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

To establish an instantaneous displacement current of $2\,A$ in the space between two parallel plates of a $1\,\mu F$ capacitor,the potential difference across the capacitor plates will have to be changed at the rate of:

An accelerated electron produces ...

$A$ 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 $C R .$ )

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).

An accelerated electric charge emits