16 May 2025
Electromagnetic Boundaries, Dielectric Waves, and Energy Flow
Boundary conditions, source-free wave equations, dielectric plane waves, and Poynting's theorem.
Let a smooth interface separate medium 1 from medium 2, and define $\hat{\mathbf n}$ to point from 1 to 2. The limiting forms of Maxwell’s integral equations determine which field components can jump.
Normal boundary conditions
Place a thin pillbox of face area $\Delta A$ across the interface. As its height tends to zero, the side flux vanishes. Gauss’s electric law leaves
\[\mathbf D_2\cdot\hat{\mathbf n}\,\Delta A -\mathbf D_1\cdot\hat{\mathbf n}\,\Delta A =\rho_s\Delta A,\]so
\[\boxed{\hat{\mathbf n}\cdot(\mathbf D_2-\mathbf D_1)=\rho_s}.\]$\rho_s$ is free surface-charge density in $\mathrm{C\,m^{-2}}$. Applying the same pillbox to magnetic flux gives
\[\boxed{\hat{\mathbf n}\cdot(\mathbf B_2-\mathbf B_1)=0}.\]Thus $B_n$ is always continuous, while $D_n$ jumps only by the free surface charge.
Tangential boundary conditions
Use a narrow rectangular loop of length $\Delta\ell$ parallel to the interface. Its area tends to zero with its height. Faraday’s flux term then vanishes for finite $\partial\mathbf B/\partial t$, leaving
\[\boxed{\hat{\mathbf n}\times(\mathbf E_2-\mathbf E_1)=\mathbf0}.\]For the Ampere-Maxwell law, a free sheet current $\mathbf K$ can remain finite as the loop height tends to zero. With $\mathbf K$ in $\mathrm{A\,m^{-1}}$,
\[\boxed{\hat{\mathbf n}\times(\mathbf H_2-\mathbf H_1)=\mathbf K}.\]If $\rho_s=0$ and $\mathbf K=\mathbf0$, the continuous combinations are $D_n$, $B_n$, $\mathbf E_t$, and $\mathbf H_t$. The fields $E_n=D_n/\epsilon$ and $B_t=\mu H_t$ may still change when the material constants change.
Wave equations in a source-free dielectric
In a homogeneous, source-free, nonconducting dielectric,
\[\rho_{\mathrm f}=0,\qquad \mathbf J=0,\qquad \mathbf D=\epsilon\mathbf E,\qquad \mathbf B=\mu\mathbf H,\]with constant $\epsilon$ and $\mu$. Maxwell’s curl equations reduce to
\[\nabla\times\mathbf E=-\mu\frac{\partial\mathbf H}{\partial t}, \qquad \nabla\times\mathbf H=\epsilon\frac{\partial\mathbf E}{\partial t}.\]Take the curl of the first equation:
\[\nabla\times(\nabla\times\mathbf E) =-\mu\frac{\partial}{\partial t}(\nabla\times\mathbf H) =-\mu\epsilon\frac{\partial^2\mathbf E}{\partial t^2}.\]The identity $\nabla\times(\nabla\times\mathbf E)=\nabla(\nabla\cdot\mathbf E)-\nabla^2\mathbf E$ and Gauss’s law $\nabla\cdot\mathbf E=0$ give
\[\boxed{\nabla^2\mathbf E-\mu\epsilon\frac{\partial^2\mathbf E}{\partial t^2}=0}.\]The same steps yield
\[\boxed{\nabla^2\mathbf H-\mu\epsilon\frac{\partial^2\mathbf H}{\partial t^2}=0}.\]Both have wave speed
\[\boxed{v=\frac1{\sqrt{\mu\epsilon}}}.\]Plane wave in a dielectric
For propagation along $+z$, take
\[\mathbf E(z,t)=E_0\cos(kz-\omega t)\,\hat{\mathbf x}.\]Substitution in the wave equation gives
\[-k^2+\mu\epsilon\omega^2=0, \qquad \boxed{k=\omega\sqrt{\mu\epsilon}}.\]For a general plane-wave phasor $\mathbf E_0e^{i(\mathbf k\cdot\mathbf r-\omega t)}$, Gauss’s law gives
\[i\mathbf k\cdot\mathbf E_0=0.\]Therefore $\mathbf k\cdot\mathbf E_0=0$: the electric field is transverse. Faraday’s law gives
\[\mathbf k\times\mathbf E_0=\omega\mu\mathbf H_0,\]so $\mathbf H_0$ is transverse to both $\mathbf k$ and $\mathbf E_0$. Its magnitude is fixed by the intrinsic impedance
\[\boxed{\eta=\frac{E_0}{H_0}=\sqrt{\frac{\mu}{\epsilon}}},\]measured in ohms. For the chosen axes,
\[\mathbf H(z,t)=\frac{E_0}{\eta}\cos(kz-\omega t)\,\hat{\mathbf y},\]and $\mathbf E\times\mathbf H$ points along $+z$.
Poynting theorem from Maxwell’s equations
Dot the Ampere-Maxwell equation with $\mathbf E$ and Faraday’s law with $\mathbf H$:
\[\mathbf E\cdot(\nabla\times\mathbf H) =\mathbf J\cdot\mathbf E+\mathbf E\cdot\frac{\partial\mathbf D}{\partial t},\] \[\mathbf H\cdot(\nabla\times\mathbf E) =-\mathbf H\cdot\frac{\partial\mathbf B}{\partial t}.\]Subtract the first relation from the second and use
\[\nabla\cdot(\mathbf E\times\mathbf H) =\mathbf H\cdot(\nabla\times\mathbf E) -\mathbf E\cdot(\nabla\times\mathbf H).\]For a linear, nondispersive medium,
\[\mathbf E\cdot\frac{\partial\mathbf D}{\partial t} =\frac{\partial}{\partial t}\left(\frac12\mathbf E\cdot\mathbf D\right), \qquad \mathbf H\cdot\frac{\partial\mathbf B}{\partial t} =\frac{\partial}{\partial t}\left(\frac12\mathbf B\cdot\mathbf H\right).\]Therefore
\[\boxed{\frac{\partial u}{\partial t}+\nabla\cdot\mathbf S+\mathbf J\cdot\mathbf E=0},\]where
\[\boxed{u=\frac12(\mathbf E\cdot\mathbf D+\mathbf B\cdot\mathbf H)}, \qquad \boxed{\mathbf S=\mathbf E\times\mathbf H}.\]$u$ is electromagnetic energy density in $\mathrm{J\,m^{-3}}$, $\mathbf S$ is energy flux in $\mathrm{W\,m^{-2}}$, and $\mathbf J\cdot\mathbf E$ is the rate per unit volume at which field energy becomes material energy. Integration over a fixed volume $V$ gives
\[\boxed{\frac{d}{dt}\int_Vu\,dV =-\oint_{\partial V}\mathbf S\cdot d\mathbf a -\int_V\mathbf J\cdot\mathbf E\,dV}.\]The outward surface normal fixes the signs: positive outward Poynting flux decreases the stored field energy.
For the dielectric plane wave, $H=E/\eta$ and $\eta^2=\mu/\epsilon$, so
\[u_E=\frac12\epsilon E^2, \qquad u_B=\frac12\mu H^2 =\frac12\mu\frac{E^2}{\eta^2} =\frac12\epsilon E^2.\]The electric and magnetic contributions are equal instantaneously. Averaging $\cos^2(kz-\omega t)$ over one cycle gives
\[\boxed{\langle\mathbf S\rangle =\frac{E_0H_0}{2}\hat{\mathbf z} =\frac{E_0^2}{2\eta}\hat{\mathbf z}}.\]Solved Problems
1. Free surface charge at a dielectric boundary
Medium 1 has $\epsilon_{r1}=2.00$ and medium 2 has $\epsilon_{r2}=5.00$. The unit normal points from 1 to 2. Immediately at the interface,
\[E_{1n}=3.00\ \mathrm{kV\,m^{-1}}, \qquad E_{2n}=1.60\ \mathrm{kV\,m^{-1}},\]with both normal components directed along $+\hat{\mathbf n}$. Find the free surface-charge density.
Solution. The signed normal boundary condition is
\[\rho_s=\hat{\mathbf n}\cdot(\mathbf D_2-\mathbf D_1) =\epsilon_0(\epsilon_{r2}E_{2n}-\epsilon_{r1}E_{1n}).\]Substitution gives
\[\rho_s=\epsilon_0[5.00(1600)-2.00(3000)] =1.771\times10^{-8}\ \mathrm{C\,m^{-2}}.\]Thus
\[\boxed{\rho_s=+17.7\ \mathrm{nC\,m^{-2}}}.\]The positive sign means free positive charge resides on the interface for the chosen normal. Each term $\epsilon E$ has unit $\mathrm{C\,m^{-2}}$. If the two normal displacement components become equal, the expression correctly tends to $\rho_s=0$ even when the electric-field components differ.
2. Energy and power in a dielectric plane wave
A nonmagnetic, lossless dielectric has $\epsilon_r=4.00$. A $+z$-travelling plane wave has peak electric field $E_0=120\ \mathrm{V\,m^{-1}}$ along $+x$. Find $H_0$, the average Poynting flux, and the average total energy density.
Solution. For $\mu_r=1$,
\[\eta=\frac{\eta_0}{\sqrt{\epsilon_r}} =\frac{376.73}{2}=188.37\ \Omega.\]The right-handed orientation $\mathbf E\times\mathbf H\parallel+\hat{\mathbf z}$ fixes $\mathbf H$ along $+y$:
\[H_0=\frac{E_0}{\eta}=0.6371\ \mathrm{A\,m^{-1}}.\]For peak amplitudes,
\[\langle S\rangle=\frac{E_0^2}{2\eta} =38.22\ \mathrm{W\,m^{-2}}.\]The electric and magnetic energies are equal, so their cycle-averaged sum is
\[\langle u\rangle=\frac{\epsilon E_0^2}{2} =2.550\times10^{-7}\ \mathrm{J\,m^{-3}}.\]Therefore
\[\boxed{\mathbf H_0=0.6371\hat{\mathbf y}\ \mathrm{A\,m^{-1}},\quad \langle\mathbf S\rangle=38.22\hat{\mathbf z}\ \mathrm{W\,m^{-2}},\quad \langle u\rangle=2.550\times10^{-7}\ \mathrm{J\,m^{-3}}}.\]The ratio $\langle S\rangle/\langle u\rangle=1.499\times10^8\ \mathrm{m\,s^{-1}}$ equals the wave speed. This also checks dimensions because $\mathrm{(W\,m^{-2})/(J\,m^{-3})=m\,s^{-1}}$. As $E_0\to0$, both power and stored energy vanish quadratically.
Descriptive Questions
- Derive the four electromagnetic boundary conditions from pillbox and rectangular-loop constructions, stating which sources can create jumps.
- Why can $E_n$ and $B_t$ change across a source-free material boundary even when $D_n$ and $H_t$ are continuous combinations?
- Explain the local and integral meanings of every term in Poynting’s theorem, including its sign convention.
- Under what assumptions are the electric and magnetic energy densities of a plane wave equal instantaneously?
Numerical Problems
- With $\hat{\mathbf n}=\hat{\mathbf z}$, $\mathbf H_1=4\hat{\mathbf x}\ \mathrm{A\,m^{-1}}$ and $\mathbf H_2=1\hat{\mathbf x}\ \mathrm{A\,m^{-1}}$. Find the free sheet current. Final answer: $\boxed{\mathbf K=-3\hat{\mathbf y}\ \mathrm{A\,m^{-1}}}$.
- No free surface charge exists between dielectrics with $\epsilon_{r1}=2.50$ and $\epsilon_{r2}=5.00$. If $E_{1n}=800\ \mathrm{V\,m^{-1}}$, find $E_{2n}$. Final answer: $\boxed{E_{2n}=400\ \mathrm{V\,m^{-1}}}$.
- Find the wave speed in a nonmagnetic dielectric with $\epsilon_r=9.00$. Final answer: $\boxed{v=9.993\times10^7\ \mathrm{m\,s^{-1}}}$.
- A lossless plane wave has peak $E_0=50.0\ \mathrm{V\,m^{-1}}$ in a medium of impedance $250\ \Omega$. Find its mean intensity. Final answer: $\boxed{\langle S\rangle=5.00\ \mathrm{W\,m^{-2}}}$.
The wave identities are checked in the Unit I Maxima worksheet, and every worked and numerical value above is checked in the MJ-8 problem-verification worksheet; every printed residual and check is zero.
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