How do I solve this?

1.

Consider a capacitor made of two parallel, circular plates of radius R. In the initial con- guration, sketched in Fig. 1(a), the capacitor carries charge Q0 on one plate and Q0 on the other. At time t = U a conducting material is placed between the plates as in Fig. 1(1)). (Note that the material lls the entire space between the plates.) The material makes good electrical contact with the plates, allowing the capacitor to discharge. At time t > 0 the charge on the plates is therefore reduced in magnitude to Q(t) and Q(t). Throughout this problem assume that the current density is UNIFORM in the conducting material, i.e., current ows uniformly throughout the material during the discharge. You can also assume that the electric eld lines between the plates point straight rightward (in other words, ignore fringing elds). (a) (b) on | | on \".0\" conducting materlal medlatlng capacltor discharge FIG. 1: a Evaluate the time derivative of the electric ux, d'p, passing through the surface of all. radius 1' shown in Fig. 1(b). (The surface is parallel to the plates.) Express your answer in terms of the current I (r, 12) passing through that surface at time t, with the convention that positive currents are directed rightward. Be very careful about signs! (b) Use the Ampere-Maxwell law to deduce the magnetic eld B generated at an arbitrary point between the plates at time if during the discharge. In this problem we will revisit the magnetic and electric field inside a charging capacitor. You have already worked on this in QP15 on HW #7. The capacitor consists of two circular metal plates of radius R separated by a distance D. To refresh your memory, the capacitor is being charged by a steady current I. Charge is building up uniformly on the plates. The magnitude of the electric field between the plates is E = ItR-co where t is time and R is the radius of the plates. The magnetic field magnitude, for r