Electric susceptibility problem

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Homework Help Overview

The discussion revolves around the concept of electric susceptibility in the context of a homogeneous linear dielectric and its relationship to bound charge density and free charge density. Participants are exploring the implications of susceptibility at boundaries, particularly regarding surface charge density.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants are questioning why bound charge density is not proportional to free charge density at the surface and are trying to understand Griffiths' assertion regarding the non-independence of susceptibility at boundaries. There are attempts to relate bound surface charge to susceptibility and discussions about the implications of homogeneity within the dielectric.

Discussion Status

The discussion is active, with participants expressing confusion over the concept of susceptibility at boundaries and seeking clarification. Some guidance has been offered regarding the relationship between bound surface charge and susceptibility, but there is no explicit consensus on the interpretation of Griffiths' statements.

Contextual Notes

Participants note that the relationship between bound charge and free charge may not hold at the surface due to the differing susceptibility at the boundary, highlighting a potential gap in understanding the material's behavior in that region.

Psi-String
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In a homogeneous linear dielectric , bound charge density [tex]\rho _b[/tex] is proportional to free charge density [tex]\rho_f[/tex] , because

[tex]\rho _b = - \nabla \cdot \vec{P} = \frac{\chi_e}{1+\chi_e} \nabla \cdot \vec{D} = \frac{\chi_e}{1+\chi_e} \rho _f[/tex]

My problem is that bound charge density is proportional to free charge density does not apply to surface charge, because susceptibility is not independent to position at the boundary, but why??

Griffiths says this is obvious, I can't see why :cry: susceptibility is not independent to position at the boundary, could someone tell me, thanks!
 
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Bound surface charge depends on P_\perp at the surface.
You can use this to relate bound surface charge to the susceptiblilty.
Griffiths is rather weak on bound surface charge.
This is unfortunate, because for a linear dielectric,
rho_bound=0 (since rho_free is usually zero), and it is only the bound surface charge that contrilbutes.
 
hmm...

[tex]\sigma_b = \vec{P} \cdot \hat{n} = \epsilon_0 \chi \vec{E}[/tex]
This the only relation between the bound surface charge and susceptibility that bumped into my head, but I don't think this is helpful ...:confused:
 
Psi-String said:
Griffiths says this is obvious, I can't see why :cry: susceptibility is not independent to position at the boundary, could someone tell me, thanks!

I don't quite understand what he means by "obvious" either.

Maybe what he means is that, inside the dielectric [itex]\chi_e[/itex] is the same because the material is homogeneous, but at either side of the surface, [itex]\chi_e[/itex] is different.
 
Last edited:
When a book uses the word obvious it means the author is confused.
 
Meir Achuz said:
When a book uses the word obvious it means the author is confused.
:smile: :smile: :smile: :smile: :smile: :smile: :smile: :smile: :smile:
 

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