Answered: In a dielectric (σ = 10−4 S/m, μr = 1,…

In a dielectric (σ = 10−4 S/m, μr = 1, εr = 4.5), the conduction current density is given as Jc = 0.4 cos(2π x 108 t) A/m2. Determine the displacement current density.

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

In a dielectric (σ = 10⁻⁴ S/m, μ_r = 1, ε_r = 4.5), the conduction current density is given as J_c = 0.4 cos(2π x 10⁸ t) A/m². Determine the displacement current density.

Expert Solution

Step 1

According to the data given

σ = 10⁻⁴ S/m

μ_r = 1

ε_r = 4.5

Also,

J_c = 0.4 cos(2π x 10⁸ t) A/m²

^{Now, the conduction current density is}

^{$J_{c} = 0.4 \cos (2 π 10^{8} t) A / m^{2} σ E = 0.4 \cos (2 π 10^{8} t) (J_{c} = σ E) (10^{- 4}) E = 0.4 \cos (2 π 10^{8} t) E = (0.4 \times 10^{- 4}) \cos (2 π 10^{8} t) V / m$}

Step 2

The displacement current density will be

$J_{D} = \frac{\partial D}{\partial t} = \frac{\partial (ε_{o} ε_{r} E)}{\partial t} Since, D = ε_{o} ε_{r} E = ε_{o} ε_{r} \frac{\partial (E)}{\partial t} = \frac{10^{- 9}}{36 π} (4.5) \frac{\partial (0.4 \times 10^{4})}{\partial t} \cos (2 π \times 10^{8} t) = \frac{10^{- 9}}{36 π} (4.5) (0.4 \times 10^{4}) \cos (2 π \times 10^{8}) (- \sin (2 π \times 10^{8} t))$

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