Permittivity values in conductors

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Discussion Overview

The discussion revolves around the relationship between electric current density and displacement current in conductors, particularly focusing on the expression for permittivity when the electric current density exceeds the displacement current. The scope includes technical electrodynamics and mathematical reasoning.

Discussion Character

  • Technical explanation, Mathematical reasoning, Debate/contested

Main Points Raised

  • ksnf3000 inquires about the condition under which the electric current density exceeds the displacement current and the resulting expression for permittivity, ε≈-jσ/ω.
  • Hassan2 provides a mathematical explanation involving the definitions of electric current density (J=σE) and displacement current density (J_{d}=jωD=jωεE), noting that the ratio ωε/σ indicates when their magnitudes become equal at high frequencies.
  • ksnf3000 questions the interpretation of the negative sign in the expression, suggesting it indicates loss in the imaginary part.
  • Another participant challenges the presence of the imaginary unit j, arguing that the parameters should be real, implying that the right-hand side must also be real.

Areas of Agreement / Disagreement

Participants express differing views on the interpretation of the negative sign in the expression for permittivity and the presence of the imaginary unit j, indicating unresolved disagreements on these points.

Contextual Notes

The discussion includes assumptions about the nature of the parameters involved and the conditions under which the electric and displacement currents are compared, which remain unspecified.

ksnf3000
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Hello All,

I am a masters student in electronics engg and reading technical electrodynamics. Please let me know when the electric current density becomes more than the displacement current (for a conductor ) why is the value of ε≈-jσ/ω

Quick replies will be highly appreciated!

Thanks and Regards,

ksnf3000
 
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ksnf3000 said:
Hello All,

I am a masters student in electronics engg and reading technical electrodynamics. Please let me know when the electric current density becomes more than the displacement current (for a conductor ) why is the value of ε≈-jσ/ω

Quick replies will be highly appreciated!

Thanks and Regards,

ksnf3000

High ksnf3000,

electric current density :

J=σE

displacement current density :

J_{d}=\frac{\partial D}{\partial t}

for a steady stead sinusoidal field,

J_{d}=jωD=jωεE

the ratio is \frac{ωε}{σ}

when ωε=σ, the magnitudes become the same. Considering typical values of the parameters, this happen at very high frequencies.

I hope that helps.
 
Thanks for your reply Hassan2.

The negative sign indicates the loss the imaginary part suffers, right??
 
ksnf3000 said:
Thanks for your reply Hassan2.

The negative sign indicates the loss the imaginary part suffers, right??

that j is not supposed to be there. The parameters are real, so, the right hand side must be real too. When comparing their magnitudes, j and also -1 becomes 1.
 

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