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Capacitance of a Fractal Surface

  1. Feb 16, 2013 #1
    I understand how to calculate the capacitance of an isolated conductor, say a sphere, and I know that the associated formula involves the surface area of the conductor. Does all of this make the assumption that the surface in question is convex, or at least "non-concave"? More specifically, what if the surface is extremely complex, like a fractal where the surface area might be extremely large, but parts of the surface are directly facing other parts or even hidden behind other portions and so on. Is it still the outright surface area that matters, or do the bits and pieces of the surface have to have external normals that point out into space (if you know what I mean). Put another way, if I was to fashion an extremely convoluted surface, could I wind up with an extremely high capacitance for the volume the object is occupying? Or perhaps here's a more concrete example: What if I fashioned an object consisting of hundreds of thin metal plates all connected electrically at their centres by a rod, but with an air gap between each. A one meter cube so built could have a surface area of 500 square meters, and one might even argue that that is an "exposed" surface area, albeit primarily perpendicular to the outside world. Would this just calculate out as 6 m^2 because its basically a six sided cube, or would it in fact behave according to its, say, 500 square meter "surface"?
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  3. Feb 16, 2013 #2

    Simon Bridge

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    Nope - you can make a capacitor out of any shaped materials - consider two bi-concave lenses facing each other.
    It will depend on the details of how the surface is arranged. You'll have to do the math.
    Most complicated surfaces can be understood in terms of a network of regular capacitors.

    Yes you can. The simplest example, I think, would be wrapping a regular parallel plate cap into a spiral.
    A network of capacitors, in other words.
    How do you have to join two capacitors together to increase the overall capacitance?

    But I think you should be using, as your basic model of a capacitor, something involving paired conductors ... one providing charge and the other receiving it. The concept of capacitance does not really make sense if there is only one conductor.
  4. Feb 16, 2013 #3

    Vanadium 50

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    I'm pretty sure that a capacitor with fractal shapes will have zero capacitance. Doesn't that mean you will have arbitrarily sharp edges, which means you'll have an arbitrarily large field near that edge, and thus have breakdown as soon as any voltage is applied.
  5. Feb 16, 2013 #4
    You said, "The concept of capacitance does not really make sense if there is only one conductor," however it was in that context that I was asking the question. I'm actually pretty comfortable and familiar with real electronic circuit capacitors and their "two plates" and how they work. But what I was asking about was situations where you have just a single object and you want to know how much charge it can hold at a given potential. For example, a VanDegraff generator with a spherical top will have some particular charge (in microcoulombs) at some particular potential (in killovolts) for some particular sphere radius. Now I'm a bit fuzzy on the following statement, but here it is: I think you pretend the other plate is another sphere of infinite radius at an infinite distance, and it becomes a math problem with some value that you take the limit as it tends to infinity or some such approach.

    In practice, I believe such measurements are with respect to ground which we assume is at a neutral charge and zero potential. So when there is 350kV wrt ground on the sphere of a VanDegraff generator, we can calculate not only the "capacitance" of the sphere, but can use Q=CV to solve for the charge on the sphere. And I've seen such "solved problems" in many text books, so I know the idea of calculating the capacitance of a free object is not unusual.

    Now I know traditionally capacitors were constructed as long strips of almost a sort of tinfoil and wax paper and were all rolled up to cram a huge area into a small volume, and I'm good with that, however when you do that, the surfaces of the plates remain parallel to one another (with the advantage that BOTH sides of BOTH plates are "looking" at each other thereby doubling the capacitance). If you "fractalled" them all up in a convoluted mess, but ensured that the surfaces were not allowed to cross the original boundary, would the capacitance change? i.e. the wax paper must remain stretched tight as original, but the surfaces of the plates can have ruts and textures and places where the material opens into huge "caverns" with tiny openings, etc. etc. I'm not certain how I'd model that. But I have a suspicion that it wouldn't make more "capacitance".

    Let me suggest a particular configuration as an example to make my point. Lets say we have a two (identical) plate capacitor, where each plate is described as follows: Each plate is a thin rectangle, but the inside has been "hollowed out." Clearly this fact would be immaterial if that were the end of it all, and the capacitance would be calculated the same as if the plate was solid. But now, lets say that we cut a sliver out in the surface of the plates so that the hollowed interior is "accessible" and each plate can "see inside" the hollowed portion of the other through this "slit". I have a lot of trouble seeing that suddenly the capacitance will now double or triple now, even though we now have the original surface, the back of the surface, and the inside surface of the opposite side, now all "accessible" because of the slit. Clearly the surface area is four times more than the area of the facing side of the plate, and three quarters of this surface area now all "face towards" the other plate. One quarter of it is on the opposite side and everyone would agree it remains excluded from discussion just as when the plate is solid.

    Now when I look at this, I tend to think that because all this new surface area is not really "line of sight" to the other plate, it doesn't really count. Its a hollow, cavernous region inside the plate with a little slit leading to the outside world.

    So that's in the context of a two plate capacitor. I could be wrong, but I think it makes no difference. However, in the context of a single free object in space that we are calculating the charge on, I'm not as convinced, because it DOES represent an area that free charge can occupy, because even if its "inside", its still "on the surface" ... i.e. the surface is still a surface, even if you have to go through a little slit to reach it. But I really don't know, which is why I'm posting this ...
    Last edited: Feb 16, 2013
  6. Feb 16, 2013 #5
    Lets say that the sharp edges have been "rounded" somehow. My point is more that we have a huge surface area and I described it using the term "fractal" because that concept allows us to create a huge surface area in a tiny volume, which is what I really wanted to emphasize.
  7. Feb 16, 2013 #6

    Simon Bridge

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    Please show the derivation for the capacitance of a spherical conductor by itself.
    Perhaps you are thinking of self capacitance ?
    Last edited: Feb 16, 2013
  8. Feb 16, 2013 #7
    If the sphere of a VDG was made of activated carbon I'm pretty sure it would have the same capacitance as a polished sphere made of metal.
    You could just write a program that simulates the behavior of electrons in metal. Just a very simple simulation that works basically like a game where every object is moved a little bit in each iteration of the game loop. Such a simulation would only need Coulomb's law to calculate the forces between the electrons and it would allow the electrons to move within arbitrarily shaped boundaries which represent the surface of the metal. Just put e.g. 1000 electrons into the simulation and let it run.
  9. Feb 17, 2013 #8
    You can google "capacitance of an isolated sphere" and "charge on an isolated sphere" for all kinds of examples, but here one I've lifted from the bunch because it actually states, "Note there is only one surface, but the formula still works" which emphasizes that there's no second plate. This example goes like this:

    If you have been told the voltage on the generator, you can compute the charge using the above formulas, however as you can see, if the voltage and charge are unknown, they cancel out when you substitute the equation for V into that for C. In other words, the value C is independent of the current charge and potential on the sphere. If you have been given the voltage (as you will find in many example hits) then you can calculate the charge as Q=CV.

    When I was first introduced to this idea of the capacitance of an isolated body, it was decades after I had been working as an electronic designer (and I have two patents) where I always thought of capacitors as the "two plate" variety, and so I had considerable trouble initially understanding the concept of the capacitance of an isolated body in space. However it does make sense when you think of the following example: Suppose you have a signal generator with its ground lead connected to ground and its hot lead connected to an isolated coin such as a dime. If you connect an oscilloscope up from ground to the coin clearly you will read the signal. This shows that the coin must have a potential that "follows" the signal generator wrt ground. However the coin is small, then the signal generator doesn't have to supply much current to "charge" and "discharge" the coin. But what if you connect the signal generator up to a huge metal object like a metal ship that's in dry dock and isolated from ground. If you connect the signal generator up to one side of the ship and run over to the opposite side with your oscilloscope, the only way you still still see the test signal is if the signal generator can supply sufficient current to "charge up" and "charge down, i.e. discharge" the body of the entire ship. Intuitively, you can see that the ship is huge and a lot of electrons need to be pushed onto and pulled off of the surface in order that the scope will actually show the test voltage. Therefore, the large ship represents a "capacitive load" to the signal generator, and in fact its a capacitive load whose value could be approximated as a huge metal sphere. If the ship is too large, it will have too much capacitance, and the signal generator will not be able to provide sufficient current to charge and discharge the ship, and the scope will not show the full amplitude of the signal. In practice, as an electronic designer I would never have considered such issues, because my circuits use small short wires to connect points to other points and the capacitance of those isolated wires are so negligible in most applications (except in high frequency microwave circuits) that we simply ignore this effect. But you can certainly see that its a completely different ball game if one of your circuit wires is the size of a ship! You can no longer just view it as a conductive path with zero impedance in that case.

    Of course you don't need such an elaborate set up to test this. It will do to connect a high resistor with a high frequency sine wave. Monitor the other end of the resistor with a scope, and then have someone electrically isolated from all the equipment touch that end of the resistor. The amplitude of the signal on the scope will drop significantly. The human body capacitance to a far ground is at least 100-200 pF. If you use a 100K resistor and set the signal generator for 10KHz. If you set the amplitude to 1vpp, then with the human touching the wire (and remember they are otherwise isolated from ground and from the circuit), the amplitude of the signal as measured by the scope will drop to about 60% of its original value. If you repeat the experiment with a small child, it will only drop to about 80%. And if you repeat it using a small coin, the amplitude will not change noticeably. You can try different objects of different sizes to see the effect. In all cases, the objects act as a capacitor to ground and have the exact same effect as though you really connected up a two plate capacitor of that value between the end of the resistor and ground. I know its weird, but I've done the experiment, so I know its true.
    Last edited: Feb 17, 2013
  10. Feb 17, 2013 #9
    Ironically, this is EXACTLY the scenario that I was wondering about - it was the inspiration for my post. So you've hit the nail right on the head. My suspicions are the same as yours, that even though the activated carbon has a surface area orders of magnitude larger than the polished metal sphere, that somehow only the "externally directly visible" parts of the carbon matter, and the surface area by that measure is identical to the polished sphere, and they will have the same capacitance. But this is just intuition, and I'm not even sure if its right. Hence, my post ...

    I'm not certain how to simulate it, but having also had decades of experience as a programmer, I think its easier to say it than to do :)
  11. Feb 17, 2013 #10

    Simon Bridge

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    @gvjt: The quoted (but not cited) example is for the "self capacitance" - potential is being measured from the "other plate" which is placed at "infinity" - the defined zero for PD.

    Area does not, directly, form part of the equation ... though, the surface area of the sphere is related to it's radius. It will be a lot harder to come up with an equivalent fractal relation (between area and some radius) so drawing an analogy would be difficult... as you have seen.

    You'd need, instead, to start from the definition of self-capacitance.
    I'll stand by my earlier comments. You'll see when you start doing the math.
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