In the context of understanding neurons as parallel circuits (see for instance here page 12), the following formulas are used:

C = \$ \frac{Q}{V} \iff \$

C * V = Q \$\iff \$

C\$\frac{dV}{dt}\$ = \$\frac{dQ}{dt} \iff \$

C*\$\frac{dV}{dt}\$ = \$I_{C} \$

With C = capacitance; V = Voltage; Q = Charge; \$I_{C}\$ = Capacitive current

My questions:

(1) As far as I know, current is defined by \$ I = \frac{Q}{t} \$ as charge (Coulomb) over time (seconds). Here, however, the change of charge over time is equated with current. How is this justified?

(2) Both charge and voltage are differentiated, capacitance remains constant, however. Why is that?

(3) What is the capacitive current in this context? Is this current flowing through the circuit, or rather movement of particles on the capacitor's plate?


1 Answer 1


(1) I'd say the notation \$I={Q \over t}\$ is confusing. As the Wikipedia article states it:

An electric current is the rate of flow of electric charge past a point or region.

Right there we should key on the phrase rate of flow. And it's pretty intuitive too -- when we think about current, it is simply moving charge from one place to another, so all that matters is the change in the amount of charge, not the absolute amount. So I'd say it would be more accurate to write \$I={\Delta Q \over \Delta t}\$, but in practice people don't do this. At any rate, in this context, \$I={dQ \over dt}\$ should hopefully make perfect sense.

(2) Simply because the capacitor value is constant. If you replaced the constant \$C\$ with a number (say, 2.5), then seeing \$2.5 * V = Q\$ differentiated into \$2.5 {dV \over dt} = {dQ \over dt}\$ would make perfect sense. So just remember that \$C\$ is a constant, not a time-varying quantity.

(3) It is the current flowing through the capacitor, in one terminal and out the other. (Sorry for the... unreasonable... size of the schematic image...)


simulate this circuit – Schematic created using CircuitLab

  • \$\begingroup\$ (1) Say, however, we have a constant current, e.g. 3 Amps = 3 Coloumbs/second. Would then dQ/dt not equal zero, as per definition the current's "strength" is not changing? (2) But the capacitance is a function of Voltage (and charge), so how comes changing the voltage is not changing it? (3) How can the current flow "through" the capacitor when its inside is an insulator? Thx \$\endgroup\$
    – user234
    Aug 17, 2019 at 3:56
  • \$\begingroup\$ (1) dQ/dt is 3 Amps. If the current's "strength" were changing, then that would be dI/dt (font doesn't do well here -- that is dee-eye-dee-tee there). To put it a different way, 3 amps means that every second 3 Coloumbs of charge are moving from point A to point B. (2) Nope, capacitance is not a function of voltage or charge. It is a relationship between them but changing the voltage (or the charge) does not change capacitance, ... \$\endgroup\$
    – Mr. Snrub
    Aug 17, 2019 at 4:46
  • \$\begingroup\$ (3) Current can flow through the capacitor, but not continuously. Imagine a pipe filled with water, and in the middle of a pipe is a rubber diaphragm which completely seals off one end of the pipe from the other. Now force some water into one side of the pipe. The diaphragm will stretch, and water will get pushed out the other side of the pipe. Some water flowed in and an equal amount came out the other end, so in the simplest sense water flowed "through". But you couldn't keep this up continuously, the diaphragm would exert too much reverse force or break. It's very similar with a capacitor. \$\endgroup\$
    – Mr. Snrub
    Aug 17, 2019 at 4:50
  • \$\begingroup\$ Thank you:-).... \$\endgroup\$
    – user234
    Aug 17, 2019 at 5:09
  • 1
    \$\begingroup\$ Strictly, current does not flow through a capacitor. Nothing goes through the dielectric, apart from a small leakage current. Current charges/discharges the plates. \$\endgroup\$
    – Chu
    Aug 17, 2019 at 16:29

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