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

In summary, the speaker is hesitant to post a question about a common problem in introductory electrodynamics courses, but believes it is appropriate to discuss in the forum. The question involves finding the capacitance of a capacitor made of concentric, cylindrical shells with varying dielectric constants. The speaker is unsure about how to handle free charge in regions with different dielectric constants and is confused by a problem in Griffiths' text. The responder clarifies that the relationship \vec{E} = \frac{1}{\varepsilon(\vec{x})}\vec{D} is valid for finding electric fields in regions with varying dielectric constants and explains that free charge is not affected by dielectric materials. They also mention that the author of
  • #1
LawlQuals
30
0
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 [tex]a[/tex] and [tex] b \, (b > a)[/tex], respectively, and a total length [tex]L[/tex]. Further, suppose the space between the conductors is filled with a material(s) of linear dielectric constant [tex]\varepsilon[/tex] such that [tex]\varepsilon [/tex] is a function of space [tex]( \varepsilon = \varepsilon (\vec{x}))[/tex]. 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 [tex]\vec{D}[/tex] field, and by inevitably computing the corresponding electric field [tex]\vec{E}[/tex]. To find the electric displacement field, place a linear free charge density [tex]\lambda_f \equiv \lambda [/tex] on the inner conductor, and a swift application of Gauss' law provides (in Griffiths' notation of cylindrical coordinates [tex](s,\phi , z)[/tex]):

[tex]\vec{D} = \frac{\lambda}{2\pi s}\hat{s}[/tex]

The question is, if I wish to translate to corresponding electric fields, [tex]\vec{E} = \frac{1}{\varepsilon (\vec{x})}\vec{D}[/tex], 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 [tex]\varepsilon_1[/tex] over a certain length in the axial direction ([tex]z[/tex]), and another over the rest described by a value [tex]\varepsilon_2[/tex], is it proper to state the [tex]\vec{E}[/tex]-fields are given by [tex]\vec{E} = \frac{1}{\varepsilon_1}\vec{D}[/tex] and [tex]\vec{E} = \frac{1}{\varepsilon_2}\vec{D}[/tex] 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 ([tex]\lambda[/tex] and [tex]\lambda'[/tex] in the vacuum and dielectric regions respectively. It looks like he imposes interface conditions on tangential [tex]\vec{E}[/tex] fields between both regions, since the field itself is only in the radial direction ([tex]\nabla\times E = 0 \Rightarrow \vec{E}_s^{region \, 1} = \vec{E}_s^{region \, 2})[/tex], where [tex]s[/tex] denotes the radial component of [tex]\vec{E}[/tex] (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 [tex]\varepsilon_r[/tex] (relative permittivity), ([tex]\lambda' = \varepsilon_r\lambda[/tex]).

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).
 
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  • #2


Hello,

Thank you for your detailed question. It is understandable that you are hesitant to post this type of question, as it is commonly assigned in introductory electrodynamics courses. However, I believe that it is appropriate to discuss this topic in this forum, as it is a general concept and not simply a homework problem.

To answer your question, yes, you may reasonably and simply use the relationship \vec{E} = \frac{1}{\varepsilon(\vec{x})}\vec{D} to find the electric field in regions with varying dielectric constants. This is a fundamental relationship in electromagnetism and is valid in all cases. In the case of two dielectric materials with different constants, it is correct to state that the electric 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.

Regarding your confusion about the placement of free charge, it is important to note that free charge is not affected by the presence of dielectric materials. Therefore, it is valid to place the same free charge on the entire length of the cylinder, regardless of the dielectric material surrounding it. This free charge is necessary to create the electric field in the first place, and it is not affected by the dielectric material.

In the problem from Griffiths' text, it appears that the author is enforcing different free charges in each region in order to match the electric fields at the interface between the two materials. This is a common technique in solving boundary value problems in electromagnetism, and it is valid in this case.

I hope this helps to clarify your understanding of this concept. If you have any further questions, please do not hesitate to ask. Good luck with your qualifying examination!
 

1. How does the dielectric constant affect the relationship between D-field and E-field?

The dielectric constant is a measure of how much a material can polarize in response to an electric field. When the dielectric constant is a function of space, it means that it varies throughout the material. This variation can affect the relationship between the D-field and E-field, as it changes the amount of polarization and therefore the strength of the electric field.

2. Is the D-field always equal to the E-field when the dielectric constant is a function of space?

No, the D-field and E-field are not always equal in this case. The D-field is equal to the E-field only in a vacuum or when the dielectric constant is constant throughout the material. When the dielectric constant is a function of space, the D-field and E-field can be different, depending on the polarization of the material at that point.

3. Can the D-field and E-field change direction when the dielectric constant is a function of space?

Yes, the D-field and E-field can change direction when the dielectric constant is a function of space. This is because the direction of the electric field is determined by the direction of the force on a positive test charge. When the dielectric constant changes, the force on the test charge can also change, causing the electric field to change direction.

4. How does the dielectric constant affect the electric potential when it is a function of space?

The dielectric constant affects the electric potential by changing the amount of work needed to move a unit charge between two points in the electric field. When the dielectric constant is a function of space, the electric potential can also vary throughout the material, as the amount of work needed to move a charge can change at different points.

5. Can the dielectric constant be negative or zero when it is a function of space?

Yes, the dielectric constant can be negative or zero when it is a function of space. This can occur in materials with certain properties, such as metals, where the polarization can be in the opposite direction of the applied electric field. In these cases, the dielectric constant can be negative, resulting in a negative D-field and E-field. A dielectric constant of zero would indicate a material with no ability to polarize, resulting in no D-field or E-field.

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