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I'm studying electromagnetism, more specifically dielectrics. However, the concepts of bound charge and free charge have been somewhat confusing for me yet. The Griffiths' book deduces one of the boundary conditions on a surface between two dielectrics as an equation of the type:

$\epsilon_{up} E_{up} - \epsilon_{down} E_{down} = \sigma_f,$

where $\epsilon$ represents the respective permissivities and $E$ the respective eletric fields above and below the surface. But $\sigma_f$, the of free surface charge density is not very clear to me.

In what situations can I consider whether or not there is free charge at the interface between two dielectrics? With the presence of bound charge, should they not appear explicitly in boundary conditions? Why?

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  • $\begingroup$ I think it's the normal component of the electric field $E_{up} = \left(-\nabla V\right)\cdot \hat{\mathbf{n}}$ that should show up here instead of the potential. (Well technically it's the displacement field $\mathbf{D}$ that should show up but assuming the two media are linear, homogeneous and isotropic, it also works as $\mathbf{D} = \varepsilon \mathbf{E}$). The usual way the boundary condition is stated is as $(\mathbf{D}_1 - \mathbf{D}_2)\cdot\hat{\mathbf{n}} = \sigma_f$, where $\hat{\mathbf{n}}$ is the normal from medium 2 to medium 1. $\endgroup$ – Tob Ernack Jul 13 at 6:22
  • $\begingroup$ You're right. It is the electric field in place of potential. I'll correct the question, thank you very much. In fact, I am considering the media as linear, isotropic and homogeneous. $\endgroup$ – heod Jul 13 at 15:34
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You can imagine that the dielectric is composed of a large number of molecules and ions. The molecules/ions consist of a group of charged particles interacting together sufficiently strongly that they stay grouped together, hence the name bound charges for the charges stuck inside a molecule/ion. The molecule/ion as a whole can move around in the dielectric (a lot of movement in liquids or gases, but very little in solids), but the charges inside the molecule/ion stay together.

On the other hand, you can have free charges, which are charged particles that can move around freely in the dielectric (or at least, are not stuck inside a molecule). You can consider the ions' net charges as contributing to the free charge density. Another possibility is electrons moving around in a conductive material, or even things like molecules ionized under very strong electric fields.

Every molecule/ion can be modeled as a small volume containing a charge density distribution $\rho_i(\mathbf{x}, t)$ depending on position and time. The time dependence comes from the motion of charges inside the molecule/ion.

The electrostatic fields generated outside of the molecule/ion can be calculated exactly using Coulomb's law, which involves an integration over $\rho_i$ with respect to $\mathbf{x}$. But you can also use an approximation given by the multipole expansion, which allows you to model the molecule/ion approximately as a combination of a point charge $q_i$ and an ideal dipole moment $\mathbf{p}_i$ both located at the center of mass of the molecule/ion. The multipole expansion is essentially like a multivariable Taylor expansion.

Under the presence of an external electric field $\mathbf{E}_\text{ext}$, the bound charges in the molecules/ions rearrange in a certain way to align with the field. This induces a net dipole moment inside the molecules/ions, which depends on the electric field. A macroscopic external electric field $\mathbf{E}_\text{ext}$ will induce a dipole moment on a large number of molecules/ions, and the total induced dipole moment density (calculated as an average over a small volume containing lots of molecules/ions) is called the polarization density $\mathbf{P}$. The induced dipoles will in turn generate an internal electric field $\mathbf{E}_\text{int}$, calculated using the multipole expansion. The total electric field is then $\mathbf{E} = \mathbf{E}_\text{ext} + \mathbf{E}_\text{int}$. For linear, homogeneous and isotropic materials, we can write $\mathbf{P} = \chi \varepsilon_0\mathbf{E}$ where $\chi$ is the susceptibility of the dielectric and $\mathbf{E}$ is the total macroscopic electric field.

Hence the presence of the bound charges is already taken into account in the dependence of $\mathbf{P}$ on $\mathbf{E}$. In particular, they affect the value of the susceptibility $\chi$. This in turn affects the dielectric constant $\varepsilon = (1 + \chi)\varepsilon_0$ of the medium. From the microscopic Poisson equation, we have $\nabla\cdot\mathbf{E} = \frac{\rho_f + \rho_b}{\varepsilon_0}$ where $\rho_b$ is the bound charge density, which is included in the total microscopic charge density. You can also show that $\nabla\cdot\mathbf{P} = -\rho_b$.

We can combine the previous relationships to obtain the macroscopic Poisson equation, $\nabla\cdot\mathbf{D} = \rho_f$ where $\mathbf{D} = \varepsilon_0\mathbf{E} + \mathbf{P} = \varepsilon\mathbf{E}$. The $\rho_f$ is the free charge density, which might be, for example, charged ions stuck inside a solid dielectric, or ions in an aqueous solution. The bound charges are not present in $\rho_f$, they are contained in $\mathbf{P}$ and $\varepsilon$.

The boundary condition $(\mathbf{D}_2 - \mathbf{D}_1)\cdot\hat{\mathbf{n}}_{12} = \sigma_f$ can be derived from Poisson's equation and the divergence theorem.

Imagine that there are two dielectrics next to each other, and that at the interface there is a very thin layer of free charge (for example, it could be a very thin layer of ions on the surface of the dielectrics). In theory you could model the charge distribution as a continuous variation of $\rho_f(\mathbf{x})$ in a small region around the interface of the two dielectrics, and solve Poisson's equation uniformly over the whole space.

But since the charges are so concentrated in the interface, we might instead choose to model this as a discontinuity with a surface charge density $\sigma_f$ right on the interface. This is like taking a limit as the width of the region where ions are present goes to $0$, but while still maintaining the total charge. Then we can take an arbitrarily small rectangular box around the interface, and calculate the surface integral of $\mathbf{D}$. By Gauss's law, this is just the total charge inside the box, which equals $\sigma_f A$ (where $A$ is the area). The integral is just $(\mathbf{D}_2 - \mathbf{D}_1)\cdot\hat{\mathbf{n}}_{12} A$. So after dividing by the area $A$, we get the boundary condition.

So you can see that $\sigma_f$ is really just the same as $\rho_f$ except we model it as a very thin layer of charge right on the interface, and the boundary condition is just a special case of Poisson's equation in the limit where the charge density becomes discontinuous.

For more detailed treatment of electrostatics and dielectrics, I recommend reading Jackson's Classical Electrodynamics Chapter 4 (Multipoles, Electrostatics of Macroscopic Media, Dielectrics). He goes in a lot of detail about how exactly we model the dielectric and how the polarization depends on the electric field.

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  • $\begingroup$ Your answer was quite enlightening in many ways. Thank you! So from what I understand, bound charges are treated microscopically, while free charges are treated macroscopically, right? For this reason, is it not considered the bound charges in the boundary conditions (since macroscopically the net bound charge is zero)... Or am I wrong? $\endgroup$ – heod Jul 14 at 19:14
  • $\begingroup$ Yes, the bound charges are microscopic effects taken into account by $\mathbf{P}$ and $\varepsilon$. We abstract away the detailed structure of the charges inside the molecules/ions using two parameters, $q_i$ and $\mathbf{p}_i$ representing the total charge and dipole moment of the molecule/ion. You can say that the free charge density $\rho_f$ comes from the nonzero $q_i$ of ions, while the bound charge density $\rho_b$ comes from the divergence of the sum of all dipole moments $\mathbf{p}_i$ per unit volume. You're "seeing" the bound charges inside the molecule/ion that get at the surface. $\endgroup$ – Tob Ernack Jul 14 at 19:17
  • $\begingroup$ As a visual illustration, you can imagine that the molecules/ions have a charge structure such as ---+-++ for example. But we look at things from far away so we don't see with this kind of resolution. Instead we just see a dipole like (-+) and some net charge (-). The net charge (-) is a free charge, but the - and + inside the dipole (-+) are bound charges. $\endgroup$ – Tob Ernack Jul 14 at 19:20
  • $\begingroup$ The surface charge $\sigma_f$ in the boundary condition comes from lots of free charges concentrated near the surface of the dielectric. It would look something like that: $\newline$$\newline$$\newline$$|| (-)(+-)(+-)(+-)...$$\newline$$\newline$$\newline$$\newline$$\newline$$\newline$ $\newline$$\newline$$\newline$$|| (-)(+-)(+-)(+)(+-)...$$\newline$$\newline$$\newline$$\newline$$\newline$$\newline$ $\newline$$\newline$$\newline$$|| (+-)(+-)(+-)(-)(+-)...$$\newline$$\newline$$\newline$$\newline$$\newline$ $\endgroup$ – Tob Ernack Jul 14 at 19:33
  • $\begingroup$ Sorry about the formatting, SE doesn't allow newlines in comments so I tried to use a clumsy workaround. Anyway, the surface charge $\sigma_f$ comes from the (+) and (-) ions right next to the interface ||. The (+-) dipoles do not contribute to $\sigma_f$ even though they have some bound charge near the surface. $\endgroup$ – Tob Ernack Jul 14 at 19:35

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