Electromagnetic Field Theory Solved Problems & Online Tutoring

This subject is usually taught in the second or third year of an engineering degree, and the exam is far more about applying vector calculus cleanly than about reciting Maxwell's equations.

Four worked problems follow, covering the question types that make up most of a paper: a field from a charge distribution using Gauss's law, a capacitance and energy calculation, a plane wave in a dielectric, and a transmission line mismatch. Each one states the formula, substitutes with units visible, and ends with the point the examiner is actually testing.

The signals side of the syllabus — Fourier analysis, convolution, LTI systems and transforms — is covered separately on the signals and systems page, because it is a different set of skills and a different set of exam questions.

This is the single highest-value habit in the subject, and almost nobody is taught it explicitly. The symmetry of the charge or current distribution decides the coordinate system, and the right choice turns a page of algebra into two lines.

The rule behind the table: pick the system in which the field magnitude is constant over the surface you intend to integrate. If it is, the integral collapses to a multiplication. If you find yourself integrating a varying magnitude over a surface in an exam, you almost certainly chose the wrong system.

The same logic chooses the Gaussian surface. For a line charge it is a cylinder coaxial with the line, because only then is the field perpendicular to the curved surface and zero through the flat ends.

Students ask why these two operations appear at all, and the textbooks tend to answer with definitions rather than reasons.

Divergence answers: does this point produce or swallow field? It is a scalar. A positive divergence means field lines begin there, so there is a source. For the electric field, divergence is non-zero wherever charge sits, which is Gauss's law in differential form. For the magnetic field, divergence is zero everywhere, which is the mathematical way of saying no isolated magnetic pole has ever been found.

Curl answers: does the field circulate around this point? It is a vector. A non-zero curl means the field loops. A static electric field has zero curl, which is why a voltage between two points is a single well-defined number and does not depend on the path. A magnetic field curls around a current, which is Ampère's law.

So the two operations are not arbitrary mathematical machinery. They are the only two questions you can ask about a vector field at a point: *does it start here*, and *does it go round here*. Maxwell's four equations are precisely those two questions asked of each of the two fields.

That is also why a static electric field is called conservative and a magnetic field is not: one has sources and no circulation, the other has circulation and no sources.

The problem. An infinite straight line carries a uniform charge density of 10 nC/m. Find the electric field magnitude at a radial distance of 0.5 m in free space.

Step 1 — symmetry and surface. The distribution is infinite and straight, so the field is radial and depends only on the radial distance. Take a coaxial cylinder of radius r and length L as the Gaussian surface. Flux through the flat ends is zero, because the field is parallel to them.

Step 2 — Gauss's law.

∮ E · dS = Q_enclosed ÷ ε₀

E × (2πrL) = λL ÷ ε₀

The length L cancels on both sides, which is why the answer does not depend on how much of the line you enclose.

Step 3 — rearrange and substitute, with ε₀ = 8.854 × 10⁻¹² F/m:

E = λ ÷ (2πε₀r) = (10 × 10⁻⁹) ÷ (2π × 8.854 × 10⁻¹² × 0.5)

Denominator = 2.7816 × 10⁻¹¹

E = 359.51 V/m, directed radially outward.

Step 4 — the check that earns the extra mark. At r = 1.0 m the same formula gives 179.75 V/m, exactly half. A line charge falls off as 1/r, not as 1/r². If your answer quartered when you doubled the distance, you have used the point-charge formula on a line.

The problem. A parallel-plate capacitor has plates of area 0.01 m² separated by 1 mm, with a dielectric of relative permittivity 4 filling the gap. Find the capacitance, and the charge and stored energy at 100 V.

Capacitance

C = ε₀εᵣA ÷ d = (8.854 × 10⁻¹² × 4 × 0.01) ÷ (1 × 10⁻³)

= 3.5416 × 10⁻¹⁰ F = 354.16 pF

Charge

Q = CV = 3.5416 × 10⁻¹⁰ × 100 = 3.5416 × 10⁻⁸ C = 35.42 nC

Stored energy

W = ½CV² = 0.5 × 3.5416 × 10⁻¹⁰ × 100² = 1.7708 × 10⁻⁶ J = 1.771 μJ

What the question is testing. Without the dielectric the same capacitor would be 88.54 pF. Inserting a dielectric of relative permittivity 4 multiplies the capacitance by exactly 4, because the polarised dielectric partly cancels the field between the plates, so the same charge sits at a lower voltage.

Two marks are routinely lost here. The separation is in millimetres and must be converted to metres before substituting. And the energy uses V squared, so doubling the voltage quadruples the stored energy rather than doubling it.

The problem. A uniform plane wave at 100 MHz travels in a lossless dielectric with εᵣ = 4 and μᵣ = 1. Find the intrinsic impedance, the phase velocity and the wavelength.

Intrinsic impedance of free space

η₀ = √(μ₀ ÷ ε₀) = 376.73 Ω

Intrinsic impedance of the medium, with μᵣ = 1:

η = η₀ ÷ √εᵣ = 376.73 ÷ 2 = 188.37 Ω

Phase velocity

v = c ÷ √εᵣ = (2.998 × 10⁸) ÷ 2 = 1.499 × 10⁸ m/s

Wavelength

λ = v ÷ f = (1.499 × 10⁸) ÷ (100 × 10⁶) = 1.50 m

In free space the same 100 MHz wave has a wavelength of 3.00 m.

The point of the question. The frequency does not change when a wave enters a material — it is set by the source. The velocity drops, so the wavelength must drop with it, here by a factor of exactly √4 = 2. This is why an antenna or a transmission line section cut for a given frequency has to be physically shorter when it sits in or on a dielectric, and why the substrate permittivity matters so much in practice.