Imagine two half-domes with an air-gap distance between their flats, of any distance. These half-
domes will be filled with crumpled metal foil to increase their inner surface area many times.

My question is, will this give the capacitor greater capacitance, even if the dielectric area is relatively
small in comparison?

 

Not sure what you are saying, but capacitance is a function of the surface area of the conductors to each other and the distance between them. The capacitance goes up as they get closer, but the break over voltage, where the dielectric fails, goes down.

Most older foil / wax paper capacitors use multiple layer to increase the surface area.

 

The capacitance would be the essentially the same if two very thin foils face each other, or if two half-domes face each, regardless of what the half-domes are filled with, air, vacuum or otherwise.

Only the opposed areas, the gap, and what the gap is filled with, matter...

Wiki

 

The capacitance would be the essentially the same if two very thin foils face each other, or if two half-domes face each, regardless of what the half-domes are filled with, air, vacuum or otherwise.

Only the opposed areas, the gap, and what the gap is filled with, matter...

Wiki

OK, maybe I didn't explain myself well so the following typed symbols will hopefully serve. This is one half of a capacitor -(| and this is the other half |)- ...together they make -(| |)- ...the distance between the two vertical lines || is dielectric and the space between the hemispheres and vertical lines -(| & |)- is filled with crumpled foil. So the plates have far more surface area, thereby holding more charge. The dielectric space is relatively small in comparison, however. Please tell me why this won't hold more charge than say a mere flat plate cap -||-

 

None of the material within the half dome plays any part in the workings of the capacitor, whether it is solid copper, crumpled aluminium or strawberry ice cream.
It is all 'shielded' from the dielectric by the skin of the dome.

Nor does the core material make any difference to the charge stored that depends upon two factors.
The applied voltage and the capacitance.
The capacitance in turn depends upon the facing separation and the facing surface area, for a given dielectric.

That is why we have multil leaved air spaced capacitors.

 

It is not a crumpled surface area issue, it is a surface area and distance issue. Anything behind the suface layer does not count. You could reduce capacitance this way though, since it is not a smooth service.

 

I can not figure out your capacitor construction.
Is English your native language?
Can you draw a diagram?
And do you understand how a capacitor works?

 

are the "flat" sides covered in metal? or are the two halvs hollow, with no covering ? if there is a metal cover on the "flat" side, then that is the plate of the cap and nothing inside matters at all. only the surface area of the sides closest to each other matter. likewise, two balls used to make a capacitor have only the closest parts as plates, the inside isnt anything to do with the capacitance. thickness of plates does not matter either.

 

In the making of super-capacitors, surface area is a factor of total capacitance. That's why I suggest crumpled foil to fill interstitial space, or perhaps say, a solid block of aluminum with glass beads added to the melt, to increase internal surface area, the two halves separated by a dielectric. Surely a larger surface area will hold more charge than a smaller one, even if the dielectric area between them is relatively small. If this is not true, please help me understand why.

 

Surely a larger surface area will hold more charge than a smaller one,

You're the one making this outrageous claim against current scientific thinking and what others have told you here.

So you are the one who needs to say why.

 

the capacitance is in the area between the plates. making a rough surface will only decrease the capacitance. the mass of the plates makes no difference at all. its like the strain is only between the plates, not in them.

 

I think you are confusing some different surface area concepts. What you have in mind is a notion akin to increasing the surface area of something, such as a sponge or a towel or the granules of a powder or some other surface, in order to expose more total surface area to a chemical reaction. But here is it a bounding surface problem.

There is not need to use hemispheres or any other complex shape. Imagine taking two pieces of cardboard (i.e., a non-conductor) and gluing a thin sheet of aluminum foil to them (i.e., the capacitor plates) and separating them by a certain distance. You would end up with a certain capacitance. Now imagine repeating this with another two pieces of cardboard but first cut lots of small grooves in it and put the aluminum foil on the surface so that it follows the ins and outs of the grooves. Then you still make the capacitor by separating the two plates by the same amount (perhaps as measured by the distance between the backs of the cardboard pieces since nothing was done to this surface). Would you expect the capacitance to go up or go down? You would expect it to go down because the effective separation has increased. The bounding surface is the one that goes across where the tips of the aluminum foil are and the capacitance will always be less than this. Conversely, a bounding surface that goes across where the vallyes of the foil are will set a lower limit on the capacitance.

Now, if you carefully match the V-shaped profile of the two sides so that the hills in one decent into the valleys of the other, then you can get higher capacitance (and note that the upper boundaries I described earlier don't apply since they would have crossed through each other in this process).

So it is the surface area of the bounding surfaces on the two plates and not the surface area of the each plate by itself that matters.

 

WBahn;

Thank you, you seem like one of the few people here who understand the nature of my question. What you say is conventionally correct, the sponge analogy is a good one. My point is that the entire metallic sponge's surface area is charged, (or say, a brillo-pad.) The pad should hold more charge than say, a mylar coin-purse of similar size. Never mind the dielectric or distance, if one pad is charged positive and the other negative, two wires connected to each and brought together, will equalize, and the larger surface area pads should pass comparatively more current because of the greater charge on said greater surface area. Is charge only a surface phenomenon? I would expect an accordion fold of plates to hold more charge than a box big enough to contain it, because the accordion folds have comparatively more surface area than the box. Whether a second set of accordion folds is in close proximity to the first is immaterial to the discussion IMO, because the charge could be allowed to discharge to ground, through a wire. In the same way a large foil sphere should be able to be crumpled up to a smaller size and have the same capacitive surface area while taking up less space.

 

Where your reasoning fails is that what counts isn't the surface area of the plate, but the surface area of the plate that is normal to the electric field. Perhaps a better analogy would be to consider the amount of energy that is imparted to a surface exposed to the sun and then try to reason that if we used a crumpled up surface that had a lot more surface area that we could somehow extract even more energy. But we can't, because it's the surface area that is normal to the sun that matters.

 

WBahn's sunshine analogy (normal surfaces) is a very good one.

But to say this

you seem like one of the few people here who understand the nature of my question

and this in the same breath

I would expect an accordion fold of plates to hold more charge

is insulting to all those who have tried to tell you that this is a false expectation that you have no reason to hold.

By refusing to answer the many people who have told you this is wrong, you have denied yourself the opportunity to find out the truth and wasted their time and effort as well.

For the last time, a capacitor holds the amount of charge you put on it (within the limits of destruction of the capacitor).

It does not have the ability to 'hold' a particular amount of charge.




If you want to know how we increase the capacitance per unit volume, all you have to do is ask

 

WBahn's sunshine analogy (normal surfaces) is a very good one.

But to say this



and this in the same breath



is insulting to all those who have tried to tell you that this is a false expectation that you have no reason to hold.

By refusing to answer the many people who have told you this is wrong, you have denied yourself the opportunity to find out the truth and wasted their time and effort as well.

For the last time, a capacitor holds the amount of charge you put on it (within the limits of destruction of the capacitor).

It does not have the ability to 'hold' a particular amount of charge.




If you want to know how we increase the capacitance per unit volume, all you have to do is ask

 

studiot:

Mine was a theoretical question as the subject stated, not a question about conventional capacitors or how they work, so if you want to feel insulted for not reading that, feel free. There is no being wrong in a search for understanding. Incidentally I'm still not satisfied because the explanations I've received haven't fully covered the theoretical model I've been proposing yet.

A mesh or weir of many layers has greater surface area than a hollow object of same size. Are you or others saying that the entire weir is not charged if/when a charge is applied? The charge can discharge to any lesser charged object via a wire, including ground, or a similar weir structure, uncharged or oppositely charged. I'm trying to understand relative volume of charge differences in the structure of non-conventional capacitive "plates." Any metallic object can be one half of a capacitor. Clouds are a good example, and they aren't even metal.

 

You have been asked by several members to clarify your physical arrangement.
If all the 'domes' shown are metal,


(0) Offers the maximum possible capacitance.

(1) Adding in crumpled foil makes absolutely no difference to the capacitance over (0).

(2) If your dome is open at the bottom then adding crumpled foil actually decreases the capacitance over (0) or (1) but increases it slightly over (3).

(3) An open bottomed dome with no foil offers the minimum capacitance.

Do you understand why this is so?
All the information needed has been provided to explain the differences and even calculate the capacitances.

 

In your crumpled plate situation, you will NOT have uniform charge distribution. The charge will become distributed along just the "points" of the surface that are nearest the other plate of the capacitor. This distribution will result as a consequence of the fact that there can't be any non-normal component of the (static) electric field at the surface of a conductor. If there is, then the charges will move in response until an arrangement is achieved that cancels it out.

 

In your crumpled plate situation, you will NOT have uniform charge distribution. The charge will become distributed along just the "points" of the surface that are nearest the other plate of the capacitor. This distribution will result as a consequence of the fact that there can't be any non-normal component of the (static) electric field at the surface of a conductor. If there is, then the charges will move in response until an arrangement is achieved that cancels it out.

What do you mean there can't be any "non-normal?" So forget a second plate for the time being, just assume a big metallic charged metal weir or sponge, floating in space. You said it yourself, the charge distribution spreads itself out. Forget about dielectric fields for the moment. If the weir is charged to say, 10kV then every part of the weir is so charged. A wire from the weir can convey that charge to any object it touches of lesser charge. My point was a weir of greater size/surface area will hold greater total volume of charge than a smaller one, just like a big cloud can hold more lightning than a little one. This would appear obvious.