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

Sea water at frequency v=4xx10^8Hz has permitivity epsilon~~80epsilon_0 , pemeability mu=mu_0 and resistivity rho=0.25Omega-m . Imagine a parallel plate capacitor immersed in sea water and driven by an alternating voltage source V(t)=V_0sin(2pivt) . What fraction of the conduction density is the dispalcement current density?

Show that the force on each plate of a parallel plate capacitor has a magnitude equal to (1/2) QE, where Q is the charge on the capacitor, and E is the magnitude of electric field between the plates. Explain the origin of the factor 1/2.

A capacitor ,made of two parallel plates,each of area A and separaion d,is being charged by an external a.c. surce.show that the displacement current inside the capacitor is the same as the current charging the capacitor.

What is the area of the plates of a parallel plate capacitor of capacity 2pF with separation between the plates 0.5 cm?

The charging current for a capacitor is 0.25 A. WHat is the displacement current across its plates?

The charging current for a capacitor of 0.28 A. What is the displacement current across its plates?

Prove that energy stored in a parallel plate capacitor can be expressed by the relation (Q^2/2C) .

The plates of a parallel plate capacitor have an area of 90 cm^2 each and are separated by 2.5 mm. The capacitor is charged by connecting it to a 400 V supply. (a) How much electrostatic energy is stored by the capacitor? (b) View this energy as stored in the electrostatic field between the plates, and obtain the energy per unit volume u. Hence arrive at a relation between u and the magnitude of electric field E between the plates.