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

When a displacement current $i_d$ flows in the region between the two plates of a capacitor,the charge on the capacitor is:

Which of the following options represents the Ampere-Maxwell Law?

$AC$ 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 $\varepsilon_0 = 8.85 \times 10^{-12} \, F/m$]

$A$ parallel plate capacitor with circular plates of radius $1\, m$ has a capacitance of $1 \;nF$. At $t=0,$ it is connected for charging in series with a resistor $R=1 \;M \Omega$ across a $2 \;V$ battery (Figure). Calculate the magnetic field at a point $P$,halfway between the centre and the periphery of the plates,after $t=10^{-3}\; s$.
(The charge on the capacitor at time $t$ is $q(t)=C V[1-\exp (-t / \tau)],$ where the time constant $\tau$ is equal to $C R .$ )

$A$ parallel plate capacitor of area $60\, cm^2$ and separation $3\, mm$ is charged initially to $90\, \mu C$. If the medium between the plates becomes slightly conducting and the plate loses the charge at the rate of $2.5\times10^{-8}\, C/s$,what is the magnetic field between the plates?