# ok this is sci fi right here

Discussion in 'General Electronics Discussion' started by donkey, Mar 16, 2013.

1. ### donkey

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Feb 26, 2011
so anyway I have had kidney stones of late and they gave me some awesome painkillers, the issue is that theyare definately affecting my judgement right now so I need someone to very simply tell me that this is not possible. I will include a picture poorly drawn to help with you guys being able to tell me why this isn't possible.

ok so this is a really bad representation of a capacitor. on the bottom you have your 2 standard plates and when voltage is applied one is full of electrons the other depleted. So I was trying to find out why we can't chain this effect.
the next row up will be placed very close to the first 2 plates and when voltage is applied to the bottom 2 the next row up also gains or looses electrons, and then when it is full the next row up.
The big thing that plays in my mind is what the other plates will be drawing the electrons from and I thought a sheet of metal might be the simplest experiment. chuck a diode on it so there is no flow back to the metal plate andhave a circuit run of it with the plate being the pos side

so tell me why the hell this won't work cos in my drug induced state I can't see it.... I probably wouldn't see it normally either but its just something that intrigues me

I also have the theory that because it has different "supply" of electrons they wouldn't react the same way.

2. ### john monks

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Mar 9, 2012
I think I see your confusion. This confused me until I studied this in physics.
The plates cannot accept or give off electrons unless something is connected to them. But then you ask,"but isn't there voltage induced in the second row of plates?" and of course there is. Now this is the trick part. The first row of plates is charged positive and negative. The negative plate has extra electrons piled up on the mostly nearest the vicinity of the positive plate. Now the second row of plates gets a charge from the first row of plates but no electrons go to them nor leave them. The plate nearest the negative first row plate becomes negatively charge because of electrostatic coupling. And the side nearest the first row negative plate has electrons on it's surface leave the surface and travel to the opposite side of the same plate. And the total number of electrons on the plate remain exactly the same. The same effect occurs the the third row plates. Pain medication messes you up. I know from experience but hopefully this clears this up. If not let's take another crack at it.
I'm willing to spend all night on this because this question you have is at the very heart of electronics.

3. ### donkey

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Feb 26, 2011
ok so lets say I hook up a plate of metal to the second set of plates. I place a diodebetween the negative (black) and the plate so electrons can flow from the metal to the plates but cannot go back that way.
now lets hook a circuit a simple light directly to theplates on the second row, no caps, diodes or any other components just wire and lightbulb.
as the electrons cannot flow back to the platedue to the diode blocking it they will follow the easiest path, through the wires to the bulbthen back to the sheet of metal. would this create any electricity or is there a part I am missing or is it just not possible

so I am connecting the plates to something common but restricting electron flow via a diode. will this create a small current/voltage?

4. ### john monks

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Mar 9, 2012
I'm not clear what you're asking about the plate and the lightbulb but let me answer what I think you're getting at.
Let's suppose you connect a lightbulb between the second row of plates. Now let's keep in mind that the initial charge of electrons on the plates, that is the net charge, is zero. That is you have excess electrons on one side of the plates and to few electrons on the other side of the plates so you have net zero charge. Upon connecting the light bulb between the plates the bulb will light as electrons pass from one plate to the other. Now you have a net charge on both plates but at the same time you have zero volts between the plates. Dispute the net charge. But how is this possible? The plate closest to the first row of plates nearest the negatively charged plate has to few electrons and the second row plate nearest the first row plate nearest the positively charge plates has an excessive number of electrons. And still both second row plates have a net zero charge. And before the light bulb was connected both individual second row plates had a net zero charge and they had voltage between the.

5. ### donkey

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Feb 26, 2011
hey john thanks for replies so far. will get pics of how i see it working in next few hours to show what I see is happening

6. ### donkey

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Feb 26, 2011

ok by the time I get to the third pic will electrons flow the way I predict?
oh and I always get the diode thingy messed up I think thats wrong again but its meant to allow flow from the plate but not to the plate.
the brown thingy at the top is a copper plate

7. ### john monks

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Mar 9, 2012
No. When you close the switch and if the negative terminal of the battery is on the right hand side current will flow through the light build and slowly come to a stop as no more electrons are available to flow from the top right black plate and nore more electrons can be observed by the upper left red plate.
Now the circuitry has reached a state of equilibrium, the lower right red plate fully charged to the battery voltage and the lower left at the opposite polarity.
Now when you open the switch the charge on the lower plates will remain the same because the electrons have no place to goe and the voltage between the plates will remain the same.
So opening the switch will do nothing to the bulb, no light will come out at that point in time.

8. ### donkey

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Feb 26, 2011
ok so if I discharged the plates on the lower level that would in turn make the electrons flow up a level?

9. ### john monks

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Mar 9, 2012
Yes. That is if you opened the switch and shorted the two lower plates together the upper red plate would turn negative and the upper right black plate would turn positive and the diode would conduct electricity until equilibrium was reached.

10. ### donkey

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Feb 26, 2011
so this is possible right? its not as far fetched as I thought?

11. ### john monks

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Mar 9, 2012
If you mean that it is this easy to transmit power through space without wires then yeas, it's that easy.

12. ### donkey

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Feb 26, 2011
ok what I am actually after is the copper plate to ionise as the electrons would flow through the bulb creating light thus using electrons. the end result is less electrons would be in the copper plate.
my theory isthat you could use a conductive sheet and makea battery of sorts from it. the cap wouldbedrawing out the electrons but as itpasses through a circuit they would not go straight back and thus the piece of metal would then become useless after a point. well when I say useless I actually have another scheme for that.

13. ### john monks

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Mar 9, 2012
Just having a copper plate standing alone in space will not supply you with much energy reserve. Keep in mind that in the top red and black plates one is giving as many electrons as the other is receiving. The best way is to have two large plates facing each other separated by a short distance. That way the electrostatic force on one plate can affect the other. And to actually conduct such an experiment would probably requires thousands of square meters of surface.

14. ### Laplace

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Apr 4, 2010
Have tried to make an equivalent circuit of the configuration as presented. However, it was uncertain just how to model a copper plate since a copper plate is indistinguishable from a copper wire. Also, the specks shown as bunched to one side of a plate, which I assume are meant to be electrons, are a physical impossibility. There are no electric fields within the interior of a conductor, and no bunching of electrons within a conductor.

Please check the attached circuit to verify that it accurately models the electrical aspects of the physical setup. The next step would be to explain the electrical effects using the circuit model rather than the physical configuration.

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15. ### donkey

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Feb 26, 2011
laplace within a capacitor there is bunching of electrons, that is how it works
one side gets more electrons the other side gets less
the copper plate may bea bad design but I still am trying to figure out what plate would be good in its place. I need something that will give afree electron hopefully is an element to make life easy.
the bunching of electrons is an effect of the capacitor so as electrons are pushed from the pos plate they go to the copperplate. they are then drawn to the neg plate of the cap. I am seeing this like a magnet but there is definately something I am missing here. it could be because I use the water analogy too much.

the way i see this working is the 2 plates on the bottom react with the 2 plates on the top (your diagram shows 4 independant caps so it is different again.)
when they react there is a small amount of electrons moving as the plate allign to neg and pos, like a magnet does. now the electrons that bunch on the plate create a "holes" which need to be filled. when the bottom plates are discharged the electrons go back to "settle" they can't go to the copper plate so they go through the light first.
john has stated this is simply a transfer of electricity and I assume this isthe same as a piece of wire with a light attached near a cb radio. what I am hoping to do is create a reaction that ionises the plate the electrons are then used to create electricity to make the light turn on (I am seriously guessing it'll be less than 1% that the voltage of the battery, but even 1% isa good start I reckkon)
I am hoping not to transfer electricity butinstead ionise plates making any conductive material a battery of sorts. I am seriously thinking it is fair fetch as after 250 years after the finding of electricity and the capacitor that no one has seen this as anything but I am hoping that it is found somewhere and I can use it to make something else. I consider it like extending battery life if it works.

16. ### john monks

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Mar 9, 2012
Laplace, you are essentially correct in saying that the punching of electrons does not occur within a conductor. But this is not absolutely true. In studying magnetics this becomes extremely relevant. But for now we will say that bunching occurs only near the surface of a conductor. This is how a capacitor works. That is one plate has to many electrons near the surface and the other plate has to few. And this bunching occurs mostly where the plates are close to each other because of the electrostatic forces.
I deliberately did not mention magnetism because this may greatly confuse the issue. Because as donkey says that the electrostatic effect is similar to the magnetic effect. The truth is the magnetic effect is the electric effect in four dimensional physics, the fourth dimension being time. That is the magnetic effect is the electric effect relativistic as Einstein pointed out in his paper, "Special Relativity". But for those who have not studied relativity this can be extremely confusing. So for now I did not want to confuse anybody with it.

Last edited: Mar 18, 2013
17. ### Laplace

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Apr 4, 2010
In the presence of an electrostatic field, electrons will migrate to one surface of a dielectric material (molecular dipoles will rotate); that is how a capacitor works. But what if instead of a dielectric material there is nothing but vacuum? There are no dielectric electrons to bunch together in the open space, and electrons do not collect on the surface of the conductive plate. I am familiar with the skin effect resulting from the interaction of electric and magnetic fields at high frequency, but that is not what is happening in this situation. Nor should it be necessary to resort to special relativity to develop an equivalent circuit model of the physical configuration posited here. If you can find something inappropriate with the circuit model, I would like to know.

18. ### john monks

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Mar 9, 2012
I see nothing inappropriate with your circuit model. And your explanation of how a capacitor with some dielectric is correct. And for the questions donkey asked I would prefer to stay away from relativity. But the only issue left is a vacuum capacitor. How does it work. If electrons do not accumulate within the conductive portion of the plate then how does it store a charge? Lets look at another analogy. Suppose a spacecraft develops a negative one million volts while traveling into outer space. Where does that charge go? Does it go to the inside of the spacecraft? They surround the spacecraft near the outer surface like a halo. Skin effect is another issue having to do with relativity.

19. ### Laplace

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Apr 4, 2010
Yes, excess electrons accumulate within the conductive portion but they are evenly distributed when the electromagnetic fields are static, as in this case; they do not bunch to one side of the plate.

20. ### donkey

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Feb 26, 2011
ok I get to a certian place there and then get lost.
don't electrons get usedto createnew things?
so in the case of a spaceship the electrons would either be stored on the cap if not used, or used to create movement or light or whatever?
I am positive thate I have thebeginners guide to what really happens but it is what happens right?

the issue I have with the diagram shown is that I don't see how the 2 extra caps are going to have any effect whatso ever on the other 2 caps. I am trying to get the plates to react with each other as the vertical plates will not be directly affected as one side won't bepos other neg I don't see how the cap would "charge" I do seethere is a possibility but I don't see it working. if we took out the 2 horizontal caps then there is more chance this will work in my mind.
have to go to work but when I get home will try to draw new diagram