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)

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

A parallel plate capacitor is charged and then isolated. On increasing the plate separation

The nagative plate of a parallel plate capacitor is given a charge of -20 X 10^(-8) C . Find the charges appearing on the four surface of the capacitor plates.

A parallel plate capacitor is charged. If the plates are pulled apart

The charge on a parallel plate capacitor varies as =q_0 cos 2pi ft . The plates are very large and close together (area=a,separation=d). Neglecting the edge effects, find the displacement current through the capacitor.

Find out the capacitance of parallel plate capacitor of plate area A and plate separation d.