The potential difference between the two plates of a parallel plate capacitor changes at a rate of $10^6 \; V/s$. If the capacitance of the capacitor is $2 \; \mu F$, what is the displacement current in the dielectric of the capacitor?

$A$ variable frequency $AC$ source is connected to a capacitor. How will the displacement current change with a decrease in frequency?

If the rate of change of electric field across the plates of a parallel plate capacitor is $E$ and the displacement current is $I$,then the area of one plate of the capacitor is ($\varepsilon_{0}$ is permittivity of free space).

The Maxwell-modified form of Ampere's circuital law is .......

$A$ time-varying potential difference is applied between the plates of a parallel plate capacitor of capacitance $2.5 \mu \text{F}$. The dielectric constant of the medium between the capacitor plates is $1$. It produces an instantaneous displacement current of $0.25 \text{ mA}$ in the intervening space between the capacitor plates. The magnitude of the rate of change of the potential difference will be . . . . . . $\text{V s}^{-1}$.

What is the phase difference between the space current and the capacitive displacement current produced when radio waves pass through the ionosphere?