Electromagnetic waves are produced by

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.

Consider a parallel plate capacitor which is maintained at a potential of $200 \, V$. The separation distance between the plates of the capacitor and the area of the plates are $1 \, mm$ and $20 \, cm^2$, respectively. Calculate the displacement current in $1 \, \mu s$. (in $ \, mA$)

$A$ parallel plate capacitor is charged to $60\,\mu C$. Due to a radioactive source,the plate loses charge at the rate of $1.8 \times 10^{-8}\, C/s$. The magnitude of the displacement current is:

$AC$ voltage $V(t) = 20 \sin \omega t$ of frequency $50 \, Hz$ is applied to a parallel plate capacitor. The separation between the plates is $2 \, mm$ and the area is $1 \, m^2$. The amplitude of the oscillating displacement current for the applied $AC$ voltage is ...... $\mu A$.
[Take $\varepsilon_0 = 8.85 \times 10^{-12} \, F/m$]

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,$\varepsilon_{0} = 8.85 \times 10^{-12} \,C^{2} N^{-1} m^{-2}$)