The potential difference between the two plates of a parallel plate capacitor changes at a rate of $10^6 \; V/s$. If the capacitance of the capacitor is $2 \; \mu F$, what is the displacement current in the dielectric of the capacitor?

What contradiction is found by using Ampere's circuital law to obtain the magnetic field during the charging of a capacitor?

Maxwell's equations describe the fundamental laws of

The voltage between the plates of a parallel plate capacitor of capacity $1 \ \mu F$ is changing at the rate of $4 \ V/s$. The displacement current in the capacitor is: (in $\mu A$)

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

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: $\varepsilon_0 = 8.85 \times 10^{-12} \text{ F/m}$,$\mu_0 = 4\pi \times 10^{-7} \text{ T} \cdot \text{m/A}$)