What are the conditions under which these equations hold for dielectric and magnetic materials?

Question

$E_0 = \frac{D}{\varepsilon_0}$, $E' = -\frac{P}{\varepsilon_0}$
$B_0 = \mu_0 \cdot H$, $B' = \mu_0 \cdot M$
（$E_0$ refers to the electric field generated by free charges, or the external electric field, while $B_0$ denotes the external magnetic field; $E'$ refers to the electric field generated by polarized charges, and $B'$ refers to the magnetic field generated by magnetization currents.）
These equations seem to hold in most cases, but there are exceptions (e.g., a dielectric sphere uniformly polarized in a uniform electric field).
My textbook tells me that these expressions hold under certain conditions, but it is not clear about the conditions. Could you please explain in detail?
(My English is poor, if there's any language mistake I apologize)

example:in a uniformly polarized dielectric ball, $E' = -\frac{P}{3ε_0}$

while in a parallel plate capacitor or a charged metal sphere immersed in an infinite dielectric medium, $E_0 = \frac{D}{ε_0}$ and $E' = -\frac{P}{ε_0}$

Answer 1

By definition:

$$\textbf{D} = \epsilon_0 \textbf{E}_0 + \textbf{P} \ \ \ (1)$$ $$\textbf{B}_0 = \mu_0 ( \textbf{H} + \textbf{M}) \ \ \ (2)$$

these statements are always true no matter what materials you have present.

Often we make the approximation that $\textbf{D} ,\ \textbf{E}_0 \ \text{and} \ \textbf{P}$ are parallel, and that $\textbf{P}$ is linearly proportional to $\textbf{E}_0$ i.e. $\textbf{P} = \chi \epsilon_0 \textbf{E}_0$ for some constant $\chi$. For most materials this turns out to be a good assumption. This allows you to write $\textbf{D} = \epsilon_0(1+\chi) \textbf{E}_0 = \epsilon \textbf{E}_0$ for some constant $\epsilon$.

The exact same logic is used for the $\textbf{B}_0$ field:

$\textbf{B}_0 = \mu_0(\textbf{H} + \chi_m \textbf{H}) = \mu \textbf{H}$ for some constant $\mu$.

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So, to answer your question: (1) and (2) are always true. If $D , \ E_0 , \ P$ or $B_0 , \ H , \ M$ are parallel and linearly proportional (which is often the case experimentally) then you can write $\textbf{D} = \epsilon \textbf{E}_0$ and $\textbf{B}_0 = \mu \textbf{H}$

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Now to your E' and B' fields: I believe (at least for the case of a sphere) they are referring to to the depolarisation field and demagnetisation field respectively. They are sometimes denoted $\textbf{E}_d$ and $\textbf{H}_d$. Your best bet is to read:

Electricity & Magnetism by Bleaney & Bleaney, 3rd Edition sections 2.4 and 4.4 e.g. here

They fully describe the whole derivation for a dielectric sphere & magnetisable sphere, and they note the result you got of $\textbf{E}' = \textbf{E}_d = -\frac{\textbf{P}_1}{3\epsilon_0}$ where $\textbf{P}_1$ is the polarisation field within the sphere.

Answer 2

I've found the answer!
My teacher didn't answer this question clearly so I asked this question a few days ago, but then the teaching assistant answered it.
if $\frac{\vec{D}}{ε_0}$ represents the electric field of free charge, then the following expressions must hold:
$ \nabla · \vec{D} = ρ_0 $ ($ρ_0$ represents the electric density of free charge)
$ \nabla × \vec{D} = 0 $
The first expression always holds; however the second doesn't. Though inside and outside the dielectric this expression is true, but not on the surface. On the surface of dielectric,
$\oint\vec{D}·d\vec{l} = ε0\oint\vec{E}·d\vec{l}+\oint\vec{P}·d\vec{l} = \oint\vec{P}·d\vec{l}$
And $\oint\vec{P}·d\vec{l} = 0$ holds only when P has no tangential component.