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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?

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  • $\begingroup$ In physics capacitance is most generally defined by capacitance coefficients, which basically stem from the total energy of a general charge distribution that live on n electrostatic equipotential surfaces (i.e. conductors). It's basically the free space equivalent of your definition. I would be careful defining it with an AC signal, though, since the general definition is meaningful for the electrostatic case, as well, and yours only works well for frequencies at which the speed of light in the circuit does not matter, yet. $\endgroup$ – CuriousOne Jan 2 '16 at 3:38
  • $\begingroup$ The latter formula is only valid for a phase angle of 90 degrees, where pure capacitive behavior is seen (in contrast to eg inductive behavior) $\endgroup$ – Steeven Aug 6 '18 at 5:27
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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

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    $\begingroup$ While this link may answer the question, it is better to include the essential parts of the answer here and provide the link for reference. Link-only answers can become invalid if the linked page changes. $\endgroup$ – lucas Jun 24 '16 at 10:03
  • $\begingroup$ @lucas thank you for telling me that. I will surely remember it now. :) $\endgroup$ – Kumar Ayush Jun 24 '16 at 10:29

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