For an ideal capacitor, charge and voltage are related by

$$q=Cv.$$

Since current is the rate of change of charge,

$$i(t)=\frac{dq}{dt}=C\frac{dv}{dt}.$$

A capacitor therefore responds to how quickly its voltage changes, not just to the voltage value itself.

Current leads voltage by 90°

Suppose the capacitor voltage is

$$v(t)=V_m\cos\omega t.$$

Then

$$i(t)=C\frac{dv}{dt} =-\omega CV_m\sin\omega t.$$

Using

$$-\sin\omega t=\cos\left(\omega t+\frac{\pi}{2}\right),$$

we obtain

$$i(t)=\omega CV_m\cos\left(\omega t+\frac{\pi}{2}\right).$$

Thus capacitor current leads capacitor voltage by $90^\circ$.

The peak current is

$$I_m=\omega C V_m,$$

so

$$\boxed{V_m=I_mX_C},$$

where

$$\boxed{X_C=\frac{1}{\omega C}}$$

is the capacitive reactance.

Reactance has units of ohms and measures the amplitude ratio between sinusoidal voltage and current for an ideal reactive element.

Frequency dependence

Because

$$X_C=\frac1{\omega C},$$

capacitive reactance decreases as frequency increases. A capacitor therefore supports larger sinusoidal current amplitudes at higher frequency for the same voltage amplitude.

In the steady-DC limit, $\omega\to0$ and the ideal capacitor eventually behaves as an open circuit after charging.

Example

For

$$C=10,\mu\mathrm F$$

at

$$f=1.00,\mathrm{kHz},$$

$$\omega=2\pi f\approx6283,\mathrm{rad/s}.$$

Thus

$$X_C=\frac{1}{(6283)(10\times10^{-6})} \approx15.9,\Omega.$$

If the current amplitude is $0.50,\mathrm A$, the voltage amplitude is

$$V_m=I_mX_C\approx7.96,\mathrm V.$$

The current waveform reaches corresponding points in its cycle one quarter-period before the voltage waveform.

An ideal capacitor stores energy in its electric field and returns it to the circuit rather than dissipating net energy over a complete sinusoidal cycle.