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| "Veritasium" (YT) - "The Big Misconception About Electricity" ? |
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| SiliconWizard:
Note that in real life, "DC" is more or less a fantasy. We're always dealing with time-varying fields. Even when the frequency is very low. (And anyway, you'll also have some high-frequency content - just with possibly very low amplitude, but not inexistent.) So in the end, it's always a matter of using an approximation that is "good enough" for a given application. There are always a ton of phenomenons that we are neglecting. :popcorn: |
| SandyCox:
To get back to Veritasium’s problem: Under static (DC) conditions, there are both magnetic and electric fields in the empty space around the conductions. Together these give rise to a non-zero Poynting vector. If there is indeed energy being transferred through this empty space, there is no way that it can be converted back to a useful form. For this we need a time-varying magnetic or electric field (or both). So, I choose to believe that there is no power transfer through empty space at DC. Coming to think of it, the lightbulb requires a flow of charge (current) to heat up its element. Since the power travelling through empty space cannot be converted to a flow of charge, this cannot be the energy that powers the bulb. The energy has to come through the wires. Since all the energy that leaves the battery is disspiated in the lightbuld there cannot be any energy transfer through open space at DC. Maybe Dave should confront Veritasium with this argument. |
| SandyCox:
--- Quote from: SiliconWizard on January 05, 2022, 05:29:22 pm ---Note that in real life, "DC" is more or less a fantasy. We're always dealing with time-varying fields. Even when the frequency is very low. (And anyway, you'll also have some high-frequency content - just with possibly very low amplitude, but not inexistent.) So in the end, it's always a matter of using an approximation that is "good enough" for a given application. There are always a ton of phenomenons that we are neglecting. :popcorn: --- End quote --- Very true, but lets break the bigger problem up into smaller subproblems and analyse one at a time. |
| rfeecs:
--- Quote from: SandyCox on January 03, 2022, 03:40:10 pm ---By looking closely at Example 11.3.1 of Haus and Melcher we can see that Veritasium is wrong. In this example, all the power is being transferred by the conductors. No power is transferred in the region outside the conductors. Haus and Melcher should have said "power seems to flow through the open space" instead of "power is seen to flow through the open space". The power entering the washer from the voltage source is: Pw = 2*pi*sigma*delta*V*V/ln(a/b) By integrating the Poynting vector over the outer surface, we find that the power that is dissipated in the washer is: Pw = 2*pi*sigma*delta*V*V/ln(a/b) The power entering the rod from the voltage source is: Pr = pi*b^2*sigma*V^2/L By integrating the Poynting vector over the outer surface, we find that the power that is dissipated in the rod is: Pr = pi*b^2*sigma*V^2/L So all the power entering the washer from the voltage source is dissipated in the washer and all the power entering the rod from the voltage source is disspiated in the rod. There is no power being transferred in the region between the washer and the rod. Will someone please check my calulations? --- End quote --- I checked. Your calculations are wrong. You are missing the directions of the vectors. For the rod, the S vector is radial, pointing in to the center axis. So at the end of the rod, S is parallel to the surface. S dot dA is zero at the end of the rod. So the power entering the rod from the end contact is zero. All the power is entering the rod from the region between washer and rod. The same is true for the disk. The power doesn't flow through the conductors, it flows in the space around the conductors. The math still works. |
| SandyCox:
--- Quote from: rfeecs on January 05, 2022, 06:39:31 pm --- --- Quote from: SandyCox on January 03, 2022, 03:40:10 pm ---By looking closely at Example 11.3.1 of Haus and Melcher we can see that Veritasium is wrong. In this example, all the power is being transferred by the conductors. No power is transferred in the region outside the conductors. Haus and Melcher should have said "power seems to flow through the open space" instead of "power is seen to flow through the open space". The power entering the washer from the voltage source is: Pw = 2*pi*sigma*delta*V*V/ln(a/b) By integrating the Poynting vector over the outer surface, we find that the power that is dissipated in the washer is: Pw = 2*pi*sigma*delta*V*V/ln(a/b) The power entering the rod from the voltage source is: Pr = pi*b^2*sigma*V^2/L By integrating the Poynting vector over the outer surface, we find that the power that is dissipated in the rod is: Pr = pi*b^2*sigma*V^2/L So all the power entering the washer from the voltage source is dissipated in the washer and all the power entering the rod from the voltage source is disspiated in the rod. There is no power being transferred in the region between the washer and the rod. Will someone please check my calulations? --- End quote --- I checked. Your calculations are wrong. You are missing the directions of the vectors. For the rod, the S vector is radial, pointing in to the center axis. So at the end of the rod, S is parallel to the surface. S dot dA is zero at the end of the rod. So the power entering the rod from the end contact is zero. All the power is entering the rod from the region between washer and rod. The same is true for the disk. The power doesn't flow through the conductors, it flows in the space around the conductors. The math still works. --- End quote --- I did take the direction of the vectors into account. So did Haus and Melcher when they calculated the integral of the Poynting vector over the outer surface of the rod and got the same answer as I did. What you are doing is exactly the misinterpretation of the Poynting vector I am referring to. You have to calculate the integral of the Poynting vector over the total outer surface of the rod. You then have to add up the contributions from the different surfaces before coming to a conclusion. You cannot conclude that Power is entering the rod through a particular surface by looking at the integral of the Poynting vector over that surface. I suggest that you do the following: Calculate the integral of sigma times E dot E o ver the volume of the rod and the washer. This is the amount of electromagnetic power that is converted to heat (dissipated) in each of them. Now look at the problem from the circuit analyses point of view and calculate the current that is entering the rod and the washer. You will see that all the power that is transferred from the voltage source to the rod is dissipated in the rod. Likewise, all the power that is transferred from the voltage source to the washer is dissipated in the washer, i.e. there is no net transfer of power between the washer and the rod. |
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