In Griffiths' book "Introduction to Electrodynamics", section $4.4$ on "Linear Dielectrics", he says that in a homogeneous linear dielectric we have

$$\boldsymbol{\nabla}\cdot\boldsymbol{D}=\rho_{\rm f} \:\:\:{\rm and}\:\:\: \boldsymbol{\nabla}\times\boldsymbol{D}=0$$

Then he concludes that

$$\boldsymbol{D}=\epsilon_{0}{\boldsymbol{E}}_{\rm vac}$$

How does he make this conclusion? Shouldn't it be $\boldsymbol{D}=\epsilon{\boldsymbol{E}}$ where $\epsilon$ is the permittivity of the linear dielectric and $\boldsymbol{E}$ is the field due to free charges and polarization?


You are not interpreting correctly what he says. Griffiths argues that if your entire space is filled with homogeneous dielectric material, i.e. $\epsilon_{r}\left(x,y,z\right)=\rm const.$, then you can ignore the dielectric material in your calculation of $\boldsymbol{D}$. He states that you can calculate $\boldsymbol{E}$ like in the vacuum case (and that's why he writes $\boldsymbol{E}_{\rm vac}$), and then your electric displacement field is just

$$\boldsymbol{D}=\epsilon_{0}\boldsymbol{E}_{\rm vac}$$

Now, if you want to find the actual electric field, you use the usual equation $\boldsymbol{D}=\epsilon\boldsymbol{E}$. Why does this all thing work at all? Because of the following correspondence

$$\boldsymbol{\nabla}\cdot\boldsymbol{D}=\rho_{\rm f} \:\:\:{\rm and}\:\:\: \boldsymbol{\nabla}\times\boldsymbol{D}=0$$ $$\updownarrow$$ $$\boldsymbol{\nabla}\cdot\boldsymbol{E}_{\rm vac}=\frac{\rho_{\rm f}}{\epsilon_{0}} \:\:\:{\rm and}\:\:\: \boldsymbol{\nabla}\times\boldsymbol{E}_{\rm vac}=0$$

In other words, $\boldsymbol{D}$ and $\epsilon_{0}\boldsymbol{E}_{\rm vac}$ satisfy the same equations and are thus the same.

  • $\begingroup$ If two functions have the same divergence and curl, it is not true that they are the same. $\endgroup$
    – Kashmiri
    Oct 4 '20 at 15:03
  • $\begingroup$ @YasirSadiq this is true, but Griffiths is not intending to give an airtight proof here. He is making an argument, that one can prove rigorously if they want by invoking the Helmholtz theorem and giving sufficient conditions for the behaviour of each vector field over the appropriate boundary. His claim is nevertheless true. $\endgroup$
    – 1729_SR
    Oct 26 '20 at 1:24

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