The space between the plates of a parallel-plate capacitor (Fig. 4.24) is filled with two slabs of linear dielectric material. Each slab has thickness a, so the total distance between the plates is 2a. Slab 1 has a dielectric constant of 2, and slab 2 has a dielectric constant of 1.5. The free charge density on the top plate is σ and on the bottom plate −σ.
12A medium is said to be isotropic if its properties (such as susceptibility) are the same in all directions. Thus Eq. 4.30 is the special case of Eq. 4.38 that holds for isotropic media. Physicists tend to be sloppy with their language, and unless otherwise indicated the term “linear dielectric” implies “isotropic linear dielectric,” and suggests “homogeneous isotropic linear dielectric.” But technically, “linear” just means that at any given point, and for E in a given direction, the components of P are proportional to E—the proportionality factor could vary with position and/or direction .
P =\epsilon_{0} \chi_{e} E (4.30)
\left.\begin{array}{l}P_{x}=\epsilon_{0}\left(\chi_{e_{x x}} E_{x}+\chi_{e_{x y}} E_{y}+\chi_{e_{x z}} E_{z}\right) \\P_{y}=\epsilon_{0}\left(\chi_{e_{y x}} E_{x}+\chi_{e_{y y}} E_{y}+\chi_{e_{y z}} E_{z}\right) \\P_{z}=\epsilon_{0}\left(\chi_{e_{z x}} E_{x}+\chi_{e_{z y}} E_{y}+\chi_{e_{z z}} E_{z}\right)\end{array}\right\} (4.38)
(a) Find the electric displacement D in each slab.
(b) Find the electric field E in each slab .
(c) Find the polarization P in each slab.
(d) Find the potential difference between the plates.
(e) Find the location and amount of all bound charge.
(f) Now that you know all the charge (free and bound), recalculate the field in each slab, and confirm your answer to (b).
