In a homogeneous linear dielectric , bound charge density [tex] \rho _b [/tex] is proportional to free charge density [tex] \rho_f [/tex] , because(adsbygoogle = window.adsbygoogle || []).push({});

[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??

Griffithssays this is obvious, I can't see why susceptibility is not independent to position at the boundary, could someone tell me, thanks!!

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# Homework Help: Electric susceptibility problem

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