The charge of a parallel plate capacitor is varying as; \(q = q_{0} \sin\omega t\). The magnitude of displacement current through the capacitor is:
(the plate Area = \(A\), separation of plates = \(d\))
1. \(q_{0}\cos \left(\omega t \right)\)
2. \(q_{0} \omega \sin\omega t\)
3. \(q_{0} \omega \cos \omega t\)
4. \(\frac{q_{0} A \omega}{d} \cos \omega t\)

A variable frequency AC source is connected to a capacitor. Then on increasing the frequency:

Instantaneous displacement current of \(2.0~\text A\) is set up in the space between two parallel plates of \(1~\mu \text{F}\)$$ capacitor. The rate of change in potential difference across the capacitor is:
1. \(3\times 10^{6}~\text{V/s}\)
2. \(4\times 10^{6}~\text{V/s}\)
3. \(2\times 10^{6}~\text{V/s}\)
4. None of these

The S.I. unit of displacement current is:
1. Henry
2. Coulomb
3. Ampere
4. Farad

A parallel plate capacitor is made of circular plates each of radius \(R=6.0~\text{cm}\) has a capacitance \(C=100~\text{pF}.\) The capacitor is connected to a \(230~\text V\) AC supply with an (angular) frequency of \(300~\text{rad/s}.\) The amplitude of \(\vec{B}\) at the point \(3~\text{cm}\) from the axis between the plate is:
          
1. \(1.12\times 10^{-11}~\text T\)
2. \(2.01\times 10^{-12}~\text T\)
3. \(1.63\times 10^{-11}~\text T\)
4. \(1.01\times 10^{-12}~\text T\)

Displacement current goes through the gap between the plates of a capacitor when the charge of the capacitor:

A compass needle is placed in the gap of a parallel plate capacitor. The capacitor is connected to a battery through a resistance. The compass needle: