# Electrostatics of Macroscopic Media

Macroscale (CM) is on the order of a meter. The micro scale (QM) is on the order of electrons or atoms, angstrom. Germs, cells, on the size of 1$μ$m, is considered macro due to the descriptions being CM. Transistors/ microchips are in the mesoscale, 1-100nm, can see both QM and CM effects. Grey area.

## Media

• Conductors: Free electron movement (We have been assuming superconductors)
• Dielectric Materials: (Insulators) restricted movement of electrons.

## Dipoles

$$\vec{d}$$. Put the origin in the middle of the $$\pm q$$ charges.

Aligning along the z-axis, $$r_+ = \sqrt{r^2 + |\vec{d}|^2/4+ r|\vec{d}|\cos\theta}$$. $$\frac{1}{r_+} = \frac{1}{r}\left(1+\frac{d^2}{2r^2}-\frac{d}{r}\cos\theta\right)^{-1}$$. When $$r\gg |\vec{d}|\equiv d$$, $$\Phi(r) = \frac{q}{4\pi\varepsilon_0}\left(\frac{1}{r_+}-\frac{1}{r_-}\right)$$. $$\frac{1}{r_+}\approx \frac{1}{r}\left(1+\frac{d}{2r}\cos\theta\right), \frac{1}{r_-}\approx \frac{1}{r}\left(1-\frac{d}{2r}\cos\theta\right)$$. $$\Phi(r\gg d) = \frac{q d}{4\pi\varepsilon_0}\frac{\cos\theta}{r^2} = \frac{p}{4\pi\varepsilon_0}\frac{\cos\theta}{r^2}$$. With $$\vec{p} = q\vec{d}$$. In general, $$\frac{pr\cos\theta}{4\pi\varepsilon_0 r^3} = \frac{\vec{p}\cdot\vec{r}}{4\pi\varepsilon_0 r^3}$$, the latter expression is independent of coordinate system.

$$\vec{p} = \int \vec{x}\rho(\vec{x})dV$$.

$$\vec{E} = -\nabla\Phi = \frac{p}{4\pi\varepsilon_0}\frac{2\cos\theta}{r^3}\hat{r} + \frac{p}{4\pi\varepsilon_0}\frac{\sin\theta}{r^3}\hat{\theta} = \frac{p}{4\pi\varepsilon_0 r^3}\left(2\cos\theta\hat{r} + \sin\theta\hat{\theta}\right)$$.

Far field. Recall, $$\Phi(\vec{x}) = \frac{1}{4\pi\varepsilon_0}\int\frac{\rho(\vec{x}')}{|\vec{x}-\vec{x}'}d^3 x'$$ and $$\frac{1}{|\vec{x}-\vec{x}'|} = \sum_{\ell,m} = \frac{4\pi}{2\ell+1}\frac{r_<^\ell}{r_>^{\ell+1}}Y_{\ell}^{m*}(\theta',\phi')Y_\ell^m(\theta,\phi)$$. So, for the far field, $$\Phi(\vec{x}) = \frac{1}{4\pi\varepsilon_0}\sum_{\ell m}\frac{4\pi}{2\ell+1}\left(\int Y_{\ell}^{m*}(\theta',\phi')r'^\ell\rho(\vec{x}')d^3 x'\right)\frac{Y_\ell^m(\theta,\phi)}{r^{\ell+1}} = \frac{1}{4\pi\varepsilon_0}\sum_{\ell m}\frac{4\pi}{2\ell+1}q_{\ell}^m\frac{Y_\ell^m(\theta,\phi)}{r^{\ell+1}} = \frac{1}{4\pi\varepsilon_0}\sum_{\ell m}\frac{4\pi}{2\ell+1}\frac{q_{\ell}^mY_\ell^m(\theta,\phi)}{r^{\ell+1}} = \frac{1}{4\pi\varepsilon_0}\sum_{\ell m}\frac{C_\ell(\theta,\phi)}{r^{\ell+1}} = \frac{1}{4\pi\varepsilon_0}\left(\frac{C_0}{r} + \frac{C_1}{r^2} + \frac{C_2}{r^3} + \cdots\right)$$. Note, $$q_{00} = \frac{q}{\sqrt{4\pi}}$$.

For the HW: $$\Phi_1 = \frac{1}{4\pi\varepsilon_0}\frac{\vec{p}\cdot\vec{x}}{r^3}$$. $$\Phi_2 = \frac{1}{4\pi\varepsilon_0}\frac{1}{2}\sum_{ij}\frac{Q_{ij}(x_ix_j)}{r^5}$$.

Say we have some charge distribution $$\rho(\vec{x})$$ and we apply an external field, $$\vec{E}_{ex} = -\nabla\Phi_{ex}$$. The work required to assemble the distribution is then, $$W = \int\Phi_{ex}(\vec{x})\rho(\vec{x})d^3x$$. Say we have a slowly varying field, thus a slowly varying potential around the distribution. $$\Phi_{ex}(\vec{x}) = \Phi_{ex}(0) + \vec{x}\cdot\nabla\Phi_{ex}(0) + \cdots = \Phi_{ex}(0) - \vec{x}\cdot\vec{E}_{ex}$$. So, $$W = \int \Phi_{ex}\rho dV - \int\vec{x}\cdot\vec{E} + \cdots\rho dV = q\Phi(0) - \vec{p}\cdot\vec{E} + \cdots$$. The $$\vec{p}\cdot\vec{E}$$ is the dominant perturbation term in Light-matter interactions.

### Polarization

Apply a uniform electric field to a dielectric. Then, we get a bunch of atomic dipoles with polarization $$\vec{p}$$ that has units Coulomb times meter. The macroscopic polarization is then $$\vec{P} = \frac{\sum_i\vec{p}_i}{\Delta V}$$ which has units Coulomb/ meter squared.

Assume, we have a potential felt by the electron to be roughly $$V(x) = \frac{1}{2}kx^2$$. This is called a linear medium and it assumes the electric field is not too strong. If the electric field is too strong, it may strip electrons. In most cases, we can assume a linear response, thus a quadratic potential. The force $$F = -kx$$ is then felt by the electron. $$\chi = \frac{e}{k}E$$, $$\vec{p} = e\chi = \frac{e^2}{k}\vec{E} = \gamma_\alpha \vec{E}$$. Remember that this relationship is just an approximation. Most general (non-isotropic materials), $$P_i=\gamma_{\alpha ij}E_j$$. Example: Calcite medium has a rectangular lattice and is a long rod along the z-axis.

Consider a small volume in the dielectric at $$\vec{x}'$$ in a medium polarized to $$\vec{P}(\vec{x}')$$. What is the potential $$\Phi(\vec{x})$$? $$d\vec{p} = \vec{P}(\vec{x}')d^3x'$$. So, $$\Phi(\vec{x}) = \frac{1}{4\pi\varepsilon_0} \int\frac{\vec{P}(\vec{x}')\cdot(\vec{x}-\vec{x}')d^3x'}{|\vec{x}-\vec{x}'|^3} = \frac{1}{4\pi\varepsilon_0} \int \vec{P}(\vec{x}')\cdot\nabla'\frac{1}{|\vec{x}-\vec{x}'|}d^3x' = \frac{1}{4\pi\varepsilon_0}\int d^3 x' \left(\nabla'\cdot\left(\frac{\vec{P}(\vec{x}')}{|\vec{x}-\vec{x}'|}\right)-\frac{\nabla'\cdot\vec{P}(\vec{x}')}{|\vec{x}-\vec{x}'|}\right) = \frac{1}{4\pi\varepsilon_0}\left(\oint_S\frac{\vec{P}(\vec{x}')\cdot\hat{n}'}{|\vec{x}-\vec{x}'|}da' - \int\frac{\nabla'\cdot\vec{P}(\vec{x}')}{|\vec{x}-\vec{x}'|}d^3x'\right)$$. From this, we can define the bound charges as $$\sigma_b(\vec{x}') = \vec{P}(\vec{x}')\cdot\hat{n}',\rho_b(\vec{x}') = \nabla'\cdot\vec{P}(\vec{x}')$$. Thus, $$\Phi(\vec{x}) = \frac{1}{4\pi\varepsilon_0}\left(\oint\frac{\sigma_b(\vec{x}')}{|\vec{x}-\vec{x}'|}da' - \int\frac{\rho_b(\vec{x}')}{|\vec{x}-\vec{x}'|}d^3x'\right)$$.

Aside: $$\nabla\cdot(f\vec{A}) = \nabla f\cdot \vec{A} + f\nabla\cdot\vec{A}$$

Aside: In a quantum model, $$\langle \vec{p}\rangle = e\langle\vec{x}\rangle\propto E$$.

### Displacement

$$\vec{D} = \varepsilon_0\vec{E}+\vec{P}$$.

$$\rho_{total} = \rho_{free} + \rho_{bound}$$.

$$\varepsilon_0\nabla\cdot\vec{E} = \rho_{total} = \rho_{f} - \nabla\cdot\vec{P}$$. So, $$\nabla\cdot(\varepsilon_0\vec{E} + \vec{P}) = \rho_f$$. Thus, $$\nabla\cdot\vec{D}=\rho_f$$.

$$\nabla\times\vec{E} = 0$$.

For polarization aligned with the electric field, $$\vec{D} = \varepsilon\vec{E} = \varepsilon_0(1+\chi_e)\vec{E}$$.

Define, $$\varepsilon = \varepsilon_0(1+\chi_e)$$.

$$\vec{P} = \varepsilon_0\chi_e\vec{E}$$ with $$\chi_e$$ called the electric susceptibility.

$$\varepsilon$$ is the electric permitivity and $$\varepsilon_0$$ is the vacuum permitivity.

We may also write $$\vec{D}=\varepsilon_0\varepsilon_r\vec{E}$$ with $$\varepsilon_r = \frac{\varepsilon}{\varepsilon_0} = 1+\chi_e$$ called the dielectric constant.

#### Maxwell in Macroscopic Media

$$\nabla\cdot\vec{D}=\rho_f$$ and $$\nabla\times E=0$$.

#### Recall - Next Lecture’s notes

$$\vec{P} = \frac{\sum_i\vec{p}_i}{V_{olume}}$$.

In a linear isotropic uniform medium, $$\vec{P} = \varepsilon_0\chi_e\vec{E}$$.

Linear means $$\vec{P}\propto\vec{E}$$.

Isotropic means that $$\vec{P}||\vec{E}$$ is parallel to.

Consider at the boundary between two dielectrics with polarizations $$\vec{P}_1$$ and $$\vec{P}_2$$. Consider a pillbox at the interface, $$\int\vec{D}\cdot\hat{n}da=q$$. Assume we have free surface charge at the interface of $$\sigma$$. Then, $$q = \sigma\Delta S$$.

$$(\vec{P}_2\cdot\hat{n}_{21}-\vec{P}_1\cdot\hat{n}_{21})\delta S = \sigma\Delta S$$ So, $$\vec{P_{2n}-P_{1n}} = \sigma$$. Then, if there is no charge the displacement is continuous at the boundary.

Consider the electric field near the interface. From $$\nabla \vec{E} = 0$$, $$\oint\vec{E}\cdot d\vec{\ell} = 0 = \vec{E_2}\Delta\vec{\ell}-\vec{E}_1\Delta\vec{\ell}=0$$. So, $$\vec{E_2}=\vec{E}_1$$. This is equivalent to saying the potential is continous at the interface.

Note: $$\nabla\cdot\vec{E}=\frac{\rho_f}{\varepsilon}$$ so $$\nabla^2\Phi = \frac{\rho}{\varepsilon}$$.

#### Example

Consider a point charge near an interface between two dielectrics, $$\varepsilon_2,\varepsilon_1$$ and $$q$$ is in the $$\varepsilon_1$$ side.

Note, we did the $$\varepsilon\to\infty,\varepsilon_1=\varepsilon_0$$ case before with the conducting plane.

In this case, we have non-constant potentials $$\Phi_2,\Phi_1$$ on either side of the interface. We can apply the image charge method separately for the two regions.

For solving the right side potential, $$\Phi_1$$. Assume $$\varepsilon_2=\varepsilon_1$$ and put a charge $$q'$$ at at the same distance away from the interface as $$q$$. This will be our trial solution. let $$R_1$$ be the distance between $$q$$ and $$\vec{x}$$ and $$R_2$$ be the distance between $$q'$$ and $$\vec{x}$$. $$\Phi_1 = \frac{1}{4\pi\varepsilon_1}\left(\frac{q}{R_1}+\frac{q'}{R_2}\right)$$

To figure out $$\Phi_2$$, assume $$\varepsilon_1=\varepsilon_2$$ and put a charge $$q''$$ at distance $$d$$. $$\Phi_2 = \frac{1}{4\pi\varepsilon_2}\frac{q''}{R_1}$$.

Our first boundary condition, the tangential components of the electric field is continous, $$E_{1t}=E_{2t}\Rightarrow \Phi_1=\Phi_2$$. Then, for $$z=0$$, $$R_1=R_2$$. So, $$\frac{1}{\varepsilon_2}\frac{q''}{R_1} = \frac{1}{\varepsilon_1}\left(\frac{q}{R_1}+\frac{q'}{R_2}\right)$$. Thus, $$q+q' = \frac{\varepsilon_1}{\varepsilon_2}q''$$.

The second boundary condition, $$\vec{D}=-\varepsilon\nabla\Phi$$, $$-4\pi D_{1z} = \frac{\partial}{\partial z}\left(\frac{q}{R_1}+\frac{q'}{R_2}\right)_{z=0}$$ and $$-4\pi D_{2z}=\frac{\partial}{\partial z}\left(\frac{q'}{R_1}\right)_{z=0}$$.

Created: 2023-06-25 Sun 02:35

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