0
$\begingroup$

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?

$\endgroup$
1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.