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The equation for capacitance is $Q=CV$ or $V={1\over C}Q$. I don't understand what is the physical meaning of this "$C$":

  1. Does the charge in a system changes linearly with voltage under all circumstances?
  2. Wouldn't it be more comfortable to define electrostatic energy storage capability of a capacitor (or any other system) in terms of max possible charge, $Q_{max}$?

(related question: Why does the area of the plates affect the capacitance?)

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    $\begingroup$ why is resistance given by $R$ instead of Amperes in Ohms law $V=IR$? (similar applies to capacitance) $\endgroup$ – Nikos M. Nov 3 '14 at 20:18
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    $\begingroup$ technically, resistance is not given by R but rather by an I-V curve, i.e. a function and not a constant. $\endgroup$ – Sparkler Nov 3 '14 at 20:20
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    $\begingroup$ exactly similar for capacitance, it changes with other parameters like temperature etc. these relations hold only in the linear region of their curves $\endgroup$ – Nikos M. Nov 3 '14 at 20:23
  • $\begingroup$ got it: in daily language, "capacity" is a max possible value, while here it means "filling coefficient". i.e. electrostatic capacitane is "chargabiliy"... $\endgroup$ – Sparkler Nov 3 '14 at 20:49
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    $\begingroup$ actually in both cases it is the slope, which means the rate at which one parameter changes wrt to the other (in the linear region) $\endgroup$ – Nikos M. Nov 3 '14 at 20:52
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The physical meaning of the capacitance is precisely given by $\mathrm{d}Q=C\cdot \mathrm{d}V$: $C$ tells you how much charge there will be in the capacitor per voltage applied.

  1. For all capacitors, the linearity holds fairly well. Generally speaking, capacitance is given by a Q-V curve, which may consist of a linear region, a saturation region and a breakdown region ("overloading" a capacitor, i.e. frying it probably kills off the linear relation...). Leakage current as well is not taken into account in the linearized $Q=CV$ equation.

  2. No, $Q_{max}$ is just one parameter of a capacitor, which depends on the breakdown voltage. In regular operation, capacitors do generally not store the "maximum possible charge" they could, precisely because they follow $Q = CV$, and knowing $Q_\text{max}$ alone wouldn't provide information about the Q(V) behaviour up to that point. The capacitance tells you how much charge the thing will store if you apply a given voltage to it. $Q_\text{max}$ just tells you when it's full.

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  • $\begingroup$ MOS capacitors for example are not linear capacitors, because their capacitance changes with gate voltage and opertating frequency (see chart). $\endgroup$ – Sparkler Nov 4 '14 at 3:51
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Worth noticing the difference between capacity and capacitance. You want the latter to have the meaning of the former - when instead it just describes how much the voltage increases when you add charge. It doesn't tell us when the capacitor is "full" - for that you need to know the rated voltage as well as the capacitance.

When a capacitor is used in a circuit you need to know what happens to the voltage when a current flows through it - just as you do for a resistor. In fact when you use complex notation, you can write $Z$, the impedance (which is a fancy word for complex resistance especially for things that are not ideal resistors) as $$Z = \frac{1}{j\omega C}$$ where $j=\sqrt{-1}$ and $\omega$ is the angular frequency $f/2\pi$.

This close similarity makes it possible to calculate the currents in a network of resistors and capacitors - use Z just like you would use R, and write down all equations; at the end, substitute for Z with the above you will find the complex impedance of the network - and as if by magic you can determine both the amplitude and phase of the currents in the network.

So yeah - it is really useful that such a quantity exists.

PS - in many languages there isn't a different word for capacity vs. capacitance. For example, Dutch and German both use the same word for both concepts (capaciteit in Dutch, and Kapazität in German). I suspect many other languages are the same. That makes it even harder to understand that there is a very real difference between the concepts!

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  • $\begingroup$ It's felt like a long time since I last took an E&M class, but I thought that impedance was only used to describe AC circuits... Especially the way you describe it $\endgroup$ – Sean Nov 4 '14 at 0:57
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    $\begingroup$ @Sean - complex impedance describes behavior "at all frequencies" including zero (DC). $\endgroup$ – Floris Nov 4 '14 at 1:20
  • $\begingroup$ For $f=0$, you would get $\omega=0$ in your denominator, which is undefined. Unless I'm missing something... $\endgroup$ – Sean Nov 4 '14 at 12:39
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The equation for capacitance is Q=CV or V=1CQ. I don't understand what is the physical meaning of this "C":

Does the charge in a system changes linearly with voltage under all circumstances?

This first part is the statement of the behavior of an "ideal" capacitor. Because it is an idealization, it is easy to characterize as a linear component whose behavior can be easily calculated. As a concept this gives it general applicability in many real-world situations, where important and valid results can be obtained by approximating the messy, imprecise, non-linear behaviors of things with idealized components given some constraints (such as limiting the system characterization to a small range of voltages and/or currents).

Non-ideal capacitors will not behave in this way, but this linear model of capacitance is so useful, and so generally applicable in modern, everyday electronic circuit theory, that the messy, general, nonlinear forms are not even considered.

As for the area of the plates... it seems intuitively obvious that the storage of charge in the "capacitor" must be directly proportional to some measure relating to "capacity". In a freely conducting medium, the electric field is uniform, so the only other parameter which can provide more capacity is the area of the plate: more charges require more area to spread out over, if the electric field is to remain uniformly at one value.

For the nature of the question, this seems like the best answer that can be afforded "intuitively" without going into great detail about Gauss's law and other aspects of electrostatics.

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protected by Qmechanic Nov 8 '15 at 20:28

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