The charge on a parallel plate capacitor varies as $q = q_0 \cos(2\pi \nu t)$. The plates are very large and close together (area $= A$,separation $= d$). Neglecting the edge effects,find the displacement current through the capacitor.

$A$ capacitor of capacitance $C$ is connected across an $AC$ source of voltage $V$,given by $V = V_{0} \sin \omega t$. The displacement current between the plates of the capacitor would then be given by:

An electromagnetic wave of frequency $100 \text{ MHz}$ propagates through a medium of conductivity $\sigma = 10 \text{ mho/m}$. The ratio of maximum conducting current density to maximum displacement current density is . . . . . . . $\left[ \text{Take } \frac{1}{4 \pi \epsilon_0} = 9 \times 10^9 \text{ Nm}^2/\text{C}^2 \right]$

The displacement current of $4.425 \,\mu A$ is developed in the space between the plates of a parallel plate capacitor when the voltage is changing at a rate of $10^{6} \,V s^{-1}$. The area of each plate of the capacitor is $40 \,cm^{2}$. The distance between the plates of the capacitor is $x \times 10^{-3} \,m$. The value of $x$ is ................ (Permittivity of free space,$\varepsilon_{0} = 8.85 \times 10^{-12} \,C^{2} N^{-1} m^{-2}$)

$A$ 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 . . . . . . $\text{V s}^{-1}$.

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?