Complex Dielectric Permittivity

  • #1
Complex Dielectric Permittivity!!!

Homework Statement


Calculate the power dissipated per unit cycle in a dielectric medium per unit volume in terms of the dielectric loss and the strength of the electric field

The previous question was:
Why does the dielectric permittivity of a material in general become complex for an alternating signal applied across it. Whats is the significance of the real part of the dielectric permittivity?

Homework Equations



Obvious ones:
[tex]W=\frac{1}{2}CV^2[/tex]
[tex]C=\frac{k\epsilon_{0}A}{d}[/tex]

Found this on wikipedia:
[tex]\hat{\epsilon}(\omega)=\epsilon'(\omega)-i\epsilon''(\omega)[/tex]

The Attempt at a Solution



[tex]W=\frac{1}{2}(\frac{k\epsilon_{0}A}{d})V^2[/tex]

Im guessing the complex part of [tex]\hat{\epsilon}(\omega)[/tex] accounts 4 the lost power so i put that in for [tex]k\epsilon_{0}[/tex]

Divided by dA to change it to per unit volume and divided by [tex]\frac{\omega}{2\pi}[/tex] to get it for one cycle. (Im presuming by cycle they mean period of the AC current)

Which gives:

[tex]-\frac{\pi\epsilon''(\omega)V^2}{d^2\omega}[/tex]

I dont no what dielectric loss is and i doubt im allowed to just put in [tex]\epsilon''(\omega)[/tex]

Ne1 no anything about dielectric permittivity?
Ne1 wanna take a ramdom guess like i just did
 
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Answers and Replies

  • #3
Gokul43201
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What does a complex dielectric constant do to the capacitive impedance? (once you express this as a phasor, you will understand what the loss angle is - for simplicity, assume the imaginary part small compared to the real part of the dielectric constant)
 
  • #4
Thanks for replying. Those websites r hard 2 make sense of but i didnt give it too long yet. Spent ages googling before i posted here and got nowhere though.

What does a complex dielectric constant do to the capacitive impedance?
It says dielectric permittivity becomes complex when an alternating signal is applied across it. I just assumed that they're talkin about a capacitor. Dont no of anythin else it could be. And the "power dissipated" definitely has to do with impedance.

Oh, and Happy Christmas!
 
  • #5
Gokul43201
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I wasn't asking you for a clarification; I was giving you a pointer on how to do the calculation (your way of doing it is fine too, but this way makes it obvious where the loss comes from). Write the Capacitance (C) using the complex dielectric function [itex] \epsilon_0 = \epsilon _1 - i \epsilon_2 [/itex], and then calculate the capacitive impedance [itex]Z_c = 1/(i \omega C) [/itex]. You'll see that the impedance turns out to be not purely reactive - there's a resisitive term as well. It is this resistance that dissipates power. The ratio of the resistive component to the reactive component (or real to imaginary terms) is called the loss tangent.
 
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  • #6
Oh, rite! I was thinking it was an uncharacteristically simple question.

I done that and got this:

[tex]Z_c = \frac{d(\epsilon ''(\omega) - j\epsilon '(\omega))}{\omega kA(\epsilon ''(\omega)^2 + \epsilon '(\omega)^2)}[/tex]

Would the I use the absolute value of the imaginary part of that to get the power dissipated or the absolute value of the entire thing minus the real part?
 
  • #7
Gokul43201
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Oh, rite! I was thinking it was an uncharacteristically simple question.

I done that and got this:

[tex]Z_c = \frac{d(\epsilon ''(\omega) - j\epsilon '(\omega))}{\omega kA(\epsilon ''(\omega)^2 + \epsilon '(\omega)^2)}[/tex]

Would the I use the absolute value of the imaginary part of that to get the power dissipated or the absolute value of the entire thing minus the real part?
Neither! Think about this carefully.

You've got the expression for the total (complex) impedance, Z. Separate Z into real and imaginary parts and rewrite it as, Z = Re{Z} +(1/j) Im{Z} . Now compare this with the total impedance of an RC series circuit. You see immediately that the real term depicts the resistor and the imaginary term depicts the reactance of the capacitor.

The mean power loss then turns out to be Re{V2/2Z} (simply from using P = Re{V}*Re{I} and averaging over a cycle). If you do this, I believe you'll get the same expression as you calculated in the OP above. So, it's a longer procedure, but it demostrates the origin of the loss in terms of something familiar (the common RC circuit).

Finally, note that you are asked for the power loss in terms of the applied electric field, not the applied voltage. How, is the field, E related to V (the potential drop across the material) and d (the thickness of the material)? Plug this in, and you are home.
 
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  • #8
Thanks. Wow. Turns out close. The way i done it was very dodgy though

[tex]\frac{\pi k \epsilon '' (\omega) V^2}{d^2}[/tex]

Puttin in Ed for V:

[tex]\pi k \epsilon '' (\omega) E^2[/tex]

Broke down quite nicely as the end
 
  • #9
Gokul43201
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Just to add a tiny correction, the complex dielectric function is actually an expression for the net permittivity, [itex]k \epsilon_0 [/itex] (sometimes denoted [itex]\epsilon_r \epsilon_0[/itex]), and not for the permittivity of free space, [itex]\epsilon_0[/itex], as I erroneously wrote down in post #5. Sometimes, it is used simply for the dielectric constant, k itself.

So, if you're using the first convention, you should have no k in the final expression. If you're using the second convention (probably more popular), then you should have an [itex]\epsilon_0[/itex] in the place of k.
 
  • #10
Ya, looking back i replaced [tex]\epsilon_0[/tex] with [tex]\epsilon ' (\omega) - j \epsilon '' (\omega)[/tex] instead of [tex]k \epsilon_0[/tex].

Thanks again!
 
  • #11
Bak
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Thank you for the mathematical explanation, but I always want to know the real physical meaning of the problem.
I will give it a try, but correct me if I'm wrong:

Permittivity is the ability of dipoles to rearrange according to the electromagnetic field:
negative sides point to the positive charge creating this field. The energy stored in this arrangement is within the real part of the solution. This energy is released once the EM field is stopped.
The imaginary part stems from relaxation behaviour: at certain freqencies the dipole is not able to move as fast as the elecromagnetic wave wants it to. The applied energy is stored as heat dissipation and is thus regarded as dielectric loss
The imaginary part has another term: dielectric loss due to ionic conductivity, which is possible in media with available charged ions or neutral particles being charged due to the high energy of the alternating EM field: electrons move to a higher level and finaly emerge out of the atom. This is only happening at certain frequencies and high energy levels.
 

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