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How would you calculate the capacitance of a capacitor when one plate is a plain flat metallic conductor but the other has increased surface area e.g. because it was enhanced it with activated carbon?

enter image description here

I think if you added the higher surface area layer to both sides of ceramic then both plates would have the same surface area and you could approximate the capacitance as if they were parallel plates right? And if you got ride of the carbon all together it would just be the parallel plate problem, but with a low capacitance because you lost most of the surface area.

So intuitively I would think the capacitance would be somewhere in between. But would it be closer to the high cap from putting the additive on both sides, or closer to the lower cap from removing it altogether, and how to calculate?

Edit: I think the carbon layer is just confusing the issue. So what if instead of increasing the surface area you just put a bunch of nano pillars on the one plate. Like this:

enter image description here

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  • $\begingroup$ Also does it matter what kind of capacitor we are talking about? Shouldn't they all be a function of surface area? Perhaps some people would like to contribute to that discussion down here.. thanks in advance. $\endgroup$
    – user273872
    Commented Sep 10, 2016 at 19:45

2 Answers 2

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Capacitors store energy in an electric field; the porous material being activated carbon, it is conductive, and one expects no internal electric field in a conductor. So, the capacitance will be a function only of the insulating part of your structure, the ceramic layer. Area of plates A, thickness T, and the ceramic's dielectric constant, k, go into the formula

Capacitance = k * A / T

For some kinds of capacitor, a conductive liquid (electrolyte) and oxide layer on a porous material (sintered tantalum, for instance) make an electric field in a very large area (the area of all the pores) and small thickness. I'm unsure how that can be applied to carbon, but 'activated' is the opposite of making a stable nonconducting layer ('passivated' is what you want).

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  • $\begingroup$ I didn't specify what kind of capacitor I was talking about, but I had just recently been reading about electrochemical capacitors. Specifically I read this: "Early electrochemical capacitors used two aluminum foils covered with activated carbon" from here: en.wikipedia.org/wiki/Supercapacitor#History which doesn't seem to jive with some of the things in your answer. I'm sure you want the carbon to be conductive because it acts to increase the surface area of the metal conductor. Perhaps my picture above is misleading there's a gap between the carbon and the metal (they're in contact). $\endgroup$
    – user273872
    Commented Sep 10, 2016 at 19:36
  • $\begingroup$ I just updated the picture to get rid of that gap.. $\endgroup$
    – user273872
    Commented Sep 10, 2016 at 19:41
  • $\begingroup$ The other element in a supercapacitor is the electrolyte; that is how the charge gets close to the sintered graphite. That electrolyte chemically reacts somehow in order to coat the exposed surfaces of the graphite with an insulating layer. For good connection to the electrolyte, a paper or perforated plastic film holds the graphite grains away from the metal (or second graphite) electrode. $\endgroup$
    – Whit3rd
    Commented Sep 11, 2016 at 1:30
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I found an answer:

Blockquote For asymmetric capacitors, the total capacitance can be taken as that of the electrode with the smaller capacitance (if C1 >> C2, then Ctotal ≈ C2). https://en.wikipedia.org/wiki/Supercapacitor#Capacitance_distribution

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