If we are having a dielectric medium with $\epsilon > \epsilon_0$ and vacuum $\epsilon_0$. I want to find the boundary conditions in this case. Therefore I do:

$$\nabla \cdot \vec D(\vec r)=\rho_{\text{free}}(\vec r)$$

By integrating this and also using Gauss's law you can find:

$$D_{n_{\text{outside}}} - D_{n_{\text{inside}}}= \eta_{\text{free}}(\vec r)$$

Then if we have no free charges: $$D_{n_{\text{outside}}} - D_{n_{\text{inside}}}= 0$$ and from here we find out: $$ D_{n_{\text{outside}}}=D_{n_{\text{inside}}}$$

where in all the above equations: $D_{n_{\text{outside}}}$ normal component of the D-Field outside the dielectric and $D_{n_{\text{outside}}}$ the normal component inside the dielectric.

I have 2 questions about two things:

  1. What is the free charge density here? The charge inside the dielectric or a charge distribution (characterized by a charge density function) in front of the dielectric?

  2. What do we mean when we say that we have no free charge density? That we lack charges inside the dielectric or outside of it?

  3. If you lack free charges, which are the source of the D-field how can you speak about the components of the D-field, when you have no D-field in the first place, since you lack the source that generates this field?


1 Answer 1

  1. Not sure what you mean by the charge inside the dielectric? The dielectric will be net neutral. The free charge density here refers to the density of mobile conduction charges on the interface between the media. It excludes the (static) charges associated with the induced dipoles in the media.

  2. That there are no mobile conduction charges at the interface and so $\eta_{\rm free}=0$.

  3. The D-field begins and ends on mobile conduction charges that are not on the interface. They could be anywhere else. It is not necessary to have a charge present at the position where you measure the D-field. It just means that the D-field must have zero divergence at that point.

  • $\begingroup$ 1. Since within the dielectric you have a Displacement field, and this field's divergence is equal to the free charge density, that means that inside the dielectric you must have these charges, which are different from the bound charges which arise from the non-uniformal polarisation inside the medium/within the volume $\endgroup$
    – imbAF
    Dec 4, 2021 at 15:46
  • $\begingroup$ 2. The mobile conduction charges, are charges that belong to what? To the dielectric, or to the charge distribution in front of the dielectric (picture a dielectric half room and we place a charge distribution in front of it) $\endgroup$
    – imbAF
    Dec 4, 2021 at 15:47
  • $\begingroup$ @imbAF A zero divergence means there is no free charge present and field lines do not begin or end. The presence of a field does not mean it has a divergence. The free charges do not "belong" to anything. In most problems there would not be a free charge at the interface. But if they have been placed there somehow then they need to be accounted for. The charge that is not at the interface is totally irrelevant to the boundary condition problem. $\endgroup$
    – ProfRob
    Dec 4, 2021 at 15:57
  • $\begingroup$ let me try and explain what I am after with an example. When you have a charge distribution/charged conductor and you apply Gauss's law, you can imagine what you are doing. basically you are enclosing the charge distribution inside a volume and from here you proceed accordingly. Now when you you apply this law for the D-Field, and let's assume that it's not zero, what are you enclosing inside the imaginary volume? A dielectric? A charge distribution with bound and free charges? And in analog way, integrating over this volume the free charge density, you'll get the free Charge $\endgroup$
    – imbAF
    Dec 4, 2021 at 16:10
  • $\begingroup$ @imbAF $\oint \vec{D}\cdot d\vec{a} = q_{\rm free}$ is Gauss's law. Not sure what you're asking about. In a dielectric you don't have free charge., so $\oint \vec{D} \cdot d\vec{a} = 0$. That doesn't mean the D-field is zero. If $\oint \vec{D} \cdot d\vec{a} \neq 0$ then it means there is a net conduction/free charge inside your enclosed volume. $\endgroup$
    – ProfRob
    Dec 4, 2021 at 16:58

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