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Look at this VIDEO to see what is supposedly happening:

A Wimshurst machine is a seemingly simple device consisting of two plastic wheels with embedded metal plates on the rim. The wheels spin in opposite direction, facing each other. Two metal bars, one on each side, span the device: They are at an angle of 90°, but separated by the two discs spinning between them. At the end of the bars, metal brushes touch the plates spinning underneath them. A slight initial charge imbalance on one of the plates (which necessarily exists) causes the bar brushing against it to polarize along its length, so the metal plate on the other end of the same disc is charged oppositely and retains that charge when the brush disconnects during the spinning. As this two-plate "dipole" spins past the bar located behind the other disc, it induces an opposite dipole in its opposing plates on the other disc, which, in turn, polarizes the bar brushing against them. The charged metal plates give off their charges to two Leyden jars which connect to the plates with brushes of their own.

However, why do the Leyden-jar-brushes collect the charges? They tend to get very negative/positive, so why don't they push the electrons/holes away instead of collecting them? Wouldn't the brushes charge the passing plates which then travel around to the other Leyden brush, and get discharged there?

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  • $\begingroup$ At the positive Leyden-jar-brush the plate has so much positive charge that some positive charge is pushed into the jar. At the negative Leyden-jar-brush the plate has so much negative charge that some negative charge is pushed into the jar. $\endgroup$ – stuffu Aug 29 '15 at 18:35
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    $\begingroup$ When the metal bar short circuits two plates, both plates lose their charges, then new charges are induced on the plates, the new charges are the opposite to the old charges. $\endgroup$ – stuffu Aug 29 '15 at 19:30
  • $\begingroup$ @user7027 You're right, the travelling plates won't transport charges from one Leyden-jar to the other, because the brushes short circuit - and you're right that the charge density on the plates must be larger than that on the Leyden jars for them to load them - but why is this so? The video doesn't explain why the charge density on the Leyden jars will necessarily be less than the charge density on the plates. $\endgroup$ – yippy_yay Aug 29 '15 at 20:02
  • $\begingroup$ I think it's like this: How much charge is induced into a pair of plates? As much as there is charge in the two inducing plates, and then some more, as there are charged plates right next to the two "inducing plates". The plate-pair into which charges are being induced sees more than two plates worth of charges on the opposite disk, so the plate-pair charges to higher charge than what two plates in the machine have on the average. $\endgroup$ – stuffu Aug 29 '15 at 20:57
  • $\begingroup$ Whomever answers: here's a very interesting link describing why you need to use both combs and brushes for a fully functioning machine: books.google.com/… $\endgroup$ – Daniel Griscom Aug 29 '15 at 21:01
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Ignore the collecting combs and Leiden jars for the moment. As you noted, the Wimhurst disk and 45° shorting bars form a powerful electrostatic pump such that the left side of both disks gets charged one way and the right side of both disks gets charged the other way.

This charge feeds back through the shorting bars and builds up until something leaks the charge, and that's where the combs and Leiden jars come in. Under one pair of combs, the disk sectors are charged highly negative, and since the two adjacent sectors on the two disks are charged the same way, the electrons wants to escape. The combs provide such an escape route, and the electrons jump to the combs and accumulate in the Leiden jar. Yes, the charge already in the jar repels the incoming electrons, but the pump is strong enough so that (to a point) the electrons keep coming.

I found an interesting article in the 1888 Proceedings of the Institution of Electrical Engineers called "The Influence Machine from 1788 to 1888". It includes an interesting explanation of why the Wimhurst Machine needs both combs and brushes to operate properly.

Brushes directly contact the underlying sectors, and so only an infinitesimal potential is required to push electrons across them. Combs are above the sectors, with an air gap, so that a significant voltage must be present before the electrons cross the gap.

The shorting bars should end in brushes, otherwise the machine won't be self-starting. With brushes, the inevitable imbalance of charge will start being amplified right away; if combs were used then there'd have to be a significant initial charge before the electrons could cross the gap and start the amplification process.

The Leiden jar collection points should end in combs, otherwise a discharge may reverse the polarity of the machine. Were brushes used, the charge on the sectors under the brushes could be completely neutralized, returning the machine to its initial state awaiting a random imbalance. With combs, however, not all of the charge on the sector can be removed by the discharge, and thus the machine will maintain its polarity and not have to start from scratch.

(Nineteenth century technology is SO cool...)

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  • $\begingroup$ I can only see snippets of your reference - I'm accessing Google Books from Germany - are you located somewhere else, or do you have a copy of the 'Proceedings'? Yeah, the 19th century is where it all came together - most people think it was the 20th century that things took off, but for me, it was the 1800s. $\endgroup$ – yippy_yay Aug 30 '15 at 22:45
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@Daniel Grissom gives a great answer, but I wanted to drill into the exact question you asked about the direction of flow. There are several points of note here:

  1. The comb is not in direct contact with the plates, but is outside it. This means that when the plates have the same charge they are trying to push that charge outwards, if it's energetically preferable to do so. This situation occurs when the combs are available to catch them.
  2. The reverse cannot happen, because the charge on the comb is still the same on both forks and so it repelling away from the plates.
  3. Where the charge goes then is a question of relative voltage.
  4. The voltage on the comb stays low because of the Leyden jars/capacitor which enables the comb/jar system to build up a lot of charge without building up too much voltage (roughly).
  5. The plates, meanwhile simply continue to build charge until the voltage is high enough to cause the transfer to occur (even a small charge creates a large voltage on the plates).
  6. That's also why you need brushes on the arms. The brushes mean that even a low voltage will be transferred, effectively amplifying the voltage on each of the plates with minimal leakage (again roughly).

It's a bit more of a hands-wavy explanation, but sometimes they help to build intuition.

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