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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"

demagnetising field in a cylinder

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.

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"H is "opposed" to M, hence the "demagnetizing field"." H is NOT a demagnetizing field. Only B acts on the atoms to polarize them. H is only used to find B. Any 'demagnetization' that occurs is because B is half as large at the end as at the middle. H being opposite to B and M is an end effect. Because B gets weaker toward the end, while M is constant, H is in the opposite direction.

The situation is different in a dielectric. In a permanent magnet, M is constant throughout the magnet. In a dielectric P is caused by E so must be in the same direction, which makes D>E, an in the same direction.

H~B-M, and D~E+P.

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  • $\begingroup$ 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. $\endgroup$ – Plaikeeaan Jul 12 at 21:31
  • $\begingroup$ 1. I have looked at the Wikipedia article. Like too many Wikipedia articles, it is just wrong. The H field has no effect on magnetization, M. Only the be field can act to increase or decrease magnetization. $\endgroup$ – Jerrold Franklin Jul 13 at 2:28
  • $\begingroup$ 2. The situation is different for a permanent polarized ferroelectric, than for a permanent ferromagnetic. In this case, is the D field that is continuous, and the E field near the end of the permanent ferroelectric will be in the opposite direction of the polarization P. The E field will act as a depolarizing field in that case. $\endgroup$ – Jerrold Franklin Jul 13 at 2:32

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