Is the electric field stronger in resistors with higher resistance?

[Mr.Bomzh]

Well my try on the field strength calculation would go like this assuming both wires leading up to the resistor from the battery terminal have something like 2 ohms os resistance and they are say 10cm each, assuming 12v on a typical battery that would be like E=12/10 which is 1.2, although it's kinda strange because I just divided the resistor length by voltage but shouldn't this also include the total resisatnce of that particular resistor?

Or maybe I have to do the above formula first and then put the number into the second formula which you actually posted first which goes like E= Ix (resisatnce per unit length) from the previous formula I just found out that the number for resisatnce for unit length is 1.2 so now I will attempt to calculate current which is I =V/R, now I= 12/2 which is 6amps of current.

So E = 6 amps x 1.2 which is 7.2, volts per metre I guess?

Now as for the resistor , assuming it is 100 ohm and 4cm long , E=12/4 =3(resisatnce oper unit length)
then I=12/100= 0.12mA , now E=0.12x3 =0.36 again volts per metre?

So in this case the field in the resistor is smaller than that in the wire?
Also in a single resistor case , I guess the total current that flows from the battery + to - is not the current I calculated through the 2 ohm wires but the current of the 100ohm resistor because in a series circuit current through each component is the same and the overall current in a series circuit is the current of the part with the highest resistance which in my case is the 100ohm resistor is that correct?

 

[thegreenlaser]

The current doesn't affect the 'Field'. It is just the Volts per Metre that counts. If a battery is supplying the volts then they are what they are, with or without a resistive path.

Current is related to electric field... E = V/L = I*R/L and J = σE are both relationships between current and electric field which are valid in the context of this thread. However, I agree that E = V/L is the more useful way to go in the case of constant voltage. I've had to backpedal a bit because I gave an answer specifically targeted at the original question (series resistances) and then that answer started getting eroneously applied to a different context.

As I said before, since we're now dealing with electric fields in general circuits, I agree that E = V/L is a better way to go.

Bad idea to use the term "Feedback mechanism" in this context. Feedback, in Science implies the existence a control loop with active amplification in the feedback path.

I'm well aware of what a feedback loop is, but maybe I'm wrong about how the electric field gets set up in a circuit. My main intention, though, was to convince MrB that it's really not worth opening up that can of worms, especially without a lot of experience with EM theory.

I totally agree with your idea that he should take a proper course about this. Privately inventing a private version of Science for yourself seldom proves a fruitful exercise. No pain no gain MrB.

Glad we agree.

@MrB: You have 3 resistors in series connected to a battery, so having a 12 V drop across each of them would be a direct violation of Kirchoff's voltage law. To be blunt, this is a really simple circuit and the fact that you're assigning 12 V to each element tells me that you don't really understand basic circuit theory.

I don't mean to belittle you; your enthusiasm for this stuff is great, but you're getting in over your head. You need to start with the basics of circuit theory and electromagnetism before trying to understand how they align with each other. You'll be much more satisfied with your understanding that way.

 

[Mr.Bomzh]

with my calculations I concluded that this current has to be 0.12mA, is that correct?
Assuming negligible resisance of the wires attached to the resistor , we can calulate the voltage drop over the resistor which is V=IxR which is 0.12x100=12volts.

I did made a mistake previously I should have just counted the wire resistanes in with the resistor because they altogether make one large resistor just with different resistivity along the way , smaller along the wire and higher along the resistor.

But my use of formulas was correct ?

So in a a series circuit with two resistors, one 4 ohm other 8ohm and a 12v battery, the current through each would be the same but voltage drops different ,For the current in the circuit I= V/R so 12/12 = 1 amp. of current , for the 4 ohm resistor V=1x4 and for the other resisor V=1x8, so the question is the current is the same through both resistor but the resistance of each differs so the electric field in the 8 ohm resisor is higher than the field in the 4 ohm resistor , assuming both have equal lengths? Ad the reason why it is higher in the 8 ohm resistor is because to force the same current through a higher resistance needs more emf which relates to higher voltage?
I also use this page to read about resistors , find it good
http://physics.bu.edu/py106/notes/Circuits.html

 

[sophiecentaur]

Assuming negligible resisance of the wires attached to the resistor , we can calulate the voltage drop over the resistor which is V=IxR which is 0.12x100=12volts.

But you started with a 12V battery, so it is hardly surprising that the voltage across it is 12V. If it were not, you would need to take it back to the shop!

 

[Mr.Bomzh]

Yes , I get it for a single resistor over the battery the voltage drop is always going to be 12 volts , because upon measuring with a voltmeter , you basically attach to the + and - leads of the battery which also happen to be the wires entering and exiting the resistor. a two resistor case would be more interesting , i made some calculation in the previous post why doesn't anybody say something about them ?

thank you . :)

 

[sophiecentaur]

Forget the meter. A battery is a Voltage Source - period.

 

[thegreenlaser]

with my calculations I concluded that this current has to be 0.12mA, is that correct?
Assuming negligible resisance of the wires attached to the resistor , we can calulate the voltage drop over the resistor which is V=IxR which is 0.12x100=12volts.

I did made a mistake previously I should have just counted the wire resistanes in with the resistor because they altogether make one large resistor just with different resistivity along the way , smaller along the wire and higher along the resistor.

But my use of formulas was correct ?

I don't think your use of formulas was completely correct, but honestly I got pretty lost in what you were doing so I don't know what exactly you did wrong.

So in a a series circuit with two resistors, one 4 ohm other 8ohm and a 12v battery, the current through each would be the same but voltage drops different ,For the current in the circuit I= V/R so 12/12 = 1 amp. of current , for the 4 ohm resistor V=1x4 and for the other resisor V=1x8, so the question is the current is the same through both resistor but the resistance of each differs so the electric field in the 8 ohm resisor is higher than the field in the 4 ohm resistor , assuming both have equal lengths? Ad the reason why it is higher in the 8 ohm resistor is because to force the same current through a higher resistance needs more emf which relates to higher voltage?

As far as I can tell, this paragraph is correct. You should have done something more like this with the 100 Ω resistor and the 2 Ω wires. Find the voltage across each resistor exactly like you did here. Then, once you have the voltages, divide those voltages by the lengths of the elements to get the electric field in volts per metre.

 

[sophiecentaur]

I already referred him to the potential divider theory but it must have been too straightforward for him to bother with.

 

[thegreenlaser]

Indeed...

Well, I hope what's been written in this thread will become more useful once MrB takes some time to learn basic circuit analysis and EM. That's what the rest of us had to do too...

 

[Simon Bridge]

To me, this seems to be the heart of the problem:

Ok I have to lay down my cards , so that you folks don't get confused for my questions.

That is a very good idea.
However, I don't think you are being entirely straight with us even so... see below:

the idea is more about that I need a high electric field from low voltage in a small area , which would be surrounded by a dielectric material ... it has to have high resistance so that I wouldn't need much current, yet I need the E field.

That's situation 1.

I was just wondering what would happen if a dielectric placed between the plates of a capacitor had a resistive layer in between which would be attached to a circuit and current would pass in it , we know that the charges in the dielectric align themselves according to the charges on the plates oppositely of course.

This is situation 2.
It is very different from the situation in the last passage I quoted.

I was just wondering what would happen to the alignment of the dielectric charges if the dielectric had a e field inside of it.

This is an different situation again ... so you are, in fact, pondering three distinct situations.

Interestingly, circuit theory is not helpful for any of these situations.
Much confusion stems from trying to overlap two different ways of understanding something.
Stop it.

In situation 1: you need a short length of high resistivity material.
There are material constraints about how short the length can be and so on that you will discover by experimentation.

In situation 2: you are describing something like a carbon rod passing perpendicularly between two plates. "wonder what would happen" is too vague though - lots of things happen - you need to ask a specific question. What sort of effect are you interested in?

eg. electric fields (i.e. due to the capacitor) that are perpendicular to a wire (or resistor etc), have no effect on the electric current or voltages along the wire.

Situation 3: this is different from situation 2 because now you want an extra field in the dielectric (in situation 2, there is a rod of carbon where the extra field is - the dielectric is outside).
This would result simply from using two sets of plates.
The polarization of the dielectric is simply aligned with the combined field.


You still have not been straight with us though: what is the application you have in mind?
What are you hoping to discover?

 

[Mr.Bomzh]

To Sophie, excuse me I did not ignored the things you gave on purpose , it's just that sometimes I miss out on some stuff you either say or give as I am too concerned about the context, I will look back on the posts as I do to reread them and understand better some stuff.

To Simon and to others , as I said my writing is confusing and vague amny times not so much because I don't undersand but because while trying to explain myself as best as I can I'm gettin nervous and stresses and my thoughts don't go together with how I express them, although there are also pure misunderstanding by my part.

And finally , well what am I trying to doscover? I have a feeling and some thoughts what it would be but I don't have the full idea of will it work. Well you see I can repair a old tv if it's not extremely complicated and I ahve done a few, I can repair a amplifier and I understand some basic stuff, it's just that I'm a lonely person sitting at my desk and having great love towards all these devices and how their work , that love and passion is affected though by my lacking maths and my mind which tells me that I'm just an ordinary little repairguy who will probably undersand just as much as all the other grey people from the huge grey masses of society.A very suicidal feeling that is I have to tell you. :D Well lacking maths but I do have one thing , I ahve the ability to visualize thing very deeply, So i;m sitting reading papers painting pictures in my head of how those phenomenon work , when I was smaller I figured out the internal combustion engine not so much from books but froms people telling me how it works and me visualizing how it works , and a few years after that I built one , in other words I took an old one from my grandfathers old car and completely restored it all new and working and put back in.There were people helping me with advice ofcourse, some old mechanics that I made very good friends with afterwards.
So with electronics it's the same , I sit read papers try to figure out stuff, I look at what's built and so on and so I came to this capacitor idea, I was thinking like for half a year non stop how could one arrange a cirucit which would use capacitors for purposes like the ones I describe.
Maybe I hate transistors , mosfets in smps applications especially , because I have never had good luck with them because it is not easy to build a device with them you have to know everything your doing very good, it's not like building an old combustion engine , I would say it's much harder in terms of intellectual power needed,
I mean I tried both and I failed at the latter one.Even though it was a really simple smps but then again a simple smps is a easy to destroy smps at the same time, taht's why I had this idea about instead of driving a transformer with direct current paths being reversed in polarity each cycle making a varying flux in the core, using electric fields to attract , repel charges which would form varying current through a primary cirucit.
i do believe , from what I know , that it;s possible , just not entirely sure of how.

Excuse me for my long memoirs , but you wanted to know what am I after and I told you.
Will write less and more precisely later one when I'm home about the progress I ahve made with you guys and the idea itself.

 

[Simon Bridge]

smps is a complicated power supply cicuit - to learn transistor circuits you have to start a LOT simpler than that.
Usually just start with a switching circuit and/or an amplifier. Maybe a flip-flop. Basic stuff to get a feel for what they do. There's a lot of resources with things for the hobbyist to try.
i.e. http://www.pcbheaven.com/wikipages/Transistor_Circuits/

The thing about mechanical engines is that the parts are big enough to see what's going on.
Electronics is a lot more spooky.

Learning about transistors is probably harder now than when I started - so much of what I used to do with them has been taken over by integrated circuits. You may like to look at simple FM receivers though.

The variable capacitor thing requires some sort of external work. You cannot get away from that.
Until you have the basics down, I don't think we can tell you any better than that.
Perhaps you'll just have to try and see.Here is the main problem for us: your goals, as you are describing them, are far too vague for anybody to really help you with them. The kind of things you want answers to keep shifting about, so nobody knows what to tell you that will do any good. We just keep guessing in the dark - that's no good to you.

I think you need to have a sit and think what you want to achieve - specifically.
The take away lesson for you in this is to keep circuit theory separate from electric field theory.
Don't mix them up.

 

[Mr.Bomzh]

Well basically , as I said earlier I wanted to make a both polarity and amplitude varying voltage source that could be fed into the transformer pirmary in order to then convert it to whatever voltage /current needed + the isolation part between primary and secondary that the transformer gives.
The transformer is the long known part and it works just okay.The thing is to make it work with dc power source.We have smps and all kinds of other topologies etc.
I was just thinking of a easier way to do it, not saying it possible just thinking, learning in the process etc.

So we concluded after all these two thread talk that a capacitor in a circuit being in series with the pirmary could pump charges back and forth through the pirmary if it's capacity could be altered , the problem here ofcourse is how does one do that , not speaking about the rotors or any mechanical stuff but in a solid satte way.

A mechanical analogy here , when I was younger i remeber an abandoned building which had elevator shafts on the top floor in the elevator engine room , there were these concrete pads on which the electric motors that pulled the cables were located, these pads stand on 4 springs , I stood on the pads and jumped and after a few jumps I could make them oscillate, now if I would just stand there doing nothing the pad would also stand still, DC is like that it just exerts a constant pressure so to speak but nothing happens , now imagine these pads and the weight on them , if one could alter the stifness of the springs while I stand there , the pads would start to oscillate just as if I would be jumping on them , only in this case I wouldn't be jumping rather standing still but altering the stifness of the springs would make the pad go lower and then higher again and so it would gain inertai and try to reach it's resonant frequency probably , which would also be largely dependant on the rate at which one altered the springs.

In my capacitor transformer case it is similar , the dc source applies a constant voltage to the circuit but altering the capacitors capacitance with the help of the dielectric one could make this oscillation in the cirucit.
Im not sure not an expert on materials science but maybe there is a nanotechnology way to make a dielectric whose dielectric constant could be altered in a large enough amplitude to cause noticable changes in the capacitance of a capacitor, ofcourse that also begs the question by what method the dielectric could be made to change , an applied voltage or a field or whatnot.

Speaking about dielectrics I have been looking and what happens to their inner workings when they are in a capacitor and the PD is applied across the plates to charge it up.Diagrams show that polarization happens inside the dielectric as molecules polarize themselves to allign according to the applied field from the plates , what would ahppen if one could alter this alignment process inside the dielectric material ? how would that affect the field between the plates, now we know that the dielectric itself affects the field between the plates , it decreases it's strength which is the reason more charge has to flow to the plates to set equilibrum condition.

 

[sophiecentaur]

Hi again. I thought you'd given up!

I'm afraid that I can't drag myself through all that stuff you just wrote because you are still missing the main point of all this. You need to supply all the energy for this scheme of yours in the system you will use to change the permittivity because none will come from the battery in your circuit diagram. There is nothing more to discuss until you accept that. If you can't realize how fundamentally flawed your idea is then you will get nowhere.
What is the point of trying to invent a hypothetical material to do a job that can't be done, your way, in any case?

 

[Mr.Bomzh]

Yes I haven't and I never will:D
I'm not attempting a PMM machine I think I made that clear in the beginning, yes ofcourse there will have to be some energy transfer involved to do the job.

Ok why not rather speak about the mechanical analogy, far easier there yet the basic principles apply.
I stand on a cement pad which is suspended in air by 4 springs.The springs are compressed because the weight is pushing on them constantly , so this is a steady situation , nothing changes , given amount of weight given amount of spring stifness.
Now there are two ways to achieve motion of the concrete slab , one way is to jump on it which would create oscillations up and down , by jumping I would apply force each time I hit the slab when I land on it.So the energy in this scenario comes from my jumping/muslces etc.
Another way would be as I said altering the stifness of the springs.In this situation ok I'll be honest it is kinda tricky to find out where the energy comes from.but it definitely comes from whatever jamms the springs or from the springs themselves depending on what kind of process or material comes into play here.
Iamgine springs that get like two times stiffer once you reach a certain compression level.Ofcourse if the slab comes down it presses on the springs which then store this gravitational potential energy the slab had before , now this energy stored in the springs could be released and with a little additional energy input could lift the slab to its previous height, is tha correct?

 

[sophiecentaur]

If you don't include a power source then you are proposing a PMM.
I understand exactly what you are saying about the equivalence of spring stiffness and permittivity but both of those things involve input of energy, when the spring / capacitor is stressed. If you do anything to a spring to increase its spring constant (to make it lift something) you will need just as much energy (more,probably, because of efficiency issues) as just lifting the object directly. Your "little additional energy" is a nonsense - it's the TOTAL energy that's needed. You are just trying to sneak in a PMM without realising it. You must realize there is no way round this one. If you stress a capacitor or spring and then use the stored energy to do a job, the next time round you need to supply the same amount of energy again.

 

[Mr.Bomzh]

nowhere did I said there shouldn't be a power source ofcourse there should.the only question is about how it would be connected and implemeted.
ok you say that once the slab or whatever falls and compresses the springs to lift it up would require as much energy input as the first time , but not necesserily so , because while the slab fell and pressed the springs it gave its energy to the springs , assuming some leakage ,and imperfections the next time you would only need to supply the missing force due to those imperfections as the stored energy would release itself lifting the slab, ofcourse I am talking about a situation where the slab is freefalling.If one would attach something to it to do work it would gain less energy and require more energy put into the springs to lift it each next time.
I guess this is the resonance principle , if a girl is thrown from a height H to a trampoline she hits the trampoline and is being thrown upwards by it , only a little less high than the heigh from which she started now putting in some additional energy she could reach the same exact height.I think I'm correct here.

I know that given a primary and a capacitor in series the batery cannot do work , well it can as long as the capacitor is charging and then it stops , a battery can do work to a circuit as long as it can make charge flow, since a capacitor blocks charge flow a battery can do no more work on the circuit.
You see Sophie the idea here isn't about so much whose doing the work , whether the battery when charging the cap or the dielectric when changing the capacitance to make the charges flow back.The idea here is about easier running conditions on the switches.Maybe I should have explained that from the very beginning to make my point better.
In high voltage dc applications a semiconductor switch in series with a transformer like in a half bridge or other topologies has the voltage drop across the switch and due to flux lag or other factors can have to work under very harsh conditions under which it can fail.
By having this special capacitor , I have the idea to achieve the same charge flow as in a directly point to point connect circuit , but only with the help of fields rather than direct current path. Ofcourse the charges still flow through the switch but atleast there isn't a PD across it, instead the PD is across the cap and I think the cap is pretty fine with that , even if the voltage is high.
i hope you get what I am talking about here. :)

 

[sophiecentaur]

So the energy conservation laws apply - but not totally? This is ridiculous. Whatever energy is stored in a spring or capacitor and then used to do work, has to be replaced - all of it. However you re-arrange the argument. Resonance has nothing to do with supplying energy.
You are still not acknowledging the basic flaw in your idea.
The last two paragraphs are just gobbledegook. Science does not work like that.

 

[Simon Bridge]

So we concluded after all these two thread talk that a capacitor in a circuit being in series with the pirmary could pump charges back and forth through the pirmary if it's capacity could be altered , the problem here ofcourse is how does one do that , not speaking about the rotors or any mechanical stuff but in a solid satte way.

No. That is not the problem!
It is a problem but not the main one from the POV of design.
The main one is finding anything that has not already achieved the aim without having to use the capacitor.

Sophie is correct: until you are prepared to acknowledge this basic flaw in the concept itself, you will make no further headway.

 

[Mr.Bomzh]

Oh again my wording problems.Ok I do believe that the energy stored in a capacitor then used to do work has to be refilled back from a power source or whatever kind of external source which is not part of the system of the capacitor and transformer.That's not the problem , never was.It's not about supplying energy , rather making that supplied DC energy ,if one can say like that, oscilate.
The question was about how does one alter the field between the plates.
Thsi is an idea and even if it's impossible theoretically or in practice or both that's not the most important part as I still learn a great deal about the circuit situation and the capacitor since i started this.
What exactly is as you said goobledigook about a capacitor that's charging or discharging used to do work on a inductive device like transformer?

Ok so basically so far I have understood that once the cap is charged up the energy that went into it is now captured there by the charges on the plates and in the electric field that has formed between the plates.
If someone would now want to alter that situation he would have to use some or the same amount of energy to change the dielectric constant and that energy would manifes itself as the change in capacitance and the result would be charge flow?
Ok let's take the mechanical capacitor for an analogy to understand better, say I have a mechanical varyable capacitor , like the old radio tuners.
Say this cap is 100uF and 100v rated.Now the cap is in a circuit situation and is charged up to its 100v and 100uF of charge.That means that the cap now has "x" amount of energy in it stored.If I wanted to drop it down to 10uF by turning it so the plate contact area decreases by 90%, how much work would it require to do so? Would it require 90% of the original amount of energy" x" that was stored in it?
In other words whenever changing the capacitance the amount of work needed to do that is equal to the amount of work you get out from the flowing charge due to the change in the capacitance?

 

[sophiecentaur]

Last sentence of that post is what I have been telling you in pretty well every post I (and others) have written here.That is the whole crux of the problem. Energy In is Energy Out.

 

[Mr.Bomzh]

but the funny part is that I agreed to that since the very first moment you told me that :) I guess I can get confusing really fast.the part that concers me in all this is just the mechanism how does one build a device that works in such a way , not the fact that you need to use neergy just the way one uses it, the specifics of the device.
So my question about the 90% work neede dis right , right? :D

 

[sophiecentaur]

100% ?

 

[Mr.Bomzh]

no i said 90% work input is required because i wanted to drop from 100uF to 10uF , so I don't have to input the whole energy stored in the cap but the amount which corresponds to the amount of change in capacitance right?

 

[sophiecentaur]

OK
But to what purpose?

 

[Simon Bridge]

Last sentence of that post is what I have been telling you in pretty well every post I (and others) have written here.That is the whole crux of the problem. Energy In is Energy Out.

OP has been repeatedly stating that too... I suspect the issue is a bit more than that - any mechanism that changes the capacitance, in the right way, using electricity, is already the kind of device he seeks to build with the capacitor - but without the capacitor.

The result of adding the capacitor is to make a weak version of the device he already has.
There's no point. He would be better to replace the capacitor with a length of wire.

He needs to drive the device using a different kind of energy for it to make any sense at all.
Even then I suspect it would only count as art.

I say: leave him to it.

 

[Mr.Bomzh]

this is the problem , I do undersatnd I am incorrect to some dregree about how this whole thing should work but you are just making a laugh about anyone who comes up with something he has been thinking about and which doesn't sound like a typical i gave you homework i can't find the right answer question.
I understand this gets boring to you guys.

You see Simon I wouldn't ask the question here if I would just use apiece of wire and a switch in series with a transformer primary, there is nothing new in that it's already been done and is still used in various power converters.
maybe i will just tell the original thought which started me wondering about this.In a typical circuit before the switch is closed it has all the PD across it, then when it closes the current suddenly rushes through it.
if you put a capacitor in series with a primary of a traffo and then put a switch in series from the negative plate of the cap to ground , you still have current flow through the switch when you close it but no PD across the switch, ofcourse there is an impulse through the primary winding when the switch is closed only as long as and as big as the charging current of the capacitor used.After that all things stop again , the system is in equilibrium.
This is as far as we all know it.from here on I was thinking how to open the switch and empty the capacitor so that it could be used one more time in a cycle like this , through a transformer to induce secondary voltage to do work.
Simply discharging the cap would be a waste of energy and the device itself.discharging it through the same transformer primary would require another switch only this time the switch would feel the PD across it so that renders my whole idea useless again.
Ah I guess I have to pray to some deity for you to understand what I'm really trying to accomplish here.
More like a long lasting power converter which is almoust fail safe.and by long lasing i mean really long lasting.
from the topologies i know of that are made and avaivable ZVS probably comes close to my idea.But its a little different.

 

[sophiecentaur]

We are not laughing at all. I think we are just getting out of patience with the way you are treating this - as if the serious aspects of Physics don't really count and you can make stuff work just when you fancy. We all accept that something along the lines you are suggesting would produce a measurable effect but that is all; it is a totally impractical idea. There are too many 'ifs' in your idea and there are better ways of transforming from one DC supply voltage to another.
I suggest you learn a bit more respect - not for us but for the subject. What you have been saying is only trivialising it.

 

[Mr.Bomzh]

Ok the idea itself aside , I want to ask some question to clear some misunderstandings.
Simon or Sophie one of you argued here several posts back that with a cpacitor nothing can happen because the attached battery cannot do any work on the system because charges cannot flow directly as capacitor blovks charge flow.But if one makes a changing electric field due to the discharge and charge of a capacitor in series wtih a trasnformer pirmary does that also cannot do work on the transformer in the way of induction , because as I know a varying electric field produces a varying magnetic field and that again induces an electric field. All we need is a varying field.
Now I'm asking this because I remembered a schematic of a circuit i built a few years back , it's a simple SMPS operated from a IC IR2153, a fairly basic IC used my many hobbyists.I built the circit and it worked , wasnt perfect nor stable under heavy load but did it's job. It has a capacitor in series with the transformer pirmary which is then connected to a middlepoint which is at half mains voltage according to the schematic.
The way I see the schematic is that the two mosfets , one at a time , charge up the cap through the primary and then when the other one switches on it reverses polarity and recharges the cap again.
But the basic idea is there cannot be any physical charge flow through the cap in this schematic , yet the carges flow back and forth from the plates of it. yes by a different mechanism than the one I was talking here but now I'm just curious about the idea of changing electric fields doing work.
I have attached the image of the schematic.

 

[sophiecentaur]

That circuit is total different from yours. You must be mis-interpreting the way that circuit functions. There is a DC path from the supply, through the transformer primary. That is the difference. You will not let this idea go, will you? I'm afraid you are on your own with this one.