Polarization vector for linear dielectrics

Question

I'm confused about the relations between electric field vector $\bar{E}$, polarization vector $\bar{P}$ and induction vector $\bar{D}$. In a parallel plate capacitor filled with a dielectric I find that $\bar{E} =\frac{\sigma_0}{\varepsilon_0}-\frac{\sigma_p}{\varepsilon_0}=\bar{E_0}-\frac{\sigma_p}{\varepsilon_0}$ where $\sigma_p=\frac{k-1}{k}\sigma_0$, $\bar{E} $ is the field modified by the presence of the dielectric. I define the vector $\bar{P}=\varepsilon_0(k-1)\bar{E_0}$.

Now I define $\bar{D} =\varepsilon_0\bar{E}+\bar{P}$, I think that the field appearing in this formula is the field generated by free charges and polarization charges. The book then writes, for linear dielectrics, $\bar{D} =\varepsilon_0k\bar{E}$; like the field appearing in the polarization vector is the same as the field appearing in induction vector. I don't understand how this could be correct if in the polarization vector the "unperturbed" electric field vector appears, and in the induction, the field generated by free charges and polarization charges appear .

Where is my error?

Answer 1

Your definition of the polarization vector is wrong. It should read

$$ \vec{P} = \epsilon_0 (k - 1) \vec{E} $$

This can be seen (for example) by using the relations you have written for the plane capacitor:

$$ E = \frac{\sigma}{\epsilon_0} - \frac{\sigma_p}{\epsilon_0} = \frac{\sigma}{\epsilon_0} - \frac{P}{\epsilon_0} $$

whence

$$ P = \sigma - \epsilon_0 E = \epsilon_0 (E_0 - E) = \epsilon_0 (kE - E) = \epsilon_0 (k - 1) E $$