I Why does this needle get an E field while this disk gets a D field?

[CrosisBH]

So I'm reviewing old lectures to prepare for an exam soon. This is about polarization. E fields, D Fields, etc. My professor labeled this diagram like so. The figures my professor drew are cavities in a dielectric if you can't read her handwriting. However, I can't seem to figure out why the needle gets an E field, and the disk gets an D field. She stated that if we turn the disk cavity 90 degrees to align with the needle, we get an E field.

$$\textbf{D} = \epsilon_0\textbf{E} + \textbf{P} = \epsilon\textbf{E}$$

This equation doesn't make it obvious to me why it is so. The D field is in the direction of the E field, since a D field is a scaled up E field with polarization in mind. My only guess is that the size of the needle being small makes some things negligible.

Thank you for any help you can give!

 

[olgerm]

E and D are different quantities, that have some values in both cavities. If there is vacuum in cavity then: $\vec{D}=\epsilon_0*\vec{E}$

 

[Lord Jestocost]

This equation doesn't make it obvious to me why it is so.

Have a look at "ELECTRIC FIELD WITHIN A CAVITY INSIDE A DIELECTRIC" (from physicspages.com):
ELECTRIC FIELD WITHIN A CAVITY INSIDE A DIELECTRIC ...

 

[vanhees71]

A great book, for some reason totally underrated in the textbook universe, is

J. Schwinger, Classical Electrodynamics

There you find a careful analysis of all standard constitutive equations using simple classical (non-relativistic though!) models.

The standard key word to look for in this and other books is the "Lorentz-Lorenz formula" or "Clausius-Mosotti Law".

 

[Paul Colby]

The way I understand this is in terms of boundary conditions. The normal component of $D$ is continuous across a boundary while the tangential $E$ field is continuous across a boundary. Both pictures neglect fringing fields near the disk edge and the cylinder ends. At the center of the disk the normal $D$ is the same inside and outside the cavity. Near the midpoint of the cylinder, it's the tangential $E$ that's the same inside and outside the cavity.

 

[vanhees71]

Hm, usually the normal component of $\vec{D}$ is discontinuous with the jump being equal to the surface charge. This follows from the macroscopic Maxwell equation,
$$\vec{\nabla} \cdot \vec{D}=\rho_{\text{free}}.$$

 

[vanhees71]

True, within dielectrics usually you have no free charges. I thought you made a general statement about the boundary conditions.

 

[Paul Colby]

Certainly true that the step in normal $D$ is the surface charge in all cases. For ideal dielectrics, $\rho_\text{free}=0$ so in most cases this step is zero. One may always inject a surface charge onto a boundary but the statement of the problem would need to include this else the boundary value problem isn't specified completely.