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
Yes, this is how the hydraulic analogy is inaccurate. Which is why the word 'analogy' is used: it's very similar, not identical.
The energy arrives 1m/c where c is the speed of sound in air.
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.
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
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 ?
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.
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).
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.
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.
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)
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.
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.
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?
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 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].
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 smellThis 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.