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, $r_{+} = \sqrt{r^{2} + | \vec{d} |^{2} / 4 + r | \vec{d} | \cos θ}$ . $\frac{1}{r_{+}} = \frac{1}{r} {(1 + \frac{d^{2}}{2 r^{2}} - \frac{d}{r} \cos θ)}^{- 1}$ . When $r ≫ | \vec{d} | \equiv d$ , $Φ (r) = \frac{q}{4 π ε_{0}} (\frac{1}{r_{+}} - \frac{1}{r_{-}})$ . $\frac{1}{r_{+}} \approx \frac{1}{r} (1 + \frac{d}{2 r} \cos θ), \frac{1}{r_{-}} \approx \frac{1}{r} (1 - \frac{d}{2 r} \cos θ)$ . $Φ (r ≫ d) = \frac{q d}{4 π ε_{0}} \frac{\cos θ}{r^{2}} = \frac{p}{4 π ε_{0}} \frac{\cos θ}{r^{2}}$ . With $\vec{p} = q \vec{d}$ . In general, $\frac{p r \cos θ}{4 π ε_{0} r^{3}} = \frac{\vec{p} \cdot \vec{r}}{4 π ε_{0} r^{3}}$ , the latter expression is independent of coordinate system.

$\vec{p} = \int \vec{x} ρ (\vec{x}) d V$ .

$\vec{E} = - \nabla Φ = \frac{p}{4 π ε_{0}} \frac{2 \cos θ}{r^{3}} \hat{r} + \frac{p}{4 π ε_{0}} \frac{\sin θ}{r^{3}} \hat{θ} = \frac{p}{4 π ε_{0} r^{3}} (2 \cos θ \hat{r} + \sin θ \hat{θ})$ .

Far field. Recall, $Φ (\vec{x}) = \frac{1}{4 π ε_{0}} \int \frac{ρ ({\vec{x}}^{'})}{| \vec{x} - {\vec{x}}^{'}} d^{3} x^{'}$ and $\frac{1}{| \vec{x} - {\vec{x}}^{'} |} = \sum_{ℓ, m} = \frac{4 π}{2 ℓ + 1} \frac{r_{<}^{ℓ}}{r_{>}^{ℓ + 1}} Y_{ℓ}^{m *} (θ^{'}, ϕ^{'}) Y_{ℓ}^{m} (θ, ϕ)$ . So, for the far field, $Φ (\vec{x}) = \frac{1}{4 π ε_{0}} \sum_{ℓ m} \frac{4 π}{2 ℓ + 1} (\int Y_{ℓ}^{m *} (θ^{'}, ϕ^{'}) r^{' ℓ} ρ ({\vec{x}}^{'}) d^{3} x^{'}) \frac{Y_{ℓ}^{m} (θ, ϕ)}{r^{ℓ + 1}} = \frac{1}{4 π ε_{0}} \sum_{ℓ m} \frac{4 π}{2 ℓ + 1} q_{ℓ}^{m} \frac{Y_{ℓ}^{m} (θ, ϕ)}{r^{ℓ + 1}} = \frac{1}{4 π ε_{0}} \sum_{ℓ m} \frac{4 π}{2 ℓ + 1} \frac{q_{ℓ}^{m} Y_{ℓ}^{m} (θ, ϕ)}{r^{ℓ + 1}} = \frac{1}{4 π ε_{0}} \sum_{ℓ m} \frac{C_{ℓ} (θ, ϕ)}{r^{ℓ + 1}} = \frac{1}{4 π ε_{0}} (\frac{C_{0}}{r} + \frac{C_{1}}{r^{2}} + \frac{C_{2}}{r^{3}} + \dots)$ . Note, $q_{00} = \frac{q}{\sqrt{4 π}}$ .

For the HW: $Φ_{1} = \frac{1}{4 π ε_{0}} \frac{\vec{p} \cdot \vec{x}}{r^{3}}$ . $Φ_{2} = \frac{1}{4 π ε_{0}} \frac{1}{2} \sum_{i j} \frac{Q_{i j} (x_{i} x_{j})}{r^{5}}$ .

Say we have some charge distribution $ρ (\vec{x})$ and we apply an external field, ${\vec{E}}_{e x} = - \nabla Φ_{e x}$ . The work required to assemble the distribution is then, $W = \int Φ_{e x} (\vec{x}) ρ (\vec{x}) d^{3} x$ . Say we have a slowly varying field, thus a slowly varying potential around the distribution. $Φ_{e x} (\vec{x}) = Φ_{e x} (0) + \vec{x} \cdot \nabla Φ_{e x} (0) + \dots = Φ_{e x} (0) - \vec{x} \cdot {\vec{E}}_{e x}$ . So, $W = \int Φ_{e x} ρ d V - \int \vec{x} \cdot \vec{E} + \dots ρ d V = q Φ (0) - \vec{p} \cdot \vec{E} + \dots$ . 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 $\vec{P} = \frac{\sum_{i} {\vec{p}}_{i}}{Δ V}$ which has units Coulomb/ meter squared.

Assume, we have a potential felt by the electron to be roughly $V (x) = \frac{1}{2} k x^{2}$ . 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 $F = - k x$ is then felt by the electron. $χ = \frac{e}{k} E$ , $\vec{p} = e χ = \frac{e^{2}}{k} \vec{E} = γ_{α} \vec{E}$ . Remember that this relationship is just an approximation. Most general (non-isotropic materials), $P_{i} = γ_{α i j} E_{j}$ . 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 $Φ (\vec{x})$ ? $d \vec{p} = \vec{P} ({\vec{x}}^{'}) d^{3} x^{'}$ . So, $Φ (\vec{x}) = \frac{1}{4 π ε_{0}} \int \frac{\vec{P} ({\vec{x}}^{'}) \cdot (\vec{x} - {\vec{x}}^{'}) d^{3} x^{'}}{| \vec{x} - {\vec{x}}^{'} |^{3}} = \frac{1}{4 π ε_{0}} \int \vec{P} ({\vec{x}}^{'}) \cdot \nabla^{'} \frac{1}{| \vec{x} - {\vec{x}}^{'} |} d^{3} x^{'} = \frac{1}{4 π ε_{0}} \int d^{3} x^{'} (\nabla^{'} \cdot (\frac{\vec{P} ({\vec{x}}^{'})}{| \vec{x} - {\vec{x}}^{'} |}) - \frac{\nabla^{'} \cdot \vec{P} ({\vec{x}}^{'})}{| \vec{x} - {\vec{x}}^{'} |}) = \frac{1}{4 π ε_{0}} (\oint_{S} \frac{\vec{P} ({\vec{x}}^{'}) \cdot {\hat{n}}^{'}}{| \vec{x} - {\vec{x}}^{'} |} d a^{'} - \int \frac{\nabla^{'} \cdot \vec{P} ({\vec{x}}^{'})}{| \vec{x} - {\vec{x}}^{'} |} d^{3} x^{'})$ . From this, we can define the bound charges as $σ_{b} ({\vec{x}}^{'}) = \vec{P} ({\vec{x}}^{'}) \cdot {\hat{n}}^{'}, ρ_{b} ({\vec{x}}^{'}) = \nabla^{'} \cdot \vec{P} ({\vec{x}}^{'})$ . Thus, $Φ (\vec{x}) = \frac{1}{4 π ε_{0}} (\oint \frac{σ_{b} ({\vec{x}}^{'})}{| \vec{x} - {\vec{x}}^{'} |} d a^{'} - \int \frac{ρ_{b} ({\vec{x}}^{'})}{| \vec{x} - {\vec{x}}^{'} |} d^{3} x^{'})$ .

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

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

Displacement

$\vec{D} = ε_{0} \vec{E} + \vec{P}$ .

$ρ_{t o t a l} = ρ_{f r e e} + ρ_{b o u n d}$ .

$ε_{0} \nabla \cdot \vec{E} = ρ_{t o t a l} = ρ_{f} - \nabla \cdot \vec{P}$ . So, $\nabla \cdot (ε_{0} \vec{E} + \vec{P}) = ρ_{f}$ . Thus, $\nabla \cdot \vec{D} = ρ_{f}$ .

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

For polarization aligned with the electric field, $\vec{D} = ε \vec{E} = ε_{0} (1 + χ_{e}) \vec{E}$ .

Define, $ε = ε_{0} (1 + χ_{e})$ .

$\vec{P} = ε_{0} χ_{e} \vec{E}$ with $χ_{e}$ called the electric susceptibility.

$ε$ is the electric permitivity and $ε_{0}$ is the vacuum permitivity.

We may also write $\vec{D} = ε_{0} ε_{r} \vec{E}$ with $ε_{r} = \frac{ε}{ε_{0}} = 1 + χ_{e}$ called the dielectric constant.

Maxwell in Macroscopic Media

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

Recall - Next Lecture’s notes

$\vec{P} = \frac{\sum_{i} {\vec{p}}_{i}}{V_{o l u m e}}$ .

In a linear isotropic uniform medium, $\vec{P} = ε_{0} χ_{e} \vec{E}$ .

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, $\int \vec{D} \cdot \hat{n} d a = q$ . Assume we have free surface charge at the interface of $σ$ . Then, $q = σ Δ S$ .

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

Consider the electric field near the interface. From $\nabla \vec{E} = 0$ , $\oint \vec{E} \cdot d \vec{ℓ} = 0 = \vec{E_{2}} Δ \vec{ℓ} - {\vec{E}}_{1} Δ \vec{ℓ} = 0$ . So, $\vec{E_{2}} = {\vec{E}}_{1}$ . This is equivalent to saying the potential is continous at the interface.

Note: $\nabla \cdot \vec{E} = \frac{ρ_{f}}{ε}$ so $\nabla^{2} Φ = \frac{ρ}{ε}$ .

Example

Consider a point charge near an interface between two dielectrics, $ε_{2}, ε_{1}$ and $q$ is in the $ε_{1}$ side.

Note, we did the $ε \to \infty, ε_{1} = ε_{0}$ case before with the conducting plane.

In this case, we have non-constant potentials $Φ_{2}, Φ_{1}$ on either side of the interface. We can apply the image charge method separately for the two regions.

For solving the right side potential, $Φ_{1}$ . Assume $ε_{2} = ε_{1}$ 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 $\vec{x}$ . $Φ_{1} = \frac{1}{4 π ε_{1}} (\frac{q}{R_{1}} + \frac{q^{'}}{R_{2}})$

To figure out $Φ_{2}$ , assume $ε_{1} = ε_{2}$ and put a charge $q^{″}$ at distance $d$ . $Φ_{2} = \frac{1}{4 π ε_{2}} \frac{q^{″}}{R_{1}}$ .

Our first boundary condition, the tangential components of the electric field is continous, $E_{1 t} = E_{2 t} \Rightarrow Φ_{1} = Φ_{2}$ . Then, for $z = 0$ , $R_{1} = R_{2}$ . So, $\frac{1}{ε_{2}} \frac{q^{″}}{R_{1}} = \frac{1}{ε_{1}} (\frac{q}{R_{1}} + \frac{q^{'}}{R_{2}})$ . Thus, $q + q^{'} = \frac{ε_{1}}{ε_{2}} q^{″}$ .

The second boundary condition, $\vec{D} = - ε \nabla Φ$ , $- 4 π D_{1 z} = \frac{\partial}{\partial z} {(\frac{q}{R_{1}} + \frac{q^{'}}{R_{2}})}_{z = 0}$ and $- 4 π D_{2 z} = \frac{\partial}{\partial z} {(\frac{q^{'}}{R_{1}})}_{z = 0}$ .