The space between the conductor plates of a paralel plate capacitor is designed withthree slabs of linear dielectric material. Slab 1 has a dielectric constant ɛɛ, = 4.9, the second slabhas a dielectric constant of ɛ, = 5.6, and finally the third slab has a dielectric constant of ɛz = 2.1.The free charge density on the top plate is o and on the bottom plate is - o.(a) Find the electric diplacement D in each slab.(b) Find the electric field E in each slab.(c) Calculate the potential difference between slabs and equivalent capacitance.|d/2

In the given question we have multiple slabs with parallel plate capacitor.

Answered: e conductor plates of a paralel

Engineering

Electrical Engineering

e conductor plates of a paralel

Delmar's Standard Textbook Of Electricity

7th Edition

ISBN:9781337900348

Author:Stephen L. Herman

Publisher:Stephen L. Herman

Chapter19: Capacitors

Section: Chapter Questions

Problem 3RQ: A capacitor uses air as a dielectric and has a capacitance of 3 F. A dielectric material is inserted...

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Concept explainers

Dielectric Breakdown

A dielectric is a material with low electrical conductivity and so is classified as an insulator. Dielectric materials are used to block the flow of electricity through them, allowing them to behave as an insulators. They can also be used to store electric charge or improve capacitance.Dielectrics have no free electrons for conduction because electrons are tightly bound to the nucleus. Because dielectrics contain a small number of electrons for electrical conductivity, they have a dipole. The two primary roles of dielectrics in electrical devices are insulation and electrical energy storage. An electric field polarizes the dielectric, and the polarization of the molecules allows dielectric materials to store charge.

Question

The space between the conductor plates of a paralel plate capacitor is designed with
three slabs of linear dielectric material. Slab 1 has a dielectric constant ɛɛ, = 4.9, the second slab
has a dielectric constant of ɛ, = 5.6, and finally the third slab has a dielectric constant of ɛz = 2.1.
The free charge density on the top plate is o and on the bottom plate is - o.
(a) Find the electric diplacement D in each slab.
(b) Find the electric field E in each slab.
(c) Calculate the potential difference between slabs and equivalent capacitance.
|d/2

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