Considering the case of a parallel plate capacitor being charged, show how one is required to generalize Ampere's circuital law of include the term due to displacement current.

Two circular plates of radius 0.1 m are used to from a parallel plate capacitor. If displacement current between the plates is 2pi .ampere, then find m agnetic filed produced by displacement current 4 cm from the axis of the plates.

The charge on a parallel plate capacitor varies as q = q_(0) "cos" 2 pit . The plates are very large and close together (area = A, separation = d). The displacement current through the capacitor is

Assertion: The force with which one plate of a parallel plate capacitor is attracted towards the other plate is equal to square of surface density per epsilon per unit area. Reason: The electric field due to one charged plate of the capacitor at the location of the other is equal to surface density per epsilon .

A parallel capacitor is being charged by an external source show that conduction current and displacement current have same value in the circuit.

A parallel plate capacitor is charged to 60 muC . Due to a radioactive source, the plate losses charge at the rate of 1.8 xx 10^(-8) Cs^(-1) . The magnitued of displacement current is

The charge of a parallel plate capacitor is varying as q = q0 sin omega t. Then find the magnitude of displacement current through the capacitor. (Plate Area = A, separation of plates = d)

A parallel - plate capacitor is being charged.Show that the displacement current across an area in the region between the plates and parallel to it is equal to the conduction current in the connecting wires.