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Time harmonic case of Gauss's Law
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[QUOTE="Charles Link, post: 6062006, member: 583509"] Let's take the divergence of both sides of this equation, but let's assume ## J=J_{free} =0 ##, and let's look at ## 0=\frac{d \nabla \cdot D}{dt}=\frac{d (\epsilon_o \nabla \cdot E+\nabla \cdot P)}{dt} ##. If we have a single material, and a single frequency, we can write ## P(\omega)=\epsilon_o \, \chi(\omega) E(\omega) ##, with ## E(t)=E(\omega)e^{i \omega t} ## and ## P(t)=P(\omega) e^{i \omega t} ##. Having a single homogeneous material means we get no polarization charges on any surface interface, (because there are no surface interfaces), and with the equation as we have it, it shows that we must have ## \nabla \cdot E=0 ##. We won't get any polarization charge inside the single uniform material. ## \\ ## I think a similar argument could be applied to the ## \nabla \cdot J_{free}=\nabla \cdot (\sigma E ) ## term. If the conductor is homogeneous, and responds linearly with ## J=\sigma E ##, so that ## J(\omega)=\sigma(\omega) \, E(\omega)##, there is no charge build-up anywhere. (If the conductor has a boundary, so that ## \sigma ## is not constant, then you will get charge build-up, and ## \nabla \cdot E \neq 0 ## ). ## \\ ## I don't know that what is found in your textbook is saying anything of any more significance than what I have just shown. [/QUOTE]
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Time harmonic case of Gauss's Law
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