$A$ parallel plate capacitor is completely immersed in a liquid of resistivity $\rho = 0.25 \, \Omega m$ and relative permittivity $\epsilon_r = 80$. If the plates of the capacitor are connected to an alternating voltage source of $V = V_0 \sin(\omega t)$ at a frequency of $f = 0.4 \, GHz$,then the ratio of the amplitudes of the displacement current $(I_d)$ and conduction current $(I_c)$ is:

Displacement current is set up between the plates of a capacitor when the potential difference across the plates is ..............

Imagine a world where free magnetic charges exist. In this world,a circuit is made with a $U$-shape wire and a rod free to slide on it. $A$ current carried by free magnetic charges can flow in the circuit. When the circuit is placed in a uniform electric field $E$ perpendicular to the plane of the circuit and the rod is pulled to the right with a constant speed $v$,the magnetic $emf$ in the circuit and the direction of the corresponding current,arising because of changing electric flux will be ($l$ is the length of the rod and $c$ is the speed of light).

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

$A$ 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 $C R .$ )

An insulator plate is passed between the plates of a capacitor. The displacement current