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 $-4\pi D_{2z}=\frac{\partial}{\partial z}\left(\frac{q'}{R_1}\right)_{z=0}$.