Question

The space between the plates of a parallel plate capacitor is filled with two slabs of linear dielectric material. Each slab is of thickness a and the separation between the plates is 2a. The dielectric constants of the slabs are k1 and k2. The free charge density on the upper plate is and that on the lower plate is .

a.Find the electric displacement in each slab.

b.Find the electric polarization in each slab.

c.Find the location and amount of all bound charges.

Answer 1

Answer:

Explanation:

.(a)  The electric displacement will be found in each slab using Gauss’

Law:

∮~D·d~a=Qf enc→D A=σA→D=σdown(2.1)Note that by properly accounting for the signs, this displacement is true for both slabs

(c).Find the location and amount of all bound charges.

∮~E·d~a=Qenc≤0→E A=σ2≤0(σ−σ2)→E=σ2≤0(2.9)Now for slab 2:E A=σ≤0(−σ+σ3)→E=2σ3≤0(2.10