an explanation of what happens in the waveform captured by Derek.

Below with red is the voltage across the 1KOhm resistor with green is the supply voltage just after that R4 but it is an ideal voltage source so is basically the 20V supplied.
You can see that from simulation "LTspice" you get the same small power to resistor that is byproduct of charging the transmission line capacitance (I only simulated half the circuit as it is symmetrical).
Then at around 32ms current electron wave will travel half the line so at this time all line capacity is charged tho the part that charged first has already discharged not fully but significantly. Then as the electron travels on the other side of the transmission line the capacitance is being discharged starting with that far end continuing to maintain that limited amount of power for the lamp/resistor until about 65ns when the electron wave that left the source has finally arrived and rump up the lamp at full power (in fact slightly over as the remaining energy in capacitors is now basically fully discharged).
The inductors are also charged but at steady state they still contain significant amount of energy that will be delivered when switch is turned off.


Below is the power graph (area under the graph represents energy).
Here you can see that significantly more power is provided by the source than what it ends up as heat on the lamp/resistor in the first 65ns and that difference is stored energy. The wires are much lower resistance so while there is some thermal energy wasted there it is insignificant compared to the 1K resistor.   

[aetherist]

I am a retired civil engineer.
That a half of the energy is lost during transfer is interesting, i don’t have a view re that, but it smells fishy.
I have kept out of capacitor to capacitor stuff here, except that i did say that almost everything said/written here has looked wrong to me. Remember i suggested u needed a 2nd switch.
I have agreed with u that the Poynting Field carries zero energy – u must be confusing me with someone else.

The gas cylinder with a membrane is a standard analogy for capacitors – something to do with waving away the need for displacement current or something.

Electrons can't cross from one plate to the other, they would have to jump through the air or whatever.
[G] If we shorted across the gap from one plate to the other then electrons would flow through the short, & would deliver energy (whatever that means). But this is never done.

[C]  If we shorted around the circuit from one plate to the other then electrons would flow through the wire, & would deliver energy (whatever that means). This is always done.

In both cases no energy is transferred via the air.

But the amount of energy transfer will usually not be the same in [G] & [C].

There is no strange smell This is a particular case where both capacitors are identical and you charge one form another. You can imagine that a real resistor has a DC series resistance and for transient/AC impedance.
Since this are equal (identical capacitors) you will build a divider when you parallel them so if you measure symmetrical you will always measure 1/2 of the charged capacitor voltage from the time you parallel them to when they are in steady state so no more current flow.

If capacitors are not identical then more or less than half of the energy will be lost as heat the exact half is for the identical capacitor case.

Both G and C will convert the same amount of energy to heat.

Heat loss in G & C will depend on initial potentials. In the simplest case the heat loss would be equal, but in most cases it would not be equal (but this is a side issue)(not important).