# Electric susceptibility problem

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In a homogeneous linear dielectric , bound charge density $$\rho _b$$ is proportional to free charge density $$\rho_f$$ , because

$$\rho _b = - \nabla \cdot \vec{P} = \frac{\chi_e}{1+\chi_e} \nabla \cdot \vec{D} = \frac{\chi_e}{1+\chi_e} \rho _f$$

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 susceptibility is not independent to position at the boundary, could someone tell me, thanks!

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...

$$\sigma_b = \vec{P} \cdot \hat{n} = \epsilon_0 \chi \vec{E}$$
This the only relation between the bound surface charge and susceptibility that bumped into my head, but I don't think this is helpful ... Griffiths says this is obvious, I can't see why 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 $\chi_e$ is the same because the material is homogeneous, but at either side of the surface, $\chi_e$ is different.

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When a book uses the word obvious it means the author is confused.

When a book uses the word obvious it means the author is confused.
:rofl: :rofl: :rofl: :rofl: :rofl: :rofl: :rofl: :rofl: :rofl: