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I know that the defining equation of current and voltage through a capacitor is

$i= C \frac{dv}{dt}$

If $v(t)$ is a sine wave, then $i(t)$ would be a cosine wave which leads sine by 90°. That is the electrical circuit theory viewpoint of looking at it.

What I am pondering over is there any other way of looking at it from physics viewpoint.

For example the 90° phase lag of current from voltage in an inductor can be explained through Faraday's law and Lenz's law.

In this electrical circuit of an inductor the induced emf $e$ is such that it generates a current which opposes the flux produced by the primary current. This induced current opposes the change in primary flux as well as primary current which causes the primary current to fall back and hence a 90° phase lag.

Is there a similar explanation for phase difference in a capacitor ( like a variant of Lenz's law )?

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  • $\begingroup$ It's the property of sine wave. When we have V = sin(t) for the time interval [0,pi/2], the magnitude of V increase, but its change rate dV/dt decreases. So by i=C dV/dt, V and dV/dt has different phase. If you apply a voltatge such as V = e^t, then you have V = dV/dt, so you get no phase difference. $\endgroup$ – user42298 Oct 8 '16 at 3:09
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In a capacitor the current is a displacement current which is proportional to the time derivative of the electric field (and thus voltage v). The time derivative of the v=v0·sin(𝜔·t) function of the voltage is dv/dt=v0·𝜔·cos(𝜔·t) which, as you observed, leads the sine function by 90°.

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Current can only flow through a capacitor when it's charging. So consider a circuit with a battery, an open switch, a resistor, a capacitor, and then back to the negative terminal of the battery. Initially, the resistor and capacitor have 0V at both ends and zero voltage drop. But then the switch is closed. At this instant, no charge has yet accumulated on the capacitor, so its voltage drop is still zero and the total voltage of the battery is dropped through the resistor. Current flows through the system according to $I=V/R$ and the cap starts charging. At some time later, the capacitor is partially charged and has some voltage drop $\Delta V>0$, meaning that there is less voltage drop across the resistor and thus it permits less current to flow as it still obeys Ohm's law $V=IR $. Much later, the capacitor is fully charged, all the battery's voltage is dropped across the cap and there is no voltage drop through the resistor, so no current flows. Thus, we see that when a capacitor has the most $\Delta V $, it has the least current. This holds when the voltage source is changed to AC and that's where the phase difference comes from.

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