A capacitor of capacitance \(C\) is connected across an AC source of voltage \(V\), given by;
\(V=V_0 \sin \omega t\)$\mathrm{}$
The displacement current between the plates of the capacitor would then be given by:
1. \( I_d=\frac{V_0}{\omega C} \sin \omega t \)
2. \( I_d=V_0 \omega C \sin \omega t \)
3. \( I_d=V_0 \omega C \cos \omega t \)
4. \( I_d=\frac{V_0}{\omega C} \cos \omega t\)

The S.I. unit of displacement current is:

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

Instantaneous displacement current of \(2.0\) 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

A larger parallel plate capacitor, whose plates have an area of \(1~\text{m}^2,\) separated from each other by \(1\) mm, is being charged at a rate of \(25.8\) V/s. If the plates have dielectric constant \(10\), then the displacement current at this instant is:
1. \(25~\mu\text{A}\)
2. \(11~\mu\text{A}\)
3. \(2.2~\mu\text{A}\)
4. \(1.1~\mu\text{A}\)

A parallel plate capacitor with plate area \(A\) and separation between the plates \(d\), is charged by a source having current \(i\) at some instant. Consider a plane surface of area \(A/2\) parallel to the plates and drawn symmetrically between the plates. The displacement current through this area is:
1. \(i\) 
2. \(i/2\) 
3. \(i/4\) 
4. \(i/8\)

The figure shows a parallel plate capacitor being charged by a battery. If \(X\) and \(Y\) are two closed curves then during charging, \(\oint \vec{B} . \vec{dl}\) is zero along the curve: