Query on Electromagnetic Theory (Dielectric Boundary Conditions)

[warhammer]

TL;DR Summary

I have a query regarding a question from Electromagnetic Theory. I wish to confirm something from the same, hence have not posted about it on the Homework Help Section. Highlighting the issue below:

The given question from Electromagnetic Theory (which is based on Dielectric Boundary Conditions) is as follows:

Interface b/w two dielectric medium has a surface charge density (suppose xyz C / (m ^ 2) ). Using boundary condition find field in 1 (relative permittivity =xyz) if field in 2 (relative permitivitty = xyz) is given as: $a \hat x + b \hat y + c \hat z$ V/mNow my query is isn't this question incomplete? How are we supposed to find field in 1 if the interface, propagation vector or incidence plane is not given, because how else would we be able to apply Boundary Conditions?!

This is something am unable to gauge. Without any other information I'm at a loss how & which components I can classify as normal and tangential. Request someone to please guide if my understanding here is correct, and also advise on the "why" if it is incorrect.

 

[Andy Resnick]

I'm not sure if I fully understand your question (also, is this a homework problem?), but when there is a surface charge, the displacement field D undergoes a 'jump condition' across the interface:

https://web.mit.edu/6.013_book/www/chapter5/5.3.html

 

[warhammer]

I'm not sure if I fully understand your question (also, is this a homework problem?), but when there is a surface charge, the displacement field D undergoes a 'jump condition' across the interface:

https://web.mit.edu/6.013_book/www/chapter5/5.3.html

Thank you for your response. No this is not a homework problem.

I do know about the jump condition due to the presence of surface charge. I will reiterate the doubt I'm facing. In the question, no extra information such as plane of incidence, interface is along which plane, propagation is along which direction etc. is NOT revealed. So we cannot apply Boundary Conditions right, since we are at a loss to denote which component is tangential and which is normal (again, because interface is along which plane is just not given).

 

[vanhees71]

I think the problem is that you have two dielectrics with different permittivities $\epsilon_1$ and $\epsilon_2$ without free surface charges on their boundary surface.

You must note that in the usual formulation of the dielectric $\rho$ is the charge distribution of the "external" charges, i.e., all charges that you put in addition to the "bound" charges making up the dielectric. That's why you introduce the "displacement field", $\vec{D}$, as an auxilliary field, fulfilling
$$\vec{\nabla} \cdot \vec{D}=\rho.$$
Restricting ourselves to the case of electrostatics, the electric field then fulfills
$$\vec{\nabla} \times \vec{E}=0$$
and there is the "constitutive equation"
$$\vec{D}=\epsilon \vec{E}.$$
In general $\epsilon$ is a function of position, because it describes a property of the dielectric material, namely how it's polarized. It's simple to understand when thinking about the microscopic structure of the dielectric: It consists of atoms, which are such that all the electrons stay quite strongly bound to the atomic nuclei, i.e., there are no electrons which are quasi free to move within the medium as, e.g., in metals, where you have "conduction electrons" providing a good electric conductivity.

So all that happens to a dielectric when you bring in additional "external" charges is that the corresponding electric field separates the bound electrons a bit from their equilibrium positions such that the medium then consists of a lot of little dipoles, described microscopically as a polarization (i.e., the dipole moment density due to the induced dipoles).

Now even if there are no external charges, if you have two different dielectrics there is a surface charge along the boundary of these dielectrics. This is easily seen as follows. The external surface-charge density is 0 in this case, i.e., the normal component of $\vec{D}$ is continuous across the surface, i.e., making $\vec{n}$ the surface-normal-unit vector pointing from medium 2 into medium 1 you have
$$\vec{n} \cdot [\vec{D}(\vec{x}_1)-\vec{D}(\vec{x}_2)]=0,$$
where $\vec{x}_1$ is a point on the surface reached as a limit taken within medium 1 and $\vec{x}_2$ is the same point on the surface reached as a limit taken within medium 2. But this implies a jump in the electric field, because it means
$$\vec{n} \cdot [\epsilon_1 \vec{E}(\vec{x}_1) - \epsilon_2 \vec{E}(\vec{x}_2)]=0. \qquad (*)$$
Now the total surface charge (i.e., the surface charge due to the presence of the dieelectric media) is
$$\sigma_{\text{pol}}=\frac{1}{\epsilon_0} \vec{n} \cdot [\vec{E}(\vec{x}_1)-\vec{E}(\vec{x}_2)]=\frac{1}{\epsilon_0} \vec{n} \cdot \vec{E}(\vec{x}_1) (1-\epsilon_1/\epsilon_2).$$
In the last step I've used (*).