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The question was due to my frustration of text books using Permittivity of free air in Maxwell's Equations while explaining situations in all mediums. Yet they don't even bother mentioning why that's the case. This question is to confirm my understanding of Gauss's law in a dielectric material.

For example, an imaginary enclosed surface within a dielectric material, and we put some free charges into the space within the imaginary enclosed surface. If we take surface integral of D or E, I believe the following two equations are equivalent:

[1] surface integral of D (calculated using free charges only) = all of the free charges you put in the enclosed surface.  

The dielectric material doesn't really matter in this case. My understanding is that, D is almost a geometrical distribution of the free charges "projected" onto the imaginary surface, so this equation is almost about geometry, not physics. That's why permittivity isn't even involved.

[2] surface integral of E (measured right at the imaginary surface) = ALL of the charges in the enclosed surface, the free ones you put in there, and the dipoles induced in the material. Divided by vacuum permittivity.

Most dipoles are fully enclosed, they don't contribute to total charges. The dipoles that are cut in half at the imaginary surface contribute to the total charges. And those half dipoles carries the opposite charges of your free charges, so the half dipoles weakens the E. E is the result of free charges and the material's resistance of the free charges' field.

P.S. I haven't think it thru yet, but I believe the permittivity and permeability in Maxwell's last equation should all be relative, not free air.

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  • \$\begingroup\$ Seems like you have largely answered your own question. What part remains unclear to you? \$\endgroup\$
    – rpm2718
    May 5, 2020 at 19:01
  • \$\begingroup\$ In the equation \$\nabla\cdot\vec{E}=\frac{\rho}{\epsilon_o}\$, the \$\epsilon_o\$ is the permittivity of free space, \$8.85\cdot 10^{-12} \frac{F}{m}\$ \$\endgroup\$
    – rpm2718
    May 5, 2020 at 19:08
  • \$\begingroup\$ @rpm2718 my apologies, "Maxwell's 4th equation" refers to the "curl of B = rate of changes in currents and E fields" \$\endgroup\$
    – eliu
    May 5, 2020 at 19:13
  • \$\begingroup\$ Yes, same -- permittivity of free space. And minor point....time derivative in the \$\nabla \times \vec{B}\$ equation is only on \$\vec{E}\$, not the current. Just a typo I imagine. \$\endgroup\$
    – rpm2718
    May 5, 2020 at 19:22
  • \$\begingroup\$ Thank you for the correction. I wasn't paying adequate attention to all details. But, regarding the free space permittivity in the Ampere's law, I have to think hard about it. Probably in another question. \$\endgroup\$
    – eliu
    May 5, 2020 at 19:28

1 Answer 1

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You are correct.

Allow me to rigorously demonstrate what you have said using the following definitions.

1.) $$\mathbf{D} = \epsilon \mathbf{E} = \epsilon_r \epsilon_0 \mathbf{E}$$ 2.) Polarization

$$P = \epsilon_0 \chi_e \mathbf{E}\ = np_a[Cm^{-2}]$$ n is the average dipoles per \$m^3\$, \$p_a [Q m]\$ is the average dipole moment.

3.) Electric Susceptibility of a Material: $$ \chi_e = \epsilon_r - 1$$ A constant of proportionality. It quantifies the material's ability to be polarized in an electric field.

4.) Polarization Charge Density:

$$\oint \mathbf{P} \cdot d\mathbf{s} = - \int_v\rho_p dv = -q_p \Longleftrightarrow \nabla \cdot \mathbf{P}= - \rho_p$$

Let \$q_f\$ be enclosed free charge. Let \$q_t\$ be enclosed total charge. Let \$q_p\$ be enclosed polarization charge.

$$q_f = \oint_s \mathbf{D} ds $$ and $$q_t = \oint_s \mathbf{\epsilon_0 E} ds $$

Then

$$q_f = \oint_s \mathbf{D} ds = \epsilon_r \oint_s \mathbf{\epsilon_0 E} ds = (\frac{P}{\epsilon_0 \mathbf{E}} + 1 ) \oint_s \mathbf{\epsilon_0 E} ds = \oint_s \mathbf{P} \cdot d\mathbf{s} + \oint_s \mathbf{\epsilon_0 E} ds = -q_p + q_t$$

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  • \$\begingroup\$ Vigorous indeed! Any thoughts on the Ampere's equation? it uses free air permittivity and permeability. I think we should use relative. \$\endgroup\$
    – eliu
    May 6, 2020 at 13:03
  • \$\begingroup\$ Can you post a new question and explain why you think it shouldn't be this way? \$\endgroup\$
    – Alex
    May 6, 2020 at 13:38
  • \$\begingroup\$ haven't thought it thru yet, unlike the Gauss's law here \$\endgroup\$
    – eliu
    May 6, 2020 at 14:23

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