Electromagnetics Tutoring

Electromagnetics asks students to picture things they cannot see. Fields fill space, charges and currents create them, and the math is vector calculus in three coordinate systems: Cartesian, cylindrical, and spherical. The equations are compact. The work is in the setup: choosing the Gaussian surface or Amperian loop, picking the coordinate system, and keeping track of which vector points where. Sessions slow that setup down. We draw the geometry first, name the symmetry, and only then decide between Gauss's law, Ampere's law, or direct integration.

A typical sequence starts with vector analysis and coordinate systems, then electrostatics, Gauss's law, potential, conductors and dielectrics, capacitance, and boundary conditions. Magnetostatics, Ampere's law, Faraday's law, and inductance follow. The course closes with Maxwell's equations, plane waves, reflection, and transmission lines with the Smith chart. Some programs split this into two courses. We follow your split and your textbook, such as Hayt, Cheng, or Ulaby.

A typical Gauss's law problem, worked

A coaxial cable has an inner conductor of radius $a = 1\ \text{mm}$ and an outer conductor with inner radius $b = 3.5\ \text{mm}$. The space between them is filled with a dielectric of relative permittivity 2.25. Find the capacitance per meter and the characteristic impedance.

Put charge $\rho_l$ per meter on the inner conductor. The symmetry is cylindrical, so use a Gaussian cylinder of radius $r$ and length $L$ between the conductors. Gauss's law gives $D \, (2 \pi r L) = \rho_l L$, so $$E = \dfrac{\rho_l}{2 \pi \varepsilon r}$$ pointing radially outward.

The voltage between the conductors is the integral of $E \, dr$ from $a$ to $b$, which is $$V = \dfrac{\rho_l \ln(b / a)}{2 \pi \varepsilon}$$ Capacitance per meter is charge per meter over voltage, so $$C' = \dfrac{2 \pi \varepsilon}{\ln(b / a)}$$

With $\varepsilon = 2.25 \times 8.854 \times 10^{-12}\ \text{F/m}$ and $\ln(3.5) = 1.2528$, $C' = 99.9\ \text{pF/m}$. The wave speed in the dielectric is $c / 1.5 = 2.0 \times 10^{8}\ \text{m/s}$, and $Z_0 = \dfrac{1}{v C'} = 50.1\ \Omega$. The formula $Z_0 = \dfrac{60}{\sqrt{\varepsilon_r}} \ln(b / a)$ gives the same answer. Notice the chain: Gauss's law gives $E$, $E$ gives $V$, $V$ gives $C'$, and $C'$ gives the line impedance.

Where students usually get stuck

• Choosing a Gaussian surface that does not match the symmetry, so $E$ cannot come out of the integral
• Using the wrong unit vectors or integration elements in cylindrical and spherical coordinates
• Sign errors when integrating $E$ to get potential, since $V$ is the negative line integral of $E$
• Treating phasors as time-domain waveforms and losing the $e^{j \omega t}$ factor
• On the Smith chart, rotating the wrong way: moving toward the generator is clockwise