# Electric displacement in a homogenous linear dielectric

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?

## 1 Answer

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.