What is the weird field in dielectrics?

[Kaguro]

We have a material of dielectric constant (or relative permittivity) k=$\epsilon_r$. I place it in an external uniform electric field pointing down $E_0(-\hat z)$.

Induced charges and net field
There will be charges induced on the near and far surfaces normal to the field (top one negative, bottom positive). This will create an internal field E1 opposite to the external field and the net field inside will be reduced but will still point in the same direction. Total $\vec E= \vec E0 + \vec E1$.
The total field is also E0/k. So the bound charge field is $$\vec E1=\vec E0(1/k - 1)$$
Opposite direction.

Polarization density P

The material is linear, so the polarization density should be $\vec P = \epsilon_0 \chi_e \vec E_0$. Pointing in the same direction as the external field.

The bound charges

The P is uniform, so no volume bound charge. The surface bound charges then should be $\vec P \cdot \hat n = \epsilon_0 (\epsilon_r -1) E_0$

The displacement vector D

$D= \epsilon_0 E + P$
Where E is the TOTAL field inside. The sum of field due to free charges and due to the bound charges.

The direction of D will be down(same as E0,E and P).

Charges at the interface between two dielectrics placed in E0:

I will pretend that those two slabs are separated by an air gap so I can consider them as basically independent.
$\sigma_b1 = \epsilon_0 (k_1 -1) E_0$
$\sigma_b2 = \epsilon_0 (1-k_2) E_0$ (bottom of 1 is +Ve and top of 2 is -ve)
Put them back together:
$\sigma_b = \epsilon_0 (k_1 -k_2) E_0$ At the interface.

The weird field:

At this point I find a page in Griffiths saying the average macroscopic field inside a dielectric is $-\frac{P}{3 \epsilon_0}$ This is supposedly only due to the dipoles and not the external field. The total field is then E0 + weird field.

What's this field?!

 

[Charles Link]

The polarization $P=\epsilon_o \chi E_{total}$, not $E_o$. Griffiths' $E_p=-\frac{P}{3 \epsilon_o }$ is only for a uniformly polarized sphere.

When an external field $E_o$ is applied, you get $E_{total}=E_o+E_p$ inside the material. Then $P=\epsilon_o \chi E_{total}$, and for some shapes $E_p$ is a simple and uniform result with a factor $D$ that is geometrically dependent. It then becomes a self-consistent problem where you write $E_p=-\frac{D P}{\epsilon_o}$, and you can then solve for $E_{total}$, $P$, and $E_p$, in terms of $E_o$, with the 3 equations listed here.

Sometimes you have a spontaneous polarization $P$, without any external $E_o$, that is unaffected by $E_{total}$. Then you can compute $E_{total}=E_p=-\frac{D P}{\epsilon_o}$. (This case is different, but not at all weird, once you see how it works).

$D=\frac{1}{3}$ for a sphere, and $D=1$ for a flat slab. $D \approx 0$ for a long cylinder, and $D=\frac{1}{2}$ for a cylinder turned sideways.

It should be noted that the source of the $E_p$ is polarization surface charge, with polarization surface charge density $\sigma_p=P \cdot \hat{n}$.

 

[Kaguro]

The polarization $P=\epsilon_o \chi E_{total}$, not $E_o$. Griffiths' $E_p=-\frac{P}{3 \epsilon_o }$ is only for a uniformly polarized sphere.

When an external field $E_o$ is applied, you get $E_{total}=E_o+E_p$ inside the material. Then $P=\epsilon_o \chi E_{total}$, and for some shapes $E_p$ is a simple and uniform result with a factor $D$ that is geometrically dependent. It then becomes a self-consistent problem where you write $E_p=-\frac{D P}{\epsilon_o}$, and you can then solve for $E_{total}$, $P$, and $E_p$, in terms of $E_o$, with the 3 equations listed here.

Sometimes you have a spontaneous polarization $P$, without any external $E_o$, that is unaffected by $E_{total}$. Then you can compute $E_{total}=E_p=-\frac{D P}{\epsilon_o}$. (That case is different, but not at all weird, once you see how it works).

$D=\frac{1}{3}$ for a sphere, and $D=1$ for a flat slab. $D \approx 0$ for a long cylinder, and $D=\frac{1}{2}$ for a cylinder turned sideways.

It should be noted that the source of the $E_p$ is polarization surface charge, with polarization surface charge density $\sigma_p=P \cdot \hat{n}$.

Okay. So the shape heavily influences the opposing field inside which then combined with the external field E0 gives Etotal from where P maybe determined.

 

[Charles Link]

Yes, it's self-consistent in that $E_{total}$ causes $P$, but you don't know what $E_{total}$ is until you compute $P$, which gives rise to $E_p$, which is part of $E_{total}$.

 

[Kaguro]

But wait! So what about the relationship Etotal=E0/k ?

From this I could find out Ep and P.

Is this valid only for a slab?

 

[Charles Link]

Yes, that is only for a slab, where $D=1$. Note: $k=1+\chi$.

 

[Kaguro]

Thank you very much for taking time to clear my doubts.