0
$\begingroup$

In Griffiths, the electric displacement is written with just volume charge density, and that is because, he says, 'we cannot apply gauss law precisely at the surface of a dielectric, for here volume charge density blows up, taking the divergence of $E$ with it'. So I understand this with idea that volume charge density is charge/volume and at the surface $v$ goes to zero and the density blows up.

However, in problem 4.15(3ed), gauss law is used to find the electric field of a spherical shell with polarization. In this process, $Q$ total includes surface charge density!

So, my question is: how do I deal with the surface charge density when I study the dielectric?

One more. There is a comment like volume charge density is made by nonuniform polarization while uniform accumulates surface charge density. Then both charge density cannot exist simultaneously, but it's not. what is wrong in this?

$\endgroup$
1
$\begingroup$

First question: I do not have Griffiths book but I hope this is what you mean.

You can use Gauss law on a surface that encloses the surface charge, you just can't use it on the surface itself, because the electric field is infinite there. Mathematically you can state: $$\iint \mathrm{d}S \, \, \vec{n} \cdot \vec{D}= \sum_i \iiint_{V_i} \mathrm{d}V \, \, \rho(\vec{r})+ \iint_{S_i} \mathrm{d}S \, \, \pi(\vec{r})$$ Here you take the sum over the different homogeneous materials and there surface.

Actually you can see this intuitively considering a 1D situation. In the case of a surface charge there will be a jump in the electric field, which can be denoted by a heaviside function $E(x)=\mathbb{H}(x)$. The derivative of this electric field becomes a delta distribution, which represents a surface charge. Integrating this will again give you a heaviside function, but at $x=0$ this integration is not simply defined.

Second question: Maxwell's laws states $\nabla \cdot \epsilon_0\vec{E}=\rho-\nabla\cdot \vec{P}$ The right hand side represents the total charge. If there is only polarization and the polarization is homogeneous the divergence becomes zero, and there is no net charge. However at the boundary of two materials there will be a jump in polarization and a polarization charge is created.

When a material is not homogeneous there will be a volume charge, but there can still be a jump in polarization at the boundary and therefore both can exist simultaneous.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.