The space between the plates of wo slabs of linear dielectric m che total distance between the constant of 2, so the slab 2 has charge density on the top plate (a) Find the electric displacem
The space between the plates of wo slabs of linear dielectric m che total distance between the constant of 2, so the slab 2 has charge density on the top plate (a) Find the electric displacem
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Transcribed Image Text:. The space between the plates of a parallel-plate capacitor 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, so the slab 2 has a dielectric constant of 1.5. The free
charge density on the top plate is o and on the bottom plate -o.
(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).
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