Capacitor+Grounded Plate+AC Current?

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The title describes my guesses as to what is happening in this problem, but I don't really understand it.

Moderators, please do not delete this... I would give a non-dangerous example if I understood the problem well enough to actually know what the physics is.

In class today, my physics professor, for a demo (please do not try this at home: it is dangerous!) plugged a pickle into an electrical outlet. It is a demo he does every year in his E&M class. The pickle starts on fire. Again, I did not do this, and will not do this, and you should not either. But it is what the professionally done demo was.

What I do not understand is the following: he said that the end connected to the "hot" terminal of the outlet is the side on which the fire always starts. I do not understand why that should be, given that this is an alternating current source. Both sides are effectively exactly the same thing. My thought, then, is that this must have something to do with the fact that neutral is held at approximately ground potential, but I can't quite figure out how this works.

Thanks!
 

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  • #2
meBigGuy
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I can't see why it always starts on that end. Was the pickle wired to the plug and he inserted the plug, or did he plug the hot side into the grounded pickle? Maybe his connection methodology is such that it favors one side (different probes). Maybe he always connects the ground to the fat side.
 
  • #3
anorlunda
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I just sent a private conversation with pickle123 with my answer to his question. I don't want to post it publicly.

pickle123 seems to think that a thread like this is a conversation between him, the monitors, and the experts who might post answers. It is not. It is part of the public record. Even children can find it and read it. In fact, immature young people could even google the phrase " please do not try this at home: it is dangerous!" seeking fun things to do. That is why we don't like subjects like this on public forums. Use closed forums, or private conversations with the mentors or other PF experts to discuss dangerous topics.
 
  • #4
DrClaude
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I just sent a private conversation with pickle123 with my answer to his question. I don't want to post it publicly.

pickle123 seems to think that a thread like this is a conversation between him, the monitors, and the experts who might post answers. It is not. It is part of the public record. Even children can find it and read it. In fact, immature young people could even google the phrase " please do not try this at home: it is dangerous!" seeking fun things to do. That is why we don't like subjects like this on public forums. Use closed forums, or private conversations with the mentors or other PF experts to discuss dangerous topics.
I'm sorry, but this is ridiculous. There is nothing inherently dangerous in discussing the phenomenon.
 
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I can't see why it always starts on that end. Was the pickle wired to the plug and he inserted the plug, or did he plug the hot side into the grounded pickle? Maybe his connection methodology is such that it favors one side (different probes). Maybe he always connects the ground to the fat side.

Unfortunately, I am quite sure that it is nothing like this. This professor loves things with simple physical explanations, and this is a question that he has stated that he likes to ask physics graduate students during their oral exam. I asked one of his graduate students about it, and they seemed quite convinced that you could understand this problem by looking at it as a huge resistor in parallel with a capacitor, where one end is grounded. He suggested that this is also why people get zapped by outlets and feel pain at the point of contact, not through the entire path to ground (i.e. there is not much of any actual current, just charging, most of which occurs on the "hot" side--this is what I don't understand).
 
  • #6
NascentOxygen
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While being cooked, was the pickle mounted so that its longitudinal axis was vertical or horizontal?
 
  • #7
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How and where did he connect the electrodes?

You feel pain at the point of contact because the current density is higher due to the small area where current can flow. If one of the electrodes has a better contact, then heating there will be lower than at the other side.
 
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He sets the pickle down horizontally on the table, then plugs the electrodes directly in to each end (effectively everything is co-linear)
 
  • #9
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Then it should not depend on the connection of the electrodes, but only on their depth in the material and their precise position.
 
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first of all reading this otherwise interesting thread I just want to say that we should really stop this paranoia of saying something that someone might sometime in someplace find worthy of repeating and killing himself in the process, people have done stupid things all the time and also died from them ,
much like Sweden's IKEA store has now banned kitchen knifes in it's stores simply because a muslim used one to stab a woman, the world we now live in is absurd.


as for the pickle, why would the pickle be modelled as a capacitor with a resistor in series? the electrodes are conducting and so is the juice in a pickle , or am i getting something wrong here ?
 
  • #11
NascentOxygen
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He sets the pickle down horizontally on the table, then plugs the electrodes directly in to each end (effectively everything is co-linear)
So why does your subject line mention "grounded plate"? :oldconfused:
 
  • #12
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I find that weird too, the only explanation that comes in mind is that there is a " hot" electrode in the pickle and then the other electrode is attached to a series capacitor which is then connected to the neutral of the mains.
 
  • #13
meBigGuy
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The reason you feel pain at the point of shock is because it gets hot due to skin resistance at a single point. If you grabbed with both hands at the same time and exactly symmetrically there is no difference.

The key is that he connects the electrodes individually. Which does he connect first, which last. I would predict that the last connected would be where the burning starts since it would be, while inserting the electrode, higher resistance.

The statement that implied that the problem was not symmetrical because of one side being ground is wholly without substance. A ground node is no different than a power node to the load. It is a linear system.

If identical pickle electrodes were securely wired to a mains plug, and then the professor randomly inserted the plug one way or the other, then I expect you would see random ends ignite.

I also agree with Salvador that the suggested capacitance model is not valid, especially at 50/60Hz. Besides that, I can't see why it would be non-symmetrical.
 
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  • #15
NascentOxygen
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In one of those demonstrations the glow started off at one end, then after a bit the glow shifted to the other end.

I'm surprised by the glow, I didn't expect it to glow near-white. I anticipated the glow would emerge slowly, spreading as a dull red attributable to heating.
 
  • #16
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The glow is not a (purely) thermal effect.
 
  • #17
PeterDonis
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he said that the end connected to the "hot" terminal of the outlet is the side on which the fire always starts. I do not understand why that should be, given that this is an alternating current source.

It's an AC source, so the direction of current flow changes, but that does not necessarily mean the direction of energy flow changes. For a purely resistive load, voltage and current switch polarity at exactly the same time, and power is the product of voltage and current, so its sign never changes; in this idealized case, energy always flows from "hot" to load, never from ground to load, so the "hot" side of the load will heat up first. The pickle is probably not a purely resistive load, but if it is fairly close to being so, the direction of energy flow should be fairly close to being constant in the direction from "hot" to load, rather than from ground to load.
 
  • #18
meBigGuy
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I have a real problem with that (I'm not saying you are wrong, but that I have a problem with that). The energy dissipation in a resistive element will be distributed evenly based on the ohmic resistance. The electrons move at drift velocity bumping into things as they go. They start moving at all points in the element when the potential is applied.

I do not see how the arbitrary assignment of hot and neutral can affect the result. For example, drive the load from a balanced isolation transformer. The load will see no difference in the driving potential if you arbitrarily ground one or the other of the isolated lines.
 
  • #19
PeterDonis
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The energy dissipation in a resistive element will be distributed evenly based on the ohmic resistance.

In the idealized case of a constant current flowing through a resistor, yes. But we're talking about starting from zero current, then applying voltage to produce non-zero current. During the transition, energy dissipation in the resistor is not the same everywhere; it starts where energy first flows into the resistor.

The electrons move at drift velocity bumping into things as they go.

The motion of electrons is not the same as the flow of energy. Energy is contained in the EM field as well as the electron motion, and it is dissipated in the resistor by thermal vibrations of the atoms, not just drift motion of electrons.

They start moving at all points in the element when the potential is applied.

No, they don't. The potential is not applied instantaneously everywhere; that would violate causality.
 
  • #20
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but practically speaking, once you connect the pickle , resistor to the load the electrons start moving from one side to the other with whatever the speed of the EM field (speed of light) is through that medium , and if i'm correct it's about 2/3 to that of c in vacuum , so even though as Peter said the electrons don't start moving everywhere at once , for simple questions like does a resistor heats up evenly regardless of it's connection or which end of a pickle lights up first , this shouldn't matter I think because the difference would be unnoticable.
 
  • #21
sophiecentaur
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the electrons start moving from one side to the other with whatever the speed of the EM field (speed of light) is through that medium
That is just plain wrong. The speed of the charge carriers is very slow and nothing like c, in a solid. What can happen in a vacuum is quite different because there is nothing to slow them down.
Just check out this link.
 
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  • #22
meBigGuy
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I think you are reading it wrong. I read it to say that the POTENTIAL, moving at the speed of light, causes electrons all along the pickle to start to move at their drift velocity. The wave of initial movements moves at C in that medium. The "power dissipation front" (my term) moves at C.

The words "start to move at C" can also mean that the starting moved at C.

Regardless of whether that is what Salvador meant, that is the way it works. The beginning of dissipation moves at C in that medium which is pretty quick relative to pickles heating. (I previously said instantaneously - my bad)

Can we focus on the below unless there are substantial disagreements with how fast the beginning of the power dissipation effects propagate through the pickle (at C in the medium).
No one has commented on my example of an isolation transformer with 1 side/other side/no side grounded and whether the pickle could tell the difference regarding which terminal is grounded.

I maintain it could not. In fact you could connect one side to 1 million volts and it could not tell.
 
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  • #23
NascentOxygen
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I'd need to see the setup and know how many times the professor has repeated the demonstration with always the same results before putting any store in the claim. Of course, were there significant capacitance between the pickle and the electricity ground then the hot electrode would deliver more current than does the neutral wire, but such is not going to be the case here.

I would hope the professor sets a good example for electrical safety by always using an isolation transformer, whether he chooses to announce it or not. And it goes without saying that he has ensured ready access to a fire blanket....responsibly following all the normal safety measures and precautions.
 
  • #24
sophiecentaur
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The "power dissipation front" (my term) moves at C.
That's a fair description. The nearest 'offoicial term' I can think of to that would be the Poynting vector, which describes the energy flow in the wave. That would advance with the wave front at switch on.
As you say, an isolation transformer is de riguer in any experiments like that. I remember my very flash Physics master giving us a demo of boiling water in a beaker with two nails, connected directly to the Mains socket on the bench. In those days they didn't have RCDs either. (He did tell us not to try it at home but it was only good luck on my part that I didn't.)
 
  • #25
sophiecentaur
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I'd need to see the setup and know how many times the professor has repeated the demonstration with always the same results before putting any store in the claim. Of course, were there significant capacitance between the pickle and the electricity ground then the hot electrode would deliver more current than does the neutral wire, but such is not going to be the case here.

I would hope the professor sets a good example for electrical safety by always using an isolation transformer, whether he chooses to announce it or not. And it goes without saying that he has ensured ready access to a fire blanket....responsibly following all the normal safety measures and precautions.
If the professor is a man of habit, he may be holding the pickles the same way round every time and connecting one particular lead to the stalk end of the pickles. That could have an effect which would need to be eliminated by doing it both ways round an equal number of times.
 

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