3 ms·

>and that this capacitance increases not linearly, but exponentially Well, perhaps I have the "but exponentially" part wrong ("I stand corrected!") -- but it c

• gus_massa 4y ago
  Nitpicking: The correct name is https://en.wikipedia.org/wiki/Proportionality_(mathematics)#Inverse_proportionality https://en.wikipedia.org/wiki/Proportionality_(mathematics)#... --- TL;DR: Back to your original question, I guess the answer is YES, but please read the rest of my comment, because it's a surprising YES. --- Back to your original question, it's a weird question. I need like 24 hours to understand it and a few hour to think a good answer. If you try it IRL, give the other person a link to the video, my comment, and 24 or 48 hours to think about it. This is not my area, so I hope my answer is correct. I'm 90% sure it's correct, but for the record I'll only claim that I'm only 70% sure it is correct. Take all the rest with a small grain of salt. --- As I said before, C = εA/d. The force in each plate if F=Q^2/εA=εAV^2/d^2 [verify]. So if you have charged capacitor and the voltage V is constant, the force goes to infinity if the distance d goes to 0. So a very small d creates a very strong capacitor with a lot of force between the plates when they are charged. If the capacitos has no charge, then F=εA0^2/d^2=0 and there is no force. Anyway, in a real capacitor, there is thermal noise. The noise is different in each plate, so you get F=εA(V+noise_1)*(V+noise_2)/d^2 where noise_1 in the noise in one plate and noise_2 in the noise in the other plate. So even if V=0 you get F=εA\noise_1*noise_2/d^2. In the easy model, both noises are not correlated, so the average is 0. My guess is that there is a correlation and when one plate is slightly positive the other is most of the times slightly negative, and you get a average force that is not 0. This is tiny when the plates are far away, but I guess the correlation is greater when the plates are very close. Also, when d is very small, makes everything else much bigger. Note that this is not the Casimir effect. This is thermal noise, that is a classic effect cause by the movements of the parts of the capacitor and the molecules around it. This can be reduced if you put the capacitor inside something that is very cold like liquid air (or liquid helium if you have more money and need less noise). --- The Casimir effect is a quantum effect. You can't avoid it even putting the capacitor at very low temperatures. A simple way to imagine quantum mechanics is that it add uncertainty everywhere, so you again get V+"noise", but it's not the usual thermal noise, but some quantum noise. The problem with this is that you can imagine that you can use a tiny multimeter to measure this. The real thing is not that the V changes from time to time. The real effect is some weird algebraic properties and superposition. You can imagine that is like the classic effect where the V actually changes an get a good intuition in some experiments, but in others experiments you get the wrong result. Quantum Mechanics is actually more weird. So now F=εA*"noise_1"*"noise_2"/d^2, but it's not real noise, just a bad approximation to the quantum fluctuations. Again, the values of both "noises" are correlated, and you get an average that is not 0. The amount of correlations is bigger when they are close and 0 far away. Also the d^2 in the denominator makes this force almos 0 when they are far away, and big when they are very close. I guess this is what they call in the video "Van der Waals" forces. I've seen the Van der Waals forces for molecules, but I guess they can be extended for some (almost?) ideal capacitor plates. --- Again, this is not my area, so I hope my answer is correct. Take all the rest with a small grain of salt. I'm using regularly quantum mechanics, but for smaller system like molecules. A capacitor is too big for me.
  • peter_d_sherman 4y ago
    It's interesting that you bring up Thermal Noise... Thermal Noise would indeed have to be considered alongside The Casimir Effect and Capacitance -- because even though Thermal Noise could be reduced by cooling -- it is unknown if it could be eliminated entirely... If these three effects are always present alongside one another in an electronic system, most notably the metal plates of a metal plate Capacitor -- then it would seem that they are indeed related, although, I'll leave it to future Scientists to state that for a fact, and to better quantify the math of those relationships if there is one... Thus, these phenomena should definitely be studied alongside one another... Anyway, thank you for your excellent addition of the concept of Thermal Noise to the discussion!
    • gus_massa 4y ago
      [Assuming I made no horrible mistake in the GP...] In general, you can make systems that show only one of the effects. * If it's big enough and the voltage is big enough, you can almost get a system that is a classical capacitor, without noise. * If you reduce the voltaje, you may see the noise. You can increase the temperature to get more noise. I wonder if it's possible to measure a Casimir-like effect here. I never heard that, but it looks possible. Anyway, if the temperature is too high you will have some nasty new things like electrons jumping like in a vacuum tube. * If you reduce the size and keep it very cold and apply no voltaje, you can get the almost pure Casimir effect. A big part of physics is understanding which simplifications you can make and how to build a device where the simplifications are correct. Each of the effects are useful, but a device that use all of them is probably too complicated to think about. I wonder if the people building microprocessors have these problems. They have hot devices that must be at room temperature, that are tiny because they must put a lot of them in a small piece. The only part they don't have is the vacuum, so the material between the plates can keep them away.