What is capacitance, in general?

Question

In circuit analysis software capacitance can be measured between any two nodes of a circuit or of a multiterminal device.

In practical terms we take $C_{ij}$, the capacitance between $i$ and $j$ as follows. A small AC signal $\Delta \widetilde{v}_j$ (in phasor notation), with frequency $\nu$,is superimposed to the voltage at $j$, then $\Delta \widetilde{i}_i$ is the small signal current response. Admittance $\widetilde{Y}_{ij}$ is given by $$\widetilde{Y}_{ij} = \frac{\Delta \widetilde{i}_i}{\Delta \widetilde{v}_j}$$ And $$C_{ij} = \frac{Im(\widetilde{Y}_{ij})}{2\pi\nu}$$

My questions are:

1. Is that a generalized version of the capacitance as $q/V$ for a generic device?

2. What is its physical meaning (where is the charge)?

3. What can we say about it? (ex. it seems that $C_{ij} = C_{ji}$, but why?).

4. Is there a physically rigorous, and meaningful, definition of capacitance?

Answer 1

the ratio of the change in an electric charge in a system to the corresponding change in its electric potential is called capacitance I.e the ability of system to store charge. You can find more info here. https://en.m.wikipedia.org/wiki/Capacitance

Answer 2

In a general physical system we generally have through variables and across variables. We also have flow and effort variables. See Vibration and Shock Handbook - Silva.

In an electrical network current is the through and flow variable and voltage is the across and effort variable. In the network picture of an electrical circuit the nodes are ideal wires (i.e. interconnects) and linear circuit elements are edges between these nodes. Each node has a corresponding voltage so we speak of the voltage across an element. Each edge has current flowing through it so we speak of the current through an element.

In a mechanical network velocity is the across and flow variable and force is the through and effort variable. In the network picture of a mechanical circuit the nodes are chunks of matter and the edges between the nodes are springs and dampers between the nodes. Each node has a velocity associated with it so, thinking of the voltage difference between the two nodes connected by an edge, we speak of the velocity difference across an element. Each element has a force acting on the two elements it connects so we speak of the force through the element (this is a little bit of a stretch but it's not too bad).

In electricity we have current $I$, its integral charge $Q$ and its derivative change in current over time $\dot{I}$.

In mechanics we have velocity $v$, its integral position $x$ and its derivative acceleration $a$.

In electricity a capacitor relates $Q$ and $V$: $$ C = \frac{Q}{V} = \frac{\int I dt}{V} = \frac{1}{s} \frac{I}{V} $$ $Q$ is the integral of the through/flow variable $I$.

The analogous element for a mechanical circuit is the Hookian spring element. $$ \frac{1}{k} = \frac{x}{F} = \frac{\int xdt}{F} = \frac{1}{s} \frac{v}{F} $$

Generically impedance is the ratio of the effort to flow variable. $$ Z \sim \frac{V}{I}\sim \frac{F}{v} \sim \frac{\text{Effort}}{\text{Flow}} $$ We see that if impedance is high a large effort results in only a small flow. So we see the electrical impedance of a capacitor is $1/Cs$ and the mechanical impedance of a spring is $k/s$.

What, then, generically, is capacitance? Capacitance is a type of impedance. In a physical system that you can parametrize as a network (i.e. a graph) which has through variables associated with each edge and across variables at each node you can also identify these variables also as effort and flow variables. Impedance is then how the effort and flow variables for an edge element are related. Capacitance is a special type of impedance which relates the INTEGRAL of the flow variable with the value of the effort variable. That is, in the frequency domain capacitance is a type of impedance that is proportional to $1/s$.

Electrical capacitance tells us how much charge can be stored at a given voltage. Mechanical capacitance tells us how much position displacement can be sustained at a given force.

Another case of a system that can be treated with a circuit analogy is a ultra high vacuum system at equilibrium pressure. The across/effort variables are chamber pressures, the through/flow variables are flow rates between chambers. Differential pumping apertures act as resistors.

Answer 3

4 Is there a physically rigorous, and meaningful, definition of capacitance? Yes. It is the amount of net charge $q$ that can be stored in a capacitor per unit voltage across the capacitor. The charge on one plate is the opposite of the charge on the other plate, by Gauss’ law.

1 Is that a generalized version of the capacitance as $q/V$ for a generic device? Yes it is. If you integrated the current during the phase it changes and matched it to voltage it would meet the definition. For resistors it is current. For inductors it is derivative of current. For capacitors it is integral of current. So they all have different phases.

2 What is its physical meaning (where is the charge)? The charge is held on two plates if it is a plate capacitor. It actually holds a build-up of electrons (and the other side is depleted). Charge is on one plate and the opposite on the other, the difference in charge is what the circuit can accept and store. It is kept there by the voltage drop across the capacitor (which makes an electric field). Capacitors won’t hold charge without a voltage drop. Only the difference in voltage affects the difference in charge. If they both had the same charge there would be no net storage of charge (or current-time $\int Idt$). The measure of how good they are is how low that needed voltage difference needs to be per unit stored charge: i.e. how low elastance can get. (Equivalently how much charge can be held per unit voltage, capacitance).

3 What can we say about it? (ex. it seems that $C_{ij} = C_{ji}$, but why?). Because the charges are equal and opposite.

It’s the ability of a capacitor to hold charge $q$, which is $\int Idt$ per unit voltage increase. That’s what capacitors do; they hold charge, let it build up. It’s complicated in AC with the voltage changing with the integral of current. (As mentioned, for resistors it is current. For inductors it is derivative of current. For capacitors it is integral of current. So they all have different phases.)

A good capacitor can hold a lot of charge without imposing a big jump in voltage to do it. How to achieve this? For one thing the distance between the plates would be low, because the field between the plates (if they are assumed large plates) is independent of separation. There is a bigger $\int E dx$ or voltage involved in holding the stored-up charges further apart. But to have low distance and avoid arcs requires smooth and perfectly parallel surfaces so costs money.

Answer 4

I really like not to overanalyse things.C=dQ/dV.It simply means that if we have have a capacitor of 1F and we put it under a voltage source of 1V the charge stored on the plates of the capacitor will be 1C.