A capacitor is made of two square parallel plates of side
PHYSICS · Multi-part · Fermi University Physics · Electricity, Magnetism & Electromagnetism · Electric fields in matter · Maxwell's equations and conservation laws
Electricity, Magnetism and Electromagnetism, Ch Displacement current, Maxwell's equations and boundary conditions
Problem
A capacitor is made of two square parallel plates of side a=10.0cm, treated as perfect conductors, separated by d=1.0mm. One half of the gap (a 10.0cm×5.0cm region of the plate area) is filled with a linear dielectric of relative permittivity εr=3.0 and relative permeability μr=1; the other half is air, for which εr=1. The flat interface between dielectric and air runs from one plate to the other, perpendicular to the plates. Edge effects are ignored, so the field in each half of the gap is uniform and perpendicular to the plates. At the instant considered the voltage across the plates is V=400V and it is rising at the constant rate dV/dt=2.0×105Vs−1.
((a)) Apply Faraday's law to a thin rectangular loop straddling the dielectric-air interface to show that the tangential electric field is the same on the two sides, and hence give E and dE/dt in each half of the gap.
((b)) Using a pillbox straddling the surface of the positive plate, find the free surface charge density σf on that plate under the air half and under the dielectric half.
((c)) Find the bound surface charge density σb on the face of the dielectric that touches the positive plate.
((d)) Find the total displacement current through a flat surface spanning the whole gap, and hence the free current in the leads.
((e)) For a circular loop encircling one lead wire, evaluate ∮H⋅dl and ∮B⋅dl, and say why a surface pierced by the wire and a surface passing through the gap give the same answer.