Capacitor
Current–voltage relation

The current I (t) through any component in an electric circuit is defined as the rate of flow of a charge Q (t) passing through it. Actual charges – electrons – cannot pass through the dielectric of an ideal capacitor. Rather, one electron accumulates on the negative plate for each one that leaves the positive plate, resulting in an electron depletion and consequent positive charge on one electrode that is equal and opposite to the accumulated negative charge on the other. Thus the charge on the electrodes is equal to the integral of the current as well as proportional to the voltage, as discussed above. As with any antiderivative, a constant of integration is added to represent the initial voltage V (t). This is the integral form of the capacitor equation: V (t) = Q (t) C = V (t 0) + 1 C ∫ t 0 t I (τ) d τ {\displaystyle V(t)={\frac {Q(t)}{C}}=V(t_{0})+{\frac {1}{C}}\int _{t_{0}}^{t}I(\tau)\,\mathrm {d} \tau }

Taking the derivative of this and multiplying by C yields the derivative form: I (t) = d Q (t) d t = C d V (t) d t {\displaystyle I(t)={\frac {\mathrm {d} Q(t)}{\mathrm {d} t}}=C{\frac {\mathrm {d} V(t)}{\mathrm {d} t}}} for C independent of time, voltage and electric charge.

The dual of the capacitor is the inductor, which stores energy in a magnetic field rather than an electric field. Its current-voltage relation is obtained by exchanging current and voltage in the capacitor equations and replacing C with the inductance L.
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