# B Electromagnetism question

1. Aug 21, 2016

### Isaac0427

Why do people ignore polerization and magnetization when rewriting maxwells equations for different fields?

Thanks!

2. Aug 21, 2016

### Staff: Mentor

I'm sorry, I don't understand what you're referring to here. Can you give an explicit example?

3. Aug 21, 2016

### Isaac0427

Note: sorry, I'm having trouble with LaTeX today. For convenience, I am also going to have the electric constant be just ε.
Gauss' law for magnetism (integral):
∫∫∂VDdS=∫∫∫VρdV
and
∫∫∂VEdS=ε-1∫∫∫VρdV

In converting the equations from D to E, "they" left out polarization, P. Shouldn't the equation be
∫∫∂VEdS=ε-1(∫∫∫VρdV-∫∫∂VPdS)

Similarly, there is no magnetization or polarization term in Ampere's law.

4. Aug 21, 2016

### blue_leaf77

The auxiliary equation reads $\mathbf D = \epsilon \mathbf E = \epsilon_0 \mathbf E + \epsilon_0 \chi \mathbf E= \epsilon_0 \mathbf E + \mathbf P$ where $\epsilon = (1+\chi)\epsilon_0$ and $\mathbf P = \epsilon_0 \chi \mathbf E$.

5. Aug 21, 2016

### Isaac0427

But what does this have to do with my question?

6. Aug 21, 2016

### Isaac0427

If D=εE+P, where is the +P term put into account in maxwells equations?

7. Aug 21, 2016

### blue_leaf77

Your last equation in post #3 is wrong, in the RHS it should be $\epsilon_0$ instead of $\epsilon$.

8. Aug 21, 2016

### Isaac0427

Why don't you read my entire second post before you respond to it. Maybe the top of the post.

9. Aug 21, 2016

### Isaac0427

Again, can people just please answer my question- where are the P and M terms in some of maxwells equations, such as Gauss' law with the E field?

10. Aug 21, 2016

### blue_leaf77

Do you mean
? So what, I don't see any possible relation between that sentence with your doubt about the absence of P in Maxwell equations.
It's unnecessary to write Gauss equation in terms of P unless you are asked to find this quantity. If you want to express it in terms of P, you would end up with the third equation but with $\epsilon$ replaced by $\epsilon_0$.

11. Aug 21, 2016

### Isaac0427

So the third equation in post number 3 is correct (keeping in mind that I expressed the electric constant without subscripts)? Why is it unnecessary? Wouldn't that be ignoring a term in the equation? Since when is it ok to leave out a term and call the equation correct?

12. Aug 21, 2016

### blue_leaf77

You have to include the subscript in the last equation to make it different from the electric constant in the second equation. The epsilon in the second equation is the one inside a matter while that in the last equation is for vacuum.
Since the dooms day later may be?? No terms are being left out, instead they are only hidden in the appearing term.
I have got the impression that you haven't seen any equivalence between these two relations:
$$\int\int \mathbf E \cdot d\mathbf S = \frac{1}{\epsilon} \oint \rho_f dV$$
and
$$\int\int \mathbf E \cdot d\mathbf S = \frac{1}{\epsilon_0} \left( \oint \rho_f dV - \int\int \mathbf P \cdot d\mathbf S \right)$$
with $\epsilon = (1+\chi)\epsilon_0$, see these epsilons are different.
Look at eq. 23 in http://www-users.aston.ac.uk/~pearcecg/Teaching/PDF/LEC2.PDF .

13. Aug 21, 2016

### Isaac0427

The Wikipedia article on maxwells equations says that the epsilon in my second equation is also with a subscript 0.

14. Aug 21, 2016

### axmls

You're dealing with different charge densities in each equation. When you write $$\oint \vec{E} \cdot d\vec{A} = \frac{1}{\epsilon_0} \iiint_\Omega \rho \ dV$$ the $\rho$ is the total charge, i.e. the free charge plus the bound charge, whereas the proper equation using the $D$ field is $$\oint \vec{D} \cdot d\vec{A} = \iiint_\Omega \rho_f \ dV$$ using only the free charge.

More explicitly, starting from the equation for the $D$ field, let $\vec{D} = \epsilon_0 \vec{E} + \vec{P}$, so substituting this in:
$$\oint \vec{D} \cdot d\vec{A} = \iiint \rho_f \ dV \Rightarrow \oint \left(\epsilon_0\vec{E} + \vec{P}\right) \cdot d\vec{A} = \iiint_\Omega \rho_f \ dV$$ Then $$\oint \epsilon_0\vec{E} \cdot d\vec{A} = \iiint_\Omega \rho_f \ dV - \oint \vec{P} \cdot d\vec{A}$$
Applying the divergence theorem and combining the right side integrals:
$$\oint \epsilon_0\vec{E} \cdot d\vec{A} = \iiint_\Omega \rho_f - \nabla \cdot \vec{P} \ dV$$
Finally, $-\nabla \cdot \vec{P}$ is what we call the bound charge density $\rho_b$, so the integrand on the right side is $\rho = \rho_f + \rho_b$, or the total charge density including both free and bound charges, so we're left with $$\oint \vec{E} \cdot d\vec{A} = \frac{1}{\epsilon_0}\iiint_\Omega \rho \ dV$$

What is bound charge density? Well, consider a conducting place in an electric field. Because of the applied external field, the atoms in the material form small dipoles, each of which cancels out the field from the adjacent dipole, except on the edges of the material. This surface charge constitutes the bound charge density. This, by the way, demonstrates the importance of always being aware of explicitly what it is that your variables represent when applying equations.

15. Aug 21, 2016

### Isaac0427

Thank you, this clears everything up!

16. Aug 21, 2016

@Isaac0427 Your question is a very good one, and I do think the E&M textbooks would do well to present a simple case illustrating the concept of polarization $P$ before presenting the more complicated cases as they often do. Let me explain in more detail: Consider a cylinder (of finite length) of uniform spontaneous polarization $P$ along its axis with no external electric fields present.. The equation $- \nabla \cdot P=\rho_p$ can be used to compute any polarization charge density $\rho_p$ that arises. For this case of uniform $P$, $\$ $\nabla \cdot P=0$ everywhere but at the surface boundaries where there will be a surface charge density $\sigma_p$ given by $\sigma_p=P \cdot \hat{n}$. For this case, the polarization surface charge simply appears on the endfaces of the cylinder. The electric field can readily be computed everywhere using Coulomb's law. This part is simple-it gets more difficult, but this part was necessary to really understand how the polarization problems work. $\\$ What is often the case is instead of a spontaneous polarization, an internal electric field $E_i$ causes the material to be polarized, often in a linear fashion. The problem is though, when an external electric field $E_o$ is applied to a sample, because of gradients in the polarization $P$ (as well as the discontinuity in $P$ at the surface), polarization charges arise which alter the applied electric field $E_o$ so that the field in the material $E_i$ that causes the $P$ will be different from $E_o$. It is a complex self-consistent mathematics that has been found to only be simple for a couple of geometries such as a sphere. $\\$ In the case of a sphere in a uniform $E_o$ detailed calculations involving Legendre polynomials shows the polarization $P$ and $E_i$ that results is uniform. Thereby the only polarization charges are surface polarization charges. Assuming a uniform $P$ inside, you can readily show that $E_p$ from the surface polarization charges at the center of the sphere (using $\sigma_p=P \cdot \hat{n}$) is $E_p=-P/(3 \epsilon_o).$ (More detailed calculations show this $E_p$ is the same everywhere inside the sphere). Now $E_i=Eo+E_p$. Given a linear relationship $P=\chi E_i$, the $E_i$ and $P$ can readily be computed in terms of $E_o$ and $\chi$. Hopefully this example proves useful. $\\$ For a dielectric slab in a uniform $E_o$, the polarization $P$ is also uniform. (Suggestion is you compute the $E_p$ , $E_i$, and $P$ for this case, given applied $E_o$ and $\chi$ for the material). Most other geometries do not give a uniform $P$ so that the self-consistent mathematics can get quite complex.

Last edited: Aug 21, 2016
17. Aug 21, 2016

### Staff: Mentor

Think of $\vec P$ as "buried" inside $\vec D$: $$\oint \vec D \cdot d \vec a = \int \rho_{free} dV \\ \oint (\varepsilon_0 \vec E + \vec P) \cdot d \vec a = \int \rho_{free} dV$$ Similarly $\vec M$ is "buried" inside $\vec H$.

18. Aug 22, 2016

19. Aug 22, 2016

### Isaac0427

I get most of it, it's a little complicated with your wording, but I got the point.

20. Aug 22, 2016