$A$ 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:

The capacitance of a capacitor is $2 \ pF$. If the electric field inside the capacitor changes at a rate of $10^{12} \ V \ s^{-1}$,the displacement current is ..... $A$.

The existence of electromagnetic waves was first predicted by the scientist .......

Who discovered for the first time that a changing electric field can produce a magnetic field?

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

Suppose that the electric flux inside a parallel plate capacitor changes at a rate of $7 \times 10^{14} \text{ V} \cdot \text{m/s}$. If the area of the plates is $1 \text{ m}^2$,calculate the magnetic field $B$ at a distance $r = 0.1 \text{ m}$ from the axis of the capacitor. (Given: $\varepsilon_0 = 8.85 \times 10^{-12} \text{ F/m}$,$\mu_0 = 4\pi \times 10^{-7} \text{ T} \cdot \text{m/A}$)