# Maximum Charge on Designed Capacitor

1. Feb 15, 2013

### gvjt

I'll spare you all the details, but suppose for some reason (too complicated to get into here) I needed to design a capacitor with certain fixed parameters and only one adjustable aspect, and I wanted the design to maximize the total charge the device could store. According to my calculations, changing the adjustable parameter described below does not seem to make any difference to the answer, and I'm suspicious that I am making a error, hence my post:

Here are the fixed parameters:
- the plate area A
- the distance between the plates d
- the space between is filled with two dielectrics as follows:
i) Strontium Titanate, (K=300, V/m=8e6)
ii) dry air @ STP, (K=1, V/m=3e6)
- the StrontiumTi is in the middle with equal sized air gaps before each plate
- the air gaps are required in the design and must be at least 0.001d each.

The question becomes, what fraction of "d" should the thickness of the StrontiumTi be to maximize the total charge the capacitor can hold. The issue is that C increases with more StontiumTi, but the air gap has a higher voltage gradient and the maximum charging voltage falls as the air gap decreases. Since Q=CV, I suspect that Q is a constant in this case, but I wanted to set up the equations to prove this, and I'm having some trouble making sense of it all. Now don't ask why the air gap is required, that's an independent issue outside the scope of the problem and is simply a design constraint. If it could be eliminated, then clearly the entire dielectric could just be the StontiumTi and then this would give the ideal situation. And this is why I'm confused, because it seems as soon as we insist on the presence of the air gap, the amount of the other material doesn't seem to affect the total charge attainable. On the one hand, this result seems plausible because the voltage gradient is much lower in the StrontiumTi leaving the bulk of the charging voltage across the thin air gap limiting the charging voltage substantially. But on the other hand it seems fishy that things would change so suddenly and drastically due to the presence of this required air gap, and that's why I'm suspicious that something is amiss. If the air gap wasn't there, Vmax would become 8e6 X d and C would be a maximum giving Q=CV the highest possible value. However, once the air gap is introduced, Vmax drops substantially. As the air gap is increased, Vmax increases, but C falls accordingly, and Q seems to be constant as a result. In my attempts at working on this, I recognized that the voltage gradient (delta V / delta d) is 300 times lower in the StontiumTi than in the air, but that the voltage is the gradient X thickness of the layer. So I think I did all that part correctly.
-gt-

2. Feb 15, 2013

### DrZoidberg

Look at the electric flux density.
D=ϵE. If there is an air gap and you assume that the air in that gap breaks down at 3MV/m then you get 3e6 V/m * 8.85e-12 C/Vm = 2.655e-5 C/m^2
So approximately 26 µC per square meter. That would mean if the capacitor had an area of one m^2 it could hold a charge of 26µC independant of the thickness of the StontiumTi.
But that must indead be flawed since we know for a fact that the maximum charge increases by a lot if the width of the air gap approaches zero.
The solution is simple. The breakdown strength of an air gap depends on it's width. Due to the way air breaks down (electron avalanche http://en.wikipedia.org/wiki/Electron_avalanche) the maximum field strength the air gap can withstand becomes a lot higher than 3MV/m if the width of that gap is very small. A microscopically small air gap can easily withstand many hundreds of MV/m.
This is known as Paschen's law.
http://www.physics.nus.edu.sg/~L3000/Level3manuals/plasma physics.pdf
Plus even if the air should break down and the charge jumps over to the surface of the StontiumTi, the capacitor would still be charged. As Walter Lewin demonstrates here with a leyden jar.
And he explains it in detail in this video

Last edited: Feb 15, 2013
3. Feb 15, 2013

### gvjt

Let me check to make sure I'm understanding what you said:

So the 3e6V/m for air is only valid until the gap gets below some threshold whereupon it actually gets significantly higher - and this was the fact that I was missing?

Finally, when you said, "But that must indeed be flawed," what this refers to is ambiguous, so can you help me with interpreting it? i.e. was the flaw just not understanding that things become different as the gap begins to close up, or was the flaw the statement that the maximum charge (for larger gaps) is a constant? I think it was the former and not the latter, but I just want to be clear. Thanks a lot for the prompt reply! I really appreciate it!
-gt-

4. Feb 15, 2013

### DrZoidberg

What I said was - The statement that the maximum charge density remains constant at 26µC/m^2 must be flawed.
The equation for calculating the maximum voltage that a capacitor can hold is V = Eds * d where Eds is the breakdown strength of the dielectric. That equation however is ignoring the gap between the dielectric and the metal plates. But since that gap is usually microscopically small, it can be ignored safely since it's breakdown strength will be higher than that of the dielectric. And I guess that's the fact that you missed.

5. Feb 15, 2013

### gvjt

What I still can't tell from your reply is if the statement is flawed generally, or if you meant it was flawed only when the gap starts to become small. So let me rephrase: If the air gap is sufficiently large so that the breakdown voltage is indeed 3e6V/m, then should the maximum charge be independent of the thickness of the Strontium Titanate as my calculations seemed to show, or am I doing something wrong there?

So far, you've cleared up nicely for me the issue of what happens when the air gap becomes small, and why this means that things don't (as I thought) "change so suddenly and drastically." So I now understand that part of all of this. But I'm still not certain so far from what you explained what happens when the air gap is large. For that case, then is the maximum charge the cap can hold independent of the various ratios of thickness of the two dielectric materials?

In my calculations, I calculated the breakdown voltage of the capacitor by determining what voltage across the combination of dielectrics would reach the point where it would arc through the air. Note this is higher than the thickness of the air X 3eV/m, because there will be some drop across the other dielectric, but since K=300 there, the additional voltage will be proportional to the thickness of the SrTiO3, but inversely proportional to its K value ... so its not straightforward to get it right. The breakdown voltage of the device will be higher than the breakdown voltage of the air gap however. In any case, C is proportional to K(air) and to K(SrTiO3) and inversely proportional to d(air) and d(SrTiO3). This is the opposite relationship to that between the K's and d's and the breakdown voltage V. So as the V for the cap rises, the C falls and vice versa, so it seems from the equations that CV is constant (again ASSUMING the air gap is not approaching zero). Does this make sense?
-gt-

6. Feb 15, 2013

### gvjt

Oh, I should add that allowing the device to have a spark arc-ing across the air gap is not acceptable, so this is why I calculated Vmax for the device based on whatever maximum total voltage would be just below the amount that would cause a breakdown of either dielectric. I guess its always going to be the air in this case. So works out to somewhat higher than 3e6Xd(air) generally.

7. Feb 16, 2013

### DrZoidberg

If the air gap is large (1 mm already counts as large here) the maximum charge you can put on there without the air breaking down remains constant. I already mentioned before that D=ϵE in air. And D * Area gives you the charge of the capacitor. The distance between the plates is not even in those equations.
Did you watch this video?

It's relatively simple.

(ag = air gap, st = Strontium Titanate, d = thickness of the dielectric)
V = Vag + Vst
Vag = Eag*dag
Vst = Est*dst
Est = Eag/300

Now putting those equations together
V = Eag*dag + Est*dst
V = Eag*dag + (Eag/300)*dst
V = (dag + dst/300)*Eag
Eag = V/(dag + dst/300)

8. Feb 16, 2013

### gvjt

I finally did watch the video. Walter Lewin is fantastic and I've watched most of the Physics8.02 video lectures that he made on that site. I always quote bits and pieces from his videos for my students. (I'm actually a math/compsci professor with an engineering background and a PhD in neuroscience ... but I teach at a small university with no physics program, and have been the only one remotely qualified in our faculty to teach the single intro to physics course we have here). So I've depended on Walter Lewin's lectures to help "fill in the gaps" for me.

Anyway, the equations you used above were indeed the same as mine, although I described them as "not straightforward" while you described them as "relatively simple" :)

Also, I hadn't stopped to realize that the final answer depends only on the plate area and that the maximum charge depends only on A and not on d, but I completely see that now.

So, thanks to your generous dedication of time, I have a much clearer idea of what's going on here, and I really appreciate your contribution to the forum. Thanks a lot!

9. Feb 16, 2013

### DrZoidberg

Glad I could help. btw. Richard Mullers lectures are also really great for a physics intro course.

Last edited by a moderator: Sep 25, 2014