# Why does a polarized material generate a depolarizing field?

I do not understand how a polarized material (steady state, no free current, no free charge) can generate a depolarizing field.

Based on https://en.wikipedia.org/wiki/Demagnetizing_field, I "intuitively understand" that since

$$div(\vec B)=0$$

$$\vec{rot}(\vec B) = \mu_0 \vec{rot} (\vec M)$$

$$\vec H = \frac {\vec B}{\mu_0} - \vec M$$

$$div(\vec H)= - div(\vec M)$$

$$\vec{rot} (\vec H) =0$$

$$\vec M$$ creates $$\vec B$$ and both of them generate $$\vec H = \frac {\vec B}{\mu_0} - \vec M$$, and ultimately in the material, $$\vec H$$ is "opposed" to $$\vec M$$, hence the "demagnetizing field"

But in the case of a polarized material

$$div(\vec E) = \frac {div(\vec P)}{\epsilon_0}$$

$$\vec{rot}(\vec E) = 0$$

$$\vec D=\epsilon_0 \vec E+\vec P$$

$$div(\vec D)=0$$

$$\vec{rot}(\vec D)=\vec{rot}(\vec P)$$

$$\vec P$$ creates $$\vec E$$ and both of them generate $$\vec D=\epsilon_0 \vec E+\vec P$$, but I don't see "intuitively" how could $$\vec D$$ be opposed to $$\vec P$$ in the material, and so generate a so called "depolarizing field"

Best regards

Edit: I am speaking about spontaneously polarized material, where $$\vec P$$ exists without any applied electric field.

• Hi, In fact I was thinking about spontaneously polarized material, like ferroelectric. In the same geometry, without applied electric field, a $\vec P$ exits and lead to a depolarizating field, I just don't see "how" ad D~E+P. If we take the same geometry, with $P_x=1$ inside, $P_y=0$, as $\rho_b=-div(\vec P)$, E will be in the same direction than P, and so D, so ... I don't understand why book are speaking about depolarizing field. – Plaikeeaan Jul 12 at 21:31