# D-field -> E-field when the dielectric constant is a function of space

1. Aug 30, 2010

### LawlQuals

Hello everyone,

I am a little reluctant to post this question because it is a common problem type assigned for introductory electrodynamics courses, so I understand any level of vagueness that should be practiced with regards to answering the question so as to not make it too obvious how to do these problems for people looking for homework solutions on the internet. If it needs to be removed, no problem either (and, in which case, sorry for not reading the forum guidelines closely enough). I think this is appropriate for this section (rather than homework help) given its generality; however, it does get specific later on which may be of some issue regarding section relevance.

This is something I have never actually gotten down properly, and need to (I am studying for a qualifying examination in my grad school, I am not seeking homework advice to be clear). To be concrete, suppose we have a capacitor fabricated of concentric, cylindrical shells of radii $$a$$ and $$b \, (b > a)$$, respectively, and a total length $$L$$. Further, suppose the space between the conductors is filled with a material(s) of linear dielectric constant $$\varepsilon$$ such that $$\varepsilon$$ is a function of space $$( \varepsilon = \varepsilon (\vec{x}))$$. The classic question is to ask what the capacitance of this system is, or say the amount of charge needed on the conductors to store a certain amount of energy.

Calculations like this are furnished by first seeking the $$\vec{D}$$ field, and by inevitably computing the corresponding electric field $$\vec{E}$$. To find the electric displacement field, place a linear free charge density $$\lambda_f \equiv \lambda$$ on the inner conductor, and a swift application of Gauss' law provides (in Griffiths' notation of cylindrical coordinates $$(s,\phi , z)$$):

$$\vec{D} = \frac{\lambda}{2\pi s}\hat{s}$$

The question is, if I wish to translate to corresponding electric fields, $$\vec{E} = \frac{1}{\varepsilon (\vec{x})}\vec{D}$$, may I reasonably, and simply, do just that? For instance, if there were two dielectric materials, one material that was characterized by a constant value $$\varepsilon_1$$ over a certain length in the axial direction ($$z$$), and another over the rest described by a value $$\varepsilon_2$$, is it proper to state the $$\vec{E}$$-fields are given by $$\vec{E} = \frac{1}{\varepsilon_1}\vec{D}$$ and $$\vec{E} = \frac{1}{\varepsilon_2}\vec{D}$$ in the corresponding regions? That sounds ok to me, but what bothers me is placing a free charge of the same value along the entire length of cylinder. I suspect this is where the problem is, my misunderstanding of this concept (free charge).

My confusion is further perpetuated by a problem in Griffiths' text. When I took the class, we were assigned problem 4.28, which has the axial variation in a cylindrical capacitor I was just speaking of in the previous paragraph (in this case, one of the regions is vacuum). Griffiths proceeded to find the electric field in the vacuum region directly, and computes the displacement in the axial region that is filled with linear dielectric. The electric field was computed from the displacement field as usual, but the difference is that it appears he enforced a free charge (linear density) of different value on each ($$\lambda$$ and $$\lambda'$$ in the vacuum and dielectric regions respectively. It looks like he imposes interface conditions on tangential $$\vec{E}$$ fields between both regions, since the field itself is only in the radial direction ($$\nabla\times E = 0 \Rightarrow \vec{E}_s^{region \, 1} = \vec{E}_s^{region \, 2})$$, where $$s$$ denotes the radial component of $$\vec{E}$$ (the only component), and the fields are evaluated at the boundary between the materials. The conclusion made after the interface matching is that the free charges are different in both regions by a factor $$\varepsilon_r$$ (relative permittivity), ($$\lambda' = \varepsilon_r\lambda$$).

On one hand, I think it makes sense to me to put the same free charge on the conductor, since this charge has nothing to do with the dielectric material that surrounds it, but on the other it looks like Griffiths did not do that in this particular problem. I am misinterpreting something, but I just cannot put my finger on it.

Any thoughts regarding this matter? (Thank you very much for reading my verbose question, and thanks to any attempts at resolution).