$P=\epsilon_o \chi E$ or $\epsilon_o \chi E_o$ Suppose the polarisation inside a dielectric is given by $P$, then is it related to the electric field as $\vec{P}=\epsilon_o \chi \vec{E}$ where $E$ is the field inside the dielectric or is is $E$ the original field that would have been present in that region in absence of the dielectric?
 A: $E$ is the field inside the dielectric. "Original field" is called "electric displacement" and usually denoted by $\mathbf D$, not $\mathbf E$.
A: It is a matter of definition. The usual definition of $\chi$ implies the electric field $\bf E$ actually present inside the dielectric.
However, one has to notice that in the special case of a dielectric completely filling a large parallel-plates capacitor at a fixed difference of potential between the plates, the resulting field $\bf E$ is the same present in the vacuum when the same difference of potential exists. In this way, it is possible to control experimentally the field $\bf E$, which in general depends on the geometry of the sample. Moreover, since $\chi$ in the usual linear regime is not depending on the field, its value can be extracted in this favorable set-up, remaining the same for more complicated geometries of the external field and of the sample. Therefore, if one is referring to this special situation it may happen that the field $\bf E$ could be called the external field $\bf E_0$, although the definition contains the field $\bf E$.
A: Good question!  $\vec{E}$ is the total field.  i.e. the external field in the absence of the dielectric plus the field due to polarisation of charges in the dieletric.
Source: Griffiths, Introduction to Electrodynamics 4Ed., p181.  (Griffiths actually says the field due to anything except the polarisation, which could include free charges inside the dielectric, plus the field due to polarisation).
