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