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Old 28th January 2008, 09:19 PM   (permalink)
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The way that it makes sense to me (Your mileage vary)...

If you leave everything else constant, and only change the dielectric, a higher value makes higher capacitance. No charge ever moves through the dielectric. But the more insulative the gap between the plates, the larger the potential to store the charge. The capacitance in Farads, C, is equal to the area in square meters multiplied by the dielectric constant, multiplied by the permittivity of free space (8.854x10^-12 Farads/Meter), multiplied by the area in square meters of the plates, divided by the distance between the plates in meters. The units of distance cancel out, leaving a number, and the Farads unit. So,

C=k * (8.854 * 10^-12) F/m * A m^2 / D m

By increasing the dielectric constant (k), you will increase the capacitance directly. Does that help? I'm not quite sure that I am quite grasping your question.

EDIT: I re-read your question. Moving the plates closer together increases the capacitance, as well as increasing the dielectric constant. Less distance between the plates makes it easier for the charges to interact. A higher-k dielectric makes for a higher potential to be stored before the dielectric breaks down. I think that I might be oversimplifying your question, but I think that you might be obfuscating the problem.

-Mike

Last edited by henrybot; 28th January 2008 at 09:36 PM.
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Old 28th January 2008, 09:39 PM   (permalink)
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Originally Posted by henrybot
but I think that you might be obfuscating the problem.
I don't know if I'm obfuscating the problem...

I did repost this same question earlier today with a slightly different slant (as a new thread).

Your post does clearly state how the change in the dielectric constant acts, in a quantitative way, but I still don't get a sense of "why" it works as it does.

I've used capacitors on a pretty regular basis all my life and have even made a few of them (rolling up tin foil and paper, etching plates in opposite sides of double-sided pc board, even made a capacitor bridge to measure fluid level once that used the variation of capacitance due to the change in the dielectric, etc.) and have often figured out the values from the formula in my little Radio Shack Electronic Data Handbook.

20 years ago I felt like I had a pretty good handle on it but, I guess I'm getting more introspective because I realize I really don't have a good feel for what's actually happening at all (and, I don'[t like it).
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Old 28th January 2008, 10:01 PM   (permalink)
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The difference between "free space" and a dielectric is that for free space charges on one plate act only on the charges on the other plate. With the dielectric in between the charges on the plate can act on the electrons of the dielectric, which are not normally free to move about. At least they are not free to move about at low voltages.

Now as the voltage increases to the levels of the energy gap in an insulator ( 6 eV ) electrons can potentially be ripped away from their atoms and create charge distributions that alter the fields and thus the capacitence for a capacitor with the same plate area and separation as the capacitor in free space. These charges are not mobile in the sense that they would be for a metal or a semiconductor, but you can move them around.

I don't know if this helps and I can't necessarily draw a picture, but the answer lies in this direction.

Have you read Feynman, Vol. II? I strongly suggest you take a stab at it.
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Last edited by Papabravo; 28th January 2008 at 10:05 PM.
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Old 28th January 2008, 10:37 PM   (permalink)
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Originally Posted by crashsite
Your post does clearly state how the change in the dielectric constant acts, in a quantitative way, but I still don't get a sense of "why" it works as it does.
All the positive charge is stored on one plate, and the negative charge stored on the other. Unlike charges attract, so they are desperately trying to balance to a neutral charge. If you increase the distance between the plates too far, it simply becomes an open circuit. If you bring the plates extremely close, the attraction between the unlike charges increases. On the other hand, if you keep the small distance constant and make the dielectric higher (more insulative), then it means you can store more charge on the plates without shorting through the insulator.

As far as what's happening on the sub-atomic level, I'm not sure (Not a physicist!! )

-Mike
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Old 29th January 2008, 02:05 AM   (permalink)
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Quote:
Originally Posted by crashsite
I don't know if I'm obfuscating the problem...

I did repost this same question earlier today with a slightly different slant (as a new thread).

Your post does clearly state how the change in the dielectric constant acts, in a quantitative way, but I still don't get a sense of "why" it works as it does.

I've used capacitors on a pretty regular basis all my life and have even made a few of them (rolling up tin foil and paper, etching plates in opposite sides of double-sided pc board, even made a capacitor bridge to measure fluid level once that used the variation of capacitance due to the change in the dielectric, etc.) and have often figured out the values from the formula in my little Radio Shack Electronic Data Handbook.

20 years ago I felt like I had a pretty good handle on it but, I guess I'm getting more introspective because I realize I really don't have a good feel for what's actually happening at all (and, I don'[t like it).

As I read through this post I can understand your questioning of the magic of dielectric constant and how it actual causes a change in the value of the total capacitance. While I can't answer it, maybe I can state the question in a simpler manner and see if anyone would like to chew on it some more.

QUESTION

Given two metal plates of a given fixed area and a fixed spacing, if the spacing is filled with two different materials, say mica and Teflon, during two separate tests with the same applied voltage, what is the cause, at the electron or atomic level, for the different materials to cause the total capacitance value to be different for the two insulating materials.

And saying the materials have different dielectric constants is not a sufficient answer

Lefty
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Old 29th January 2008, 05:06 AM   (permalink)
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Lefty-

For mica, C=7 * 8.85 * 10^-12 F/m * m^2 / m
For teflon, C=2 * 8.85 * 10^-12 F/m * m^2 / m

Since everything else but the dielectric constant is the same, the capacitance for mica will be 3.5 times greater-- the mica will support 3.5 times the charge imbalance before it breaks down and becomes a conductor.

It doesn't seem to me that anything on the atomic level other than the unlike forces attracting one another really matters. Am I oversimplifying? Probably.

-Mike
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Old 29th January 2008, 06:40 AM   (permalink)
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Originally Posted by Papabravo
...With the dielectric in between the charges on the plate can act on the electrons of the dielectric...

Have you read Feynman, Vol. II? I strongly suggest you take a stab at it.
You're talking to a guy who's outstanding accomplishment in mathematics consists of the distinction of flunking algebra I in high school more often than any other student! So, I really need to beat things down to super basic levels to understand them. Interestingly, the great theoretical physisits like, Einstein and Hawking are really pretty good at doing this along with their obviously exceptional math abilities.

Since more electrons can go into a capacitor with a higher dielectric constant insulator (all else being equal) there is obviously some mechanism by which the capacitor's capacity increases. I figure it must be either that the electrons go into the dielectric somehow or onto the plate somehow.

I considered that the electrons might push "deeper" into the plate and mused about what might happen if the plate was only one atom thick. Certainly the capacitor formula doesn't suggest that plate volume is a factor.

As for storing the charge in the dielectric, I kind of liked the concept that the electron orbits are distorted (does that violate quantum physics?). But, me "liking it" is hardly a testamonial to its accuracy. It seems that anything that would involve energy gaps in the electron orbits would add some sort of voltage-sensitive component to the capacitor formula (you mention 6ev). Perhaps the very definition of the dielectric constant for a material automatically addresses it, making it transparent to the tech and engineer who is designing and using the devices themselves?

I mention that because I also wonder if there are colors that are "illegal" in the universe. We know that there are a LOT of colors as electrons and photons in the elements exchange energies. But, there are only about 100 elements and definitely a finite number of possible energy states for the electrons. So, unless there is a "tolerance" in the gaps that allows colors to overlap sufficiently to allow a truly continuous spectrum, there would be illegal colors. I suppose the same tolerance effect could smooth out voltage steps in a dielectric material.

While lasers are monochromatic, single-color LEDs are not and I've wondered about how precise those energy gaps are for some time. Of course, that's really another topic that I may post some time.
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Old 30th January 2008, 12:23 AM   (permalink)
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It's why I suggested Feynmann, you can skip past the math and go straight to the explantions.
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Old 30th January 2008, 03:59 AM   (permalink)
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Quote:
Originally Posted by henrybot
Lefty-

For mica, C=7 * 8.85 * 10^-12 F/m * m^2 / m
For teflon, C=2 * 8.85 * 10^-12 F/m * m^2 / m

Since everything else but the dielectric constant is the same, the capacitance for mica will be 3.5 times greater-- the mica will support 3.5 times the charge imbalance before it breaks down and becomes a conductor.

It doesn't seem to me that anything on the atomic level other than the unlike forces attracting one another really matters. Am I oversimplifying? Probably.

-Mike
Again, with a constant fixed DC voltage and fixed plate area and spacing, two different insulating materials cause two different value of capacitance, why?

What is this thing called dielectric constant that causes the change in capacitance? The spacing didn't change, the applied voltage didn't change, nor the plate area. Something about the properties of the insulating material causes a difference in total stored charge. Obviously something electrical is interacting with the insulating material, what and how?

Lefty
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Old 30th January 2008, 05:29 AM   (permalink)
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Others are grappling with the dielectric question but, not always looking for the exact same data.

http://www.goiit.com/posts/list/elec...55-b-36640.htm

I did a search for Feynmann and discovered some references to lectures that seem to relate to the topic but, haven't found any specific text. If he has an understandable, non-math-laden explanation of how the dielectric "works" perhaps a synopsis of it could be posted here?
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Old 30th January 2008, 10:13 AM   (permalink)
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Quote:
Originally Posted by Leftyretro
The spacing didn't change, the applied voltage didn't change, nor the plate area. Something about the properties of the insulating material causes a difference in total stored charge. Obviously something electrical is interacting with the insulating material, what and how?

Lefty

It's not something about the insulating properties. It's the fact that it's an insulator. There aren't many free electrons in an insulator.

So capacitance is the is the ratio of stored charge to potential, right?
(C = Q/V)
So if you keep the voltage the same, but increase the amount of charge, capacitance goes up. One way to increase the charge is to have bigger plates. Another way is to bring the plates closer together. And another way is to make the dielectric a better insulator. You can store more charge without breaking down the insulator.

-Mike
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Old 30th January 2008, 12:46 PM   (permalink)
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Originally Posted by henrybot
You can store more charge without breaking down the insulator.
I'm not quite sure what you mean by "breaking down the insulator". The capacitor formula does not suggest that the capacitance of a capacitor has anything to do with the applied voltage and states that the mere construction of a capacitor with specified plate area, spacing and dielectric constant defines its capacity.

We're not having a problem with the plate area or spacing or the fact that the dielectric constant affects the capacity but, we're still missing the "how" of the dielectric.

The link I posted earlier seems to suggest that there's some sort of (electrostatic") attraction between the plates and dielectric that must be accounted for and that takes energy. Thinking along those lines might be a step in resolving the dielectric question but, it hardly explains it.
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Old 30th January 2008, 04:31 PM   (permalink)
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Originally Posted by crashsite
The link I posted earlier seems to suggest that there's some sort of (electrostatic") attraction between the plates and dielectric that must be accounted for and that takes energy. Thinking along those lines might be a step in resolving the dielectric question but, it hardly explains it.
It has been decades since I studied the physics of capacitors, so I have stayed out of this discussion. Maybe, however, I can help with the question of how the dielectric acts.

The capacitor equation as you all know is:

C = Q/V

Q = charge
V = potential difference (volts)
C = capacitance

There are other equalities to consider as well (See: FW Sears and MW Zemansky, University Physics, 3rd edition (yes, third ), 1964, pages 594-596).

(1) V = El and

(2) El = (1/ε) (Ql/A)

E = electric intensity or potential gradient
l = separation of the plates
A = area of plates
ε = dielectric constant

Substituting equation (2) in the capacitor equation gives:

C = ε A/l

Thus, if one took a charged air-gap capacitor and inserted something of higher dielectric constant between the plates, the voltage would decrease as the capacitance increased. That is the same effect one would get by reducing the plate separation.

So, now for a physical model. Assume the plates of a capacitor are 1 mm apart. The charge between the plates must go from + to zero to – in that distance. What if you can decrease the effective distance over which the charge goes from + to - ? According to equation (1), you would increase the capacitance and reduce the voltage. You can do that by inserting something with a greater ability to dissipate charge, i.e., it can dissipate charge over a shorter distance. That is basically what “dielectric strength” tells you. Materials with a high dielectric strength are able to dissipate charge over a shorter distance, i.e., they increase the attainable potential gradient.

I find it easiest to visualize the mechanism with solutions. Polar molecules have dipole moments. The moments are not necessarily fixed charges, nor does that mean they are electrical conductors. In fact, dielectrics are non-conductors. As an example, take a simple diatomic molecule such as carbon monoxide, CO, which has a dipole moment along its axis. Oxygen is more electronegative than carbon and can be represented as pulling electrons from the carbon giving the carbon a partial positive charge and the oxygen a partial negative charge. There are lots of similar examples. Elements to the right of carbon in the periodic table are more electronegative than carbon. Thus, polychlorinated biphenyls (PCBs), polybrominated biphenyls (PBBs), acetonitrile (CH3CN), and HCN are polar with the electrons being more localized on the electronegative atoms. They have high dielectric constants.

Now, do the following thought experiment. Consider a charged plate in a liquid dielectric. At the surface of the plate, the first layer of molecules will be lined up with opposite dipoles facing the plate. The next layer will repeat the process, but show more randomization due to thermal effects and entropy, and so forth. Eventually, the molecules will be completely random, and one could say the charge has been dissipated over whatever distance it took. That distance is shorter for compounds with high dielectric constants than it is for those with lower dielectric constants. Changing to a solid phase doesn’t really change what is happening, the alignments may not change as easily, but the forces on the dipoles and the mechanism for dissipation of the energy is the same.

If you look at a list of dielectric constants, you can explain the observed order by the chemistry just described. You may ask, why is Teflon just 2.1, if fluorine is so electronegative? The perfluorination makes the molecule symmetrical. For other examples, compare Plexiglas (acrylic, polar, non-symmetrical) with a value of 3.4 with polyethylene (non-polar) with a value of 2.25. Or, compare benzene, which is very symmetrical and has a value of 2.28 with isoprene, which is much less symmetrical and has a value of 6.7.

Sorry for the length of this comment. I hope it helps.

John
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Old 30th January 2008, 07:37 PM   (permalink)
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Wow, there's a lot to digest there. I'm no chemist but, I do know that the dipole thing is very common and even has a hand in something like soap where the electrical interaction between the lye and oils pulls the atoms into such an alighnment as to relase them from a pot or pan.

Your disertation is going to take some mulling but, if I'm reading it correctly, it takes electrons (charges) to pull the dipoles into alignment and a dielectric material that takes more electrons to do that has a higher dielectric constant. Since more electrons are needed, more will need to enter the capacitor to effect the dipole alignment resulting in a larger capacitor.

I guess it really takes an understanding of physics, atomic theory, chemistry and perhaps a little alchemy and astrology to truly understand the action and ignert bastids, such as myownself, aint gonna get it without a lot more thought and study. There doesn't seem to be a simple mechanical model (like the bucket for capacitors or springs for inductors).
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