# Thread: capacitor throught experiment

1. normally in a parallel plate capacitor we have two plates and a dielectric material inbetween the plates , the plate area and the dielectric permittivity influences the specific capacitance of each of the capaciors.

Now as forum members have said a third meal plate insulated by the same dielectric from the outer plates would act as if it is not there , if the thiord plate in the middle is not attached to anything , but since the third foil that is located between the outer plates it is now insulated with a dielectric at each side which makes the dielectric twice as big, assuming a conducting plate between two plates of a capcitor acts as a mirror for the charges can we say that the capacitance has doubled due to the fact that there are now two layers of dielectric in the cap instead of one ?

Also what would happen if we had a capacitor of for example 8 or any other number of plates , each seperated from the next and previous one by an insulating dielectric , and the two outermost plates charged , the middle plates would act as if they are not there right ? but if we would connect the middle plates alltogether electrically that would change the capacitance because when they are seperated , each of the layers iof dielectric counts but once they are connected they become like one plate , so when I have any number of plates in a capacitor and dielectrics inbetween to get the total capacitance i have to take into account each of the dielectric layers seperating each of the plates , even those in the middle that are not attached right ?

thank you , excuse me for the complicated writing style.

2.

3. Originally Posted by Crazymechanic
Now as forum members have said a third meal plate insulated by the same dielectric from the outer plates would act as if it is not there , if the thiord plate in the middle is not attached to anything , but since the third foil that is located between the outer plates it is now insulated with a dielectric at each side which makes the dielectric twice as big, assuming a conducting plate between two plates of a capcitor acts as a mirror for the charges can we say that the capacitance has doubled due to the fact that there are now two layers of dielectric in the cap instead of one ?
You don't compute capacitance by counting the number of dielectric layers.

If I take a simple parallel-plate capacitor and slice the dielectric in half, insert a conducting plate that is not connected to anything, the capacitance between the two outer plates remains unchanged (I believe I actually said as much in an earlier post, but perhaps not). You can think of it this way: The capacitance between one outer plate and the newly added plate will be twice the capacitance of the original arrangement (because the distance between the outer plate and the center plate is halved; capacitance is proportional to the ratio of plate area to plate separation). But you have its mirror image in series with it, so that doubling is halved, leaving you with no net change in capacitance.

Alternatively, you may note that the added central plate does not perturb the electric field. As long as it is inserted in parallel with the outer plates, it will always lie in a plane that is an equipotential slice. Since the field is unchanged, so is the capacitance.

Also what would happen if we had a capacitor of for example 8 or any other number of plates , each seperated from the next and previous one by an insulating dielectric , and the two outermost plates charged , the middle plates would act as if they are not there right ? but if we would connect the middle plates alltogether electrically that would change the capacitance because when they are seperated , each of the layers iof dielectric counts but once they are connected they become like one plate , so when I have any number of plates in a capacitor and dielectrics inbetween to get the total capacitance i have to take into account each of the dielectric layers seperating each of the plates , even those in the middle that are not attached right ?
I'm sorry; I read the above a couple of times, and couldn't quite figure out what you are trying to say. Would you try again, maybe with fewer words, more carefully chosen?

4. .

5. Eight isolated plates, 1/7 capacitance.
Short the middle six, 1/2 capacitance.
Capacitance is derived from plate area, dielectric constant, and dielectric thickness across which a field is applied.
What are you getting at so obliquely?

6. .

7. Originally Posted by Crazymechanic
Ok thanks for the reply , so in such a cap with many plates the thing from all the ones you mentioned changes is the thickness of the dielectric as whne all the middle plates stand bythemselves not connected to anything they each act as if they would not be there but each of the dielectric layers act as if it would add up to one big layer of dielectric between two plates and so the capacitance decreases right ?
You really should get in the habit of simply drawing your circuit. Label each pair of plates as a capacitance of value epsilon*A/d. Then use parallel and series capacitance formulas to compute the capacitance of any arrangement you want. Without doing a few exercises of this sort, you'll never see why what you're seeking is flawed here.

Now futher I wonder what happens whne we take such a capacitor with 8 plates for example , charge it up to some fixed value , the capacitor reaches a capacitance of say "1/7 X" .
The capacitance is a property of the geometry (which includes their interconnection) as I've said before. Therefore, capacitance doesn't depend on how you charge or discharge the plates. So the capacitor does not "reach" a capacitance of 1/7 X (or whatever the value is). The capacitance is that value all the time (unless you change the topology by reconnecting things or moving plates around). In this case, you haven't defined the connection of the plates, so I can't agree or disagree with your assertion that the capacitance of the 8-plate arrangement is 1/7 X.

We then short the middle plates
Ok, so you seem to be saying that the middle plates are not connected to anything above. In that case, the capacitance above should be X, not 1/7 X, where X is the capacitance between the two outermost plates with no intermediate plates (but dielectric is filling the whole space between).

so that they effectively become like one plate inbetween with only two layers of dielectric , the capacitance now rises to " 1/2 X" ,
If you are shorting the 6 central plates together, and computing the capacitance between the two remaining plates, the capacitance will be 3.5 X. The capacitance between one outer plate and the next plate will be 7X (because the spacing is 1/7). You have two such capacitances in series, so there's a halving of that 7X.

when this happens , we then switch off the connection between the inner plates so that they each become seperate again , what happens now? What happens to the charge on the outr plates and the capacitance itself as the capacitance should drop once again and return to 1/7 X ?
The capacitance goes back to X. Charge is conserved. V = Q/C, so the voltage rises by 3.5X. This behaviour is precisely what happens in the parametric amplifier I described to you at the beginning of this endeavour.

I am getting at something very unique here , just a bit of your hel would be great and I will explain myself later on.
You should do it now. "Hide and seek" is more than a bit irritating.

8. .

9. Originally Posted by Crazymechanic
thanks for the replies tk , I wonder why most of the smart guys always get this sense of " something's wrong" when a guy like me asks questions like these. I really do learn and understand a few of my own ideas and approches this way.Hope you don't get bothered by this.
The reason that one gets a sense that something is, if not wrong, at least not what it appears, is simply that you have explicitly said that you are building up to a reveal of some sort. That's somewhat irritating, and certainly counterproductive, because without knowing the context and overall goal of the interchange, those who reply cannot focus their explanations on the bits that would be most relevant.

10. .

11. Originally Posted by Crazymechanic
well i already answered sometime ago i guess but can do once more. I am interested in a dc circuit with a voltage source say a battery and a transformer but for a transformer to work it needs a changing flux which can only be generated by a time and or polarity varying current , signal etc.
So I was interested in ways to have this varying (time/polarity) current.
CM, I've already answered your question about this at least twice. You're not going to build a better switch this way. A magnetic amplifier is the best option if ruggedness is the primary requirement, but won't work at DC. A transformer doesn't, either.

In a varying capacitance capacitor the current would be amplitude varying but that too does the job.
What job?

Taht's why I asked the question about the inner plates and what happens when we connect them together etc, so if the capacitance changes in a capacitor that is in a circuit the charges are able to go aka drain from the plates so instead of rising voltage maybe we could see current flow to and from the outer plates that are atached to the cirucit?

Also would there be current flow in the inner switch that connecs or disconnects the inner plates to vary the capacitance ? for more details of this question please refer to my previous post.
You don't seem particularly interested in my answers, so I'm going to back out now. Perhaps others would be willing to engage with you.

12. I am interested in your answers just that sometimes i ask in my own style and sometimes maybe i ask twice if i dont get i the first way around.

Anyways could you aleast please answer the few last question i had , like when a switch would connect or disconnect the inner plates would there be current through the switch ?

I guess i have to ask you because noone else seems either interested or wise about this thread.

13. If I understand you correctly, you are attempting to design a crude inverter, so that you can use a transformer to step up the voltage from a battery. You are trying to do this by switching capacitors in and out to create a varying voltage or current.

There are a lot easier ways to make an inverter, some of which do involve mechanical switches. See here:
Power inverter - Wikipedia, the free encyclopedia

14. something in the lines of that Harold , it's just that I am really interested what happens in a capacitor when one does something like I proposed, it's not so much about building a crude inverter as understanding what happens in a capacitor of such desighn I proposed , sadly some of my question never got answered

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