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.


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


\(\vec{d}\). 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.


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, \(\langle \vec{p}\rangle = e\langle\vec{x}\rangle\propto E\).


\(\vec{D} = \varepsilon_0\vec{E}+\vec{P}\).

\(\rho_{total} = \rho_{free} + \rho_{bound}\).

\(\varepsilon_0\nabla\cdot\vec{E} = \rho_{total} = \rho_{f} - \nabla\cdot\vec{P}\). 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 \(\nabla^2\Phi = \frac{\rho}{\varepsilon}\).


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

Author: Christian Cunningham

Created: 2023-06-25 Sun 02:35