A parallel-plate capacitor is constructed by filling the space between two square plates with a block of three dielectric materials, as shown in the figure below. You may assume that ℓ ≫ d.A parallel-plate capacitor is comprised of two square plates with sides of length ℓ and separation distance d. There are three dielectric materials in the space between the capacitor plates. A material with length ℓ⁄2, height d, and dielectric constant ?1 fills the left half of the space. A material with length ℓ⁄2, height d⁄2, and dielectric constant ?2 fills the top right quarter of the space. A material with length ℓ⁄2, height d⁄2, and dielectric constant ?3 fills the bottom right quarter of the space. (a) Find an expression for the capacitance of the device in terms of the plate area A and d, ?1, ?2, and ?3. (Use the following as necessary: ?1, ?2, ?3, ?0, A and d.)C = (b) Calculate the capacitance using the values A = 1.30 cm2, d = 2.00 mm, ?1 = 4.90, ?2 = 5.60, and ?3 = 2.50. d/2dK1K3e/2-

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Answered: A parallel-plate capacitor is…

College Physics

11th Edition

ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

Section: Chapter Questions

Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...

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A parallel-plate capacitor is constructed by filling the space between two square plates with a block of three dielectric materials, as shown in the figure below. You may assume that ℓ ≫ d.

A parallel-plate capacitor is comprised of two square plates with sides of length ℓ and separation distance d. There are three dielectric materials in the space between the capacitor plates. A material with length ℓ⁄2, height d, and dielectric constant ?₁ fills the left half of the space. A material with length ℓ⁄2, height d⁄2, and dielectric constant ?₂ fills the top right quarter of the space. A material with length ℓ⁄2, height d⁄2, and dielectric constant ?₃ fills the bottom right quarter of the space.

(a) Find an expression for the capacitance of the device in terms of the plate area A and d, ?₁, ?₂, and ?₃. (Use the following as necessary: ?₁, ?₂, ?₃, ?₀, A and d.)
C =

(b) Calculate the capacitance using the values A = 1.30 cm², d = 2.00 mm, ?₁ = 4.90, ?₂ = 5.60, and ?₃ = 2.50.