Calculating Electric Flux Density in a Dielectric Material

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Homework Help Overview

The discussion revolves around calculating electric flux density in a dielectric material within a capacitor setup. The original poster presents a scenario involving an electric field and the introduction of a dielectric, questioning how this affects the electric flux density.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to determine the correct formula for electric flux density D when a dielectric is introduced, considering two different electric field values. Some participants discuss the relationship between electric displacement field D and free charge, questioning how the presence of a dielectric influences this relationship.

Discussion Status

Participants are exploring the implications of dielectric materials on electric flux density and discussing the conditions under which D remains constant. There is an acknowledgment of varying charge densities in different capacitor configurations, indicating a productive exploration of the topic.

Contextual Notes

Participants note that the surface charge density may differ in certain capacitor types, such as spherical capacitors, while the total charge remains the same. This introduces complexity in understanding the relationship between charge density and electric flux density.

Nikitin
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Hey. So if I have a condenser, where the electric field between the plates is equivalent to 50kV/m, the electric flux density D= ε0*50kV/m.

If I insert a dielectic between the plates, the electric field will decrease to 10kV/m, and kappa (the relative permittivity) equals 5. So, will the flux density D = ε0κ*10kV/m, or D = ε0κ*50kV/m?

The first one makes more sense to me, but I am unsure.
 
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Hey Nikitin! :smile:
Nikitin said:
If I insert a dialectic between the plates, the electric field will decrease to 10kV/m, and kappa (the relative permitivity) equals 5. So, will the flux density D = ε0κ*10kV/m, or D = ε0κ*50kV/m?

The first one makes more sense to me, but I am unsure.

(is this a marxist dialectic? :wink:)

D (the electric displacement field) is related only to free charge

(as opposed to E, which is related to total charge, and P, which is only related to bound charge)

so the presence of the dielectric (and the bound charge in it) makes no difference …

if there is a volume density of free charge, divD is always equal to it

if there is a surface density of free charge, |D| is always equal to it :smile:

(and in particular, D for a parallel plate capacitor depends only on the (free) charge on the plates, not on the dielectric between the plates:

D = σn, and E is then calculated from D and from the dielectric)​
 
So D should be equal for the condenser regardless if there is a dielectric present or not?

PS: Sorry about my spelling!
 
Yes (assuming the surface charge density on the plates is the same). :smile:
 
Uhm, you can have condensers where the surface-charge density isn't equal among the plates?

Well, at least the charge Q on both the plates is the same. Right?
 
Nikitin said:
Well, at least the charge Q on both the plates is the same. Right?


Yes, the charge on both plates is the same.

But on eg a spherical capacitor, although the charge is the same, obviously the charge density is different.
 

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