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Problem 4.18 The space between the plates of a parallel-plate capacitor (Fig. 4.24) is fiIIed with two slabs of linear dielectric material. Each slab has thickness a, so the total distance between the plates is 2a. Slab I 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 O'

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FIGURE4.24

- (~) Find the electric displacement Din 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. ½ -------· (e) Find the location and amount of all bound charge. __

______ l._e_(_ /5_ .. ~---- (f) Now that you know all the charge (free and bound), recalculate the field in each; slab, and confirm your answer to (b).

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