Polarization and D field

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[math]\mathbf{p} = \alpha \mathbf{E}[/math]
Torque on a dipole
[math]\mathbf{\tau} = (\mathbf{r}_+\times\mathbf{F}_+) + (\mathbf{r}_- \times\mathbf{F}_-)[/math]
[math]\mathbf{\tau} = \mathbf{p}\times \mathbf{E}[/math]
Force on a dipole
[math]\mathbf{F} = q(\mathbf{E}_+ -\mathbf{E}_-) = q(\mathbf{d}\cdot\nabla)\mathbf{E}[/math]
[math]\mathbf{F} = (\mathbf{p}\cdot\nabla)\mathbf{E}[/math]

Polarization (dielectric)

P is defined as dipole moment per unit volume

Then, since dipole moment in each volume element is [math]\mathbf{p} = \mathbf{P}d\tau' [/math], the potential due to material polarization is:

[math]V(\mathbf{r}) = {1\over 4\pi\epsilon_0} \int{\mathbf{P}(\mathbf{r}')\cdot (\mathbf{r}-\mathbf{r}')\over |\mathbf{r}-\mathbf{r}'|^3} d\tau'[/math]

By using [math]\nabla'\left({1 \over |\mathbf{r}-\mathbf{r}'|} \right) = {\mathbf{r}-\mathbf{r}' \over |\mathbf{r}-\mathbf{r}'|^3}[/math]

[math] \begin{align} V &= {1\over 4\pi\epsilon_0} \int_V \mathbf{P}\cdot\nabla'\left({1 \over |\mathbf{r}-\mathbf{r}'|} \right) d\tau' \\ &= {1\over 4\pi\epsilon_0}\left[ \int_V \nabla'\cdot\left( {\mathbf{P}\over|\mathbf{r}-\mathbf{r}'|} \right)d\tau' - \int_V {\nabla'\cdot\mathbf{P} \over |\mathbf{r}-\mathbf{r}'|} d\tau' \right] \\ &= {1\over 4\pi\epsilon_0} \oint_S {\mathbf{P}\over |\mathbf{r}-\mathbf{r}'|} \cdot d\mathbf{a}' -{1\over 4\pi\epsilon_0}\int_V {\nabla'\cdot\mathbf{P} \over |\mathbf{r}-\mathbf{r}'|} d\tau' \end{align} [/math]

It is natural to define surface and volume charge density due to material polarization as follows:

[math]\sigma_b \equiv \mathbf{P}\cdot\hat{\mathbf{n}}[/math]
[math]\rho_b \equiv -\nabla\cdot\mathbf{P} [/math]

so that potential can be expressed as

[math] V = {1\over 4\pi\epsilon_0} \oint_S {\sigma_b \over |\mathbf{r}-\mathbf{r}'|} \cdot da' +{1\over 4\pi\epsilon_0}\int_V {\rho_b \over |\mathbf{r}-\mathbf{r}'|} d\tau' [/math]

D field (displacement)

Since total charge can be regarded as sum of bound charge and free charges,

[math]\rho = \rho_b + \rho_f[/math]

When it is plugged in the r.h.s. of Gauss's law, we can write

[math]\epsilon_0 \nabla\cdot\textbf{E} = \rho_b + \rho_f = -\nabla\cdot\textbf{P} + \rho_f [/math]

So it is convenient to define a D field as

[math]\textbf{D} \equiv \epsilon_0\textbf{E}+\textbf{P}[/math]

so that we can write the Gauss's law as

[math]\nabla\cdot\textbf{D} = \rho_f[/math]

Boundary conditions

From [math]\nabla\cdot\textbf{D} = \rho_f[/math], we have

[math]D^{\perp}_{above}-D^{\perp}_{below} = \sigma_f[/math]

Also, for the parallel components,

[math]\mathbf{D}^{\parallel}_{above}-\mathbf{D}^{\parallel}_{below} = \mathbf{P}^{\parallel}_{above}-\mathbf{P}^{\parallel}_{below}[/math]

Electric susceptibility

For many substances, polarization is proportional to the electric field

[math]\mathbf{P} = \epsilon_0 \chi_e \mathbf{E}[/math]

where [math]\chi_e[/math] is called electric susceptibility, and is dimensionless.

In linear medium, displacement field then becomes

[math]\mathbf{D} = \epsilon_0 \mathbf{E} + \mathbf{P} = \epsilon_0(1+\chi_e)\mathbf{E} \equiv \epsilon\mathbf{E}[/math]

where [math]\epsilon[/math] is defined to be the permittivity of the material.

The relative permittivity [math]\epsilon/\epsilon_0 = 1+\chi_e [/math] is also called dielectric constant of the material.

Example 2 & 3 of Griffiths pp174-178

Uniformly polarized sphere of radius R

Surface charge approach

[math]\sigma_b = \mathbf{P}\cdot\hat{\mathbf{n}} = P \cos\theta[/math]
[math]V(r,\theta) = \frac{P}{3\epsilon_0}r\cos\theta [/math]
[math]\mathbf{E} = -\nabla V = -\frac{P}{3\epsilon_0}\hat{\mathbf{z}} = -\frac1{3\epsilon_0}\mathbf{P}[/math]
[math]V(r,\theta) = \frac{P}{3\epsilon_0} \frac{R^3}{r^2} \cos\theta = \frac1{4\pi\epsilon_0}\frac{\mathbf{p}\cdot\hat{\mathbf{r}}}{r^2} [/math]
[math]\mathbf{E} = -\nabla V = \frac1{4\pi\epsilon_0}\frac{3(\mathbf{p}\cdot\mathbf{\hat{r}})\mathbf{\hat{r}}-\mathbf{p}}{r^3} [/math]

Total charge approach

A positive sphere and a negative sphere with total charge q each.


Field between two uniformly charged sphere

[math]\mathbf{E} = -\frac1{4\pi\epsilon_0}\frac{q\mathbf{d}}{R^3} = -\frac1{3\epsilon_0}\mathbf{P}[/math]

Treat the system as two point charges separated by distance d, which makes a dipole