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As a student, learning about a capacitor after understanding what a resistor is, it was quite surprising to note that the capacitance does not depend on the nature of the plates used, at least in any type of capacitor I have known.

I am guided, "it makes no difference as long as the plates are conducting." Is that true?

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    \$\begingroup\$ Yes, it makes no difference as long as the plates are able to store charge. The differentiating factor is the area of the plate (More charge) to increase the capacitance and the separation between the two plates is a limiting factor. If you are new, I HIGHLY recommend you read the ENTIRE wikipedia page about capacitors and try to really think about why the equations are formulated as they are. Look at each parameter and try to relate them to each other. \$\endgroup\$ – Sorenp Aug 5 '19 at 11:22
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    \$\begingroup\$ What you are learning about is an ideal capacitor, made from a material with zero electrical resistance. Of course such a thing doesn't exist, but if the resistance is small, it is a pretty good approximation. In real-world applications capacitors are affected by their electrical resistance (even if they are made of good conductors like metal sheets, the sheets are very thin) and can be considered as an ideal capacitor in series with a resistor. The effective value of the resistor is specified in the capacitor's data sheet, and for some uses of capacitors its value is very important. \$\endgroup\$ – alephzero Aug 5 '19 at 19:58
  • \$\begingroup\$ Think of capacitance as a surface effect in the adjacent (conducting) surfaces; and which is dependent on the (dielectric) material in between these surface charge layers (as this affects how strongly one layer sees the other) - then nothing else about the conductor matters. \$\endgroup\$ – FlatEarther Aug 6 '19 at 13:33
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    \$\begingroup\$ As you gain experience you'll get a better feel for the limits of the idealized models of circuit elements. For example, while an RC circuit may in theory work just as well with a 1-ohm R and 1-mF C, you'll develop an intuition that it's better to use a 1-kohm R and 1-uF C. \$\endgroup\$ – MooseBoys Aug 6 '19 at 15:36
  • \$\begingroup\$ in design of capacitors on silicon, the ohms/square of the plates becomes a big deal for transient behavior, and the design of the metal-polysilicon contact patterns is one degree-of-freedom for the engineer. I used the contacts as part of dampening for LC power/VDD cleanliness. \$\endgroup\$ – analogsystemsrf Aug 7 '19 at 6:22
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Yes that is true, capacitance is:

\$C = \frac q V\$

where q is the charge and V the voltage between the plates.

As long as the charge \$q\$ can be "hold in place" this relation applies. I mean, there is no need to have a "good" conductor as the charge is static, it does not move.

So as long as for a certain voltage \$V\$ is applied resulting in a certain charge \$q\$ to be present on the capacitor's plates then \$C\$ can be determined.

It does not matter if the plates are bad conductors (high resistance) as it will then simply take longer for all charge to reach its final location. In the final state there will be no difference compared to a capacitor with well conducting plates as the amount of charge will be the same.

Only if you look at the dynamic behavior of a capacitor (how does it respond to quick voltage changes) would you see an influence of the conductivity of the plates. In first order the capacitor would exhibit additional series resistance.

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    \$\begingroup\$ please clearly state that you are talking about a theoretical model which has its "acceptance range" but its limits as well. \$\endgroup\$ – Christian B. Aug 5 '19 at 12:23
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    \$\begingroup\$ @ChristianB. If you take "everything" into account then sure, I get your point (and your answer as well). However when learning new things it is in my opinion much easier to just simplify things and only look at the "first order" phenomena. That way you don't confuse those who are new to the subject, like OP who just wants to understand the basic concept. When my "limited and simplified world" answer is understood it becomes easier to go deeper and consider your answer. \$\endgroup\$ – Bimpelrekkie Aug 5 '19 at 12:36
  • \$\begingroup\$ I am fine with simplification and modelling as long as one clearifies it is a model after all. We already have to many ppl who "believe" that models are the reality. This can easily leads to religious like behavior and hinder progress (see history of relativity theory and quantum physics) on the one hand and even lose of trust in scientific knowledge if ppl realize that they found a "logic hole" in one of the theories although they might just "reached" the limitation of a model (compare the increasing amount of flat earthers). \$\endgroup\$ – Christian B. Aug 5 '19 at 18:07
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    \$\begingroup\$ Please dont get me wrong. I am totally fine with simple models but one has to make sure that they are perceived as that. Especially if someone ask if a model is the whole story like OP did. \$\endgroup\$ – Christian B. Aug 5 '19 at 18:08
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The active part of a capacitor is the dielectric. That's where the energy is stored, that's what the voltage is developed across. The plates just transport current to the right places. A high resistance here could make the capacitor lossy, but will not change the capacitance.

In much the same way, the resistance of a resistor depends on the material and geometry of the resistive part, not the leads.

The active part of an inductor is the iron, ferrite or air-space within the coils, because that's where the energy is stored. High resistance wires will make the inductor lossy, but won't change the inductance.

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    \$\begingroup\$ This should be the accepted answer! \$\endgroup\$ – nigel222 Aug 8 '19 at 14:49
  • \$\begingroup\$ Yes, without a doubt, this is currently the best answer here. \$\endgroup\$ – Dawood ibn Kareem Aug 9 '19 at 4:21
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Typical capacitor plates are made of conductors (metals) which have a huge number of charge carriers. Consider that (very roughly) \$N_A = 6×10^{23}\$, while \$C = 6×10^{18}e\$, so 1 mol of metal has enough charge carriers for 100000 C, assuming one mobile electron per atom. In a capacitor of 1000μF at 100V with Aluminium plates, only 27μg of Aluminium atoms have to donate/accept a single electron to hold the charge, the rest of the atoms stay neutral. Assuming the plates weight 5g, that's 99,9995% of neutral atoms plus 0,0005% of atoms missing one electron. Clearly, a typical capacitor will fail due to breakdown long before the lack of charge carriers in the plates will become apparent.

Things change in semiconductors, where the amount of free carriers is much smaller and depends on the doping. Even then, it's often easier to calculate the capacitance as a static approximation, assuming that the plates stay perfectly conductive and only the distance between them changes as the depletion region grows. It's not always possible though: in fast dynamic processes junction capacitance can only be adequately described using equations for charge flow (e.g. this one), and the solutions indeed depend on the material of the plates.

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To the best of my knowledge the chose of material DOES matter - even for the static case. If not it would imply that most insulators could be used as electrode as well due to the residual chance of existance of charge carriers within it. Some reasoning and scientific works why choice of electrode materials matter: DOI: 10.1109/16.753713 and doi.org/10.1063/1.1713297 to name just a few. Thing is that the models you learn are a good approximation. Not more not less. Main reason why the electrode material matters is that the EM field reachs into conductors as well even for static case.

LT;DR know your model's limits: It does matter but can be often neglected.

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  • \$\begingroup\$ Not for the static capacitance, it doesn't. \$\endgroup\$ – Carl Witthoft Aug 6 '19 at 13:24
  • \$\begingroup\$ great. let's start a typical flat earther discussion: it matters, no it does not, yes it does, etc. There is your scientific sound proof or source? If you actually do a little research you will find a lot of hints that models are models and "good conductor means equipotential everywhere" is good but not perfect assumption: en.wikipedia.org/wiki/Electric-field_screening physics.stackexchange.com/questions/14927/… tf.uni-kiel.de/matwis/amat/elmat_en/kap_2/backbone/r2_4_2.html \$\endgroup\$ – Christian B. Aug 6 '19 at 17:03
  • \$\begingroup\$ But lets do a little thought experiments: lets assume that the material does NOT matter at all. Implication would be that non perfect isolators would act as electrode material as well giving us a capacitance approaching infinity as the distance d would be very small (or even not existant?). So you are right. This is mostlikely a silly assumption. So lets say what only "good" conductors behave like perfect electrodes. But what is the critical value then? 10^6 S/m? What if you modify the material? Would we see an "on-off" behaviour? If there is evidence for such I would be happy to see it. \$\endgroup\$ – Christian B. Aug 6 '19 at 17:10
  • \$\begingroup\$ Good physics, bad engineering. Remember that the plate material also has a certain dielectric constant if it is a bad conductor, and even if it is a good conductor. So any penetration of the electric field into the plate material would cause the capacitance to depend on that dielectric constant. But not very much. \$\endgroup\$ – richard1941 Aug 9 '19 at 2:15
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It's the same for an inductor - the value of inductance remains constant irrespective of the wire's conductivity. Take it to extremes and consider the speed of radio waves and how they propagate through space.

The impedance of free space is determined by the permeability and the permittivity of free space and these are measured in henries per metre and farads per meter respectively. Yet there are no conductors in free space.

enter image description here

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In a typical capacitor, charges will be concentrated in thin layers on the portions of each electrode that are nearest the oppositely-charged electrode. Although this layer essentially always has non-zero thickness, and the distance between each charged particle and the surface will affect the potential difference resulting from a that charge, in practice the effect is almost always small enough to be dwarfed by measurement uncertainties or other confounding effects.

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Many practical capacitors have very weak dependence on the conductor material. Capacitor Equivalent Series Resistance (ESR) will be affected by plate material and thickness/routing and is a significant limiting factor in power applications. This also affects peak discharge currents for pulsed applications.

On a practical level, many power film capacitors have fusible links in the metallization so that failed portions of the capacitor are removed from circuit (and capacitance drops). This is a major practical consideration tied to the capacitor plate.

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