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In electromagnetism class we had a problem proposed which led me to some thoughts about the relationship between $\vec{D}$ and $\vec{E}$. I know it certainly $\textbf{is}$ a definition so it must be true always but let me explain:

In this excercise we have a spherical capacitor (inner radius $r_1 < r_2$ outer radius) with a dielectric permittivity $\varepsilon = \varepsilon_0 \left( k_1 + k_2 \cos^2 \theta \right)$, which is obviously dependent on the spherical coordinate $\theta$. A conductor is an equipotential surface so the electric potential $V$ can't depend on $\theta$. If we assume the inner terminal is positive charged with $Q$, we can compute: as $\vec{\nabla}\cdot \vec{D} = \rho$, we have $\vec{D} = \frac{Q}{4\pi r^2}$ and $Q = \int \int_S \vec{D} \cdot \hat{n} \mathrm{d}S'$ = $\int \int_S \varepsilon \left(\theta\right) E(r) \mathrm{d}S' = 4 \pi \varepsilon_0 E(r) r^2 \left( k_2 + k_2/3 \right)$, so $\vec{E} = \frac{Q}{4\pi \varepsilon_0 r^2} \frac{1}{\left( k_2 + k_2/3 \right)} \neq \varepsilon\left(\theta \right)^{-1} \vec{D}$. I know this makes sense, as because of the simmetry of this problem, the displacement field cannot depend on $\theta$ and, because of the fact that $V$ doesn't depend on $\theta$, $\vec{E}$ has only a radial component and also independent on $\theta$.

What is happening? Why is $\vec{D} \neq \varepsilon(\theta) \vec{E}$?

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    $\begingroup$ Unless you are told that the medium is Linear, Isotropic and Homogeneous, then you cannot assume that $\mathbf {D} = \epsilon \mathbf{E}$. Your medium is certainly not homogeneous. $\endgroup$
    – ProfRob
    Commented Feb 22, 2020 at 13:41
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    $\begingroup$ Also, why do you assume that the D-field cannot be a function of $\theta$ ? The situation clearly is not spherically symmetric. $\endgroup$
    – ProfRob
    Commented Feb 22, 2020 at 13:52
  • $\begingroup$ @RobJeffries as far as I know, the displacement field depends on free charges, given by the relation $\vec{\nabla} \cdot \vec{D} = \rho$, by definition. So from $\vec{D}$'s point of view it is certainly symmetric. $\endgroup$
    – Pablo
    Commented Feb 22, 2020 at 17:54
  • $\begingroup$ You also cannot assume that the free charges are distributed uniformly on the conductors. $\endgroup$
    – ProfRob
    Commented Feb 22, 2020 at 18:14
  • $\begingroup$ I retract my comment about homogenity. Linear and isotropic is sufficient. $\endgroup$
    – ProfRob
    Commented Feb 22, 2020 at 19:19

3 Answers 3

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Whilst I think you can assume that the electric field is only radial and with a magnitude that depends only on radius (because of the equipotentials and the requirement that the E-field is normal to the conducting surfaces), I don't think you can assume the same for the D-field.

If we assume that $\vec{E} = E_r(r) \hat{r}$ and a linear dielctric, then the integral form of Gauss's law applied to a spherical surface of radius $R$ passing through the dielectric, tells us $$Q = \oint \vec{D} \cdot d\vec{A} = \int^{2\pi}_0 \int^{\pi}_0 \epsilon(\theta) E_r(R) R^2 \sin \theta\ d\theta\ d\phi$$

In this case $\epsilon(\theta) = \epsilon_0 (k_1 + k_2\cos\theta)$. So $$Q = 2\pi \epsilon_0 E_r(R)R^2 \int^{\pi}_0 k_1 \sin \theta + \frac{1}{2}k_2\sin(2\theta)\ d\theta = 4\pi k_1 \epsilon_0 E_r(R)R^2 $$

Thus $$\vec{E} = \frac{Q}{4\pi k_1 \epsilon_0 r^2}\ \hat{r}$$

Assuming a linear, isotropic medium, then $$\vec{D} = \epsilon \vec{E} = \frac{(k_1 + k_2 \cos \theta)Q}{4\pi k_1 r^2}\ \hat{r}$$

If you integrate this over a spherical surface, you will of course find that it comes to $Q$, but the fact that $\vec{D}$ depends on $\theta$ is telling you that the charge is not uniformly distributed over the surfaces.

If we zoom in on a piece of the inner sphere (so close that it looks flat) and draw a Guassian pillbox straddling the inner conductor surface, then we can say that, since there is no D-field inside the conductor, then if the charge per unit area is $\sigma$, $$\vec{D}(r_1) \cdot d\vec{A} = \sigma \ dA$$ i.e. $$ \sigma = D(r_1) = \epsilon E = \frac{(k_1 + k_2 \cos \theta)Q}{4\pi k_1 r_1^2}$$

Thus I think the root cause of your difficulties is assuming that the D-field has no $\theta$ dependence and that the charge is uniformly distributed on the conductors.

As to what circumstances you can assume $\vec{D} = \epsilon \vec{E}$, though that is a red herring in the case of this question, it is that the medium must be linear and isotropic. Note the comment by @jamison , you can still use the linear relationship at each point even in an inhomogeneous medium.

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  • $\begingroup$ I'm actually not even sure that you can assume that $\vec{E}$ obeys the spherical symmetry. The asymmetry of the permittivity means that there could be an asymmetric bound charge distribution, and $\vec{E}$ responds to both free charges and bound charges. $\endgroup$ Commented Apr 23, 2021 at 19:35
  • $\begingroup$ @MichaelSeifert Hmm, but how would you arrange for an asymmetric E-field given the two spherically symmetric equipotentials it is sandwiched by? $\endgroup$
    – ProfRob
    Commented Apr 23, 2021 at 20:16
  • $\begingroup$ The potential would obey Poisson's equation (not Laplace's equation) with an asymmetric source term. It'd be like putting a charged sphere between two infinite conducting sheets — I wouldn't expect the electric field to be orthogonal to the plates anymore. $\endgroup$ Commented Apr 23, 2021 at 20:35
  • $\begingroup$ (Note that $\vec{D}$ doesn't have an associated potential, since $\vec{\nabla} \times \vec{D} \neq 0$ in general; so we can't argue that $\vec{D}$ is strictly radial either.) $\endgroup$ Commented Apr 23, 2021 at 20:47
  • $\begingroup$ I take the point about Poisson's eqn. But if the E-field is radial it still reduces to Laplace, because E is perpendicular to $\nabla \epsilon$. E must be radial at $\theta=0$. To get a theta component elsewhere then implies that $E_\theta =f(\theta)$ and there would be a corresponding $D_\theta$. But this would give D a non-zero divergence in a region with no free charge? @MichaelSeifert $\endgroup$
    – ProfRob
    Commented Apr 24, 2021 at 8:05
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$\mathbf D=\epsilon \mathbf E$ is only valid for when there material is a linear dielectric that is isotropic and homogeneous. In your case this is not valid with your $\theta$ dependence, so this is why you do not see this relationship.

Remember, $\mathbf D$ is defined as $$\mathbf D=\epsilon_0\mathbf E+\mathbf P$$

When you have the properties described above, you then have $\mathbf P=\epsilon_0\chi\mathbf E$ which gets you to what you were proposing $$\mathbf D=\epsilon_0\mathbf E+\epsilon_0\chi\mathbf E=\epsilon\mathbf E$$

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    $\begingroup$ Even if the medium is not homogeneous, the constitutive relation $\mathbf{D}(\mathbf{r}) = \epsilon(\mathbf{r})\mathbf{E}(\mathbf{r})$ still holds for each point $\mathbf{r}$ (provided that the medium is isotropic and that the electromagnetic response is local). Thus, in this problem, the constitutive relationship must hold for each point in the dielectric layer. Any violation of this indicates a problem with the way in which the fields were computed. $\endgroup$
    – jsloan
    Commented Feb 22, 2020 at 18:29
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    $\begingroup$ This is incorrect. ${\mathbf D}= \epsilon {\mathbf E}$ works fine for ths question. $\endgroup$
    – ProfRob
    Commented Mar 10, 2020 at 17:52
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    $\begingroup$ @RobJeffries I agree; I was hasty in the posting of my answer. Unfortunately, I cannot delete it as the OP has already accepted the answer. $\endgroup$ Commented Mar 10, 2020 at 17:56
  • $\begingroup$ Hmm. Awkward... $\endgroup$
    – ProfRob
    Commented Mar 10, 2020 at 17:58
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Since $\vec D$ is a response of a medium to the applied electric field it can differ along each direction i.e the response is not isotropic. Thus in general $\vec D$ is given by the following: $$\vec D =\epsilon_0\overleftrightarrow \chi \vec E$$ where $\overleftrightarrow \chi$ is the susceptibility tensor. This is still under the linear response regime. For non-linear regime you may have complicated dependence of $\vec D$ on $\vec E$.

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  • $\begingroup$ Even for linear case your expression is not 'general' - spatial/temporal dispersion could be present. $\endgroup$
    – Cryo
    Commented Feb 23, 2020 at 0:26
  • $\begingroup$ Yes. But the tensor can have a functional dependence to express that. $\endgroup$ Commented Feb 23, 2020 at 4:34
  • $\begingroup$ It could, thats my point, but this is not what you wrote in the post. $\endgroup$
    – Cryo
    Commented Feb 23, 2020 at 10:25

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