Electromagnetics

4% of exam

Electrostatics, magnetostatics, Maxwell's equations, wave propagation, transmission lines.

3 concepts

Exam areas

A. Electrostatics/magnetostatics

Electrostatics and Magnetostatics

Coulomb and Gauss for E-fields, potential and capacitance, then Biot-Savart and Ampere for B-fields, inductance, and the energy stored in each field.

Static fields are the foundation the entire Electromagnetics topic is built on: get the constants, the geometry, and the energy bookkeeping right here and the dynamics and transmission-line questions become bookkeeping on top of them. The handbook gives you a compact toolbox under "Electrostatic Fields" and "Magnetic Fields" (FE Reference Handbook, Electrical and Computer Engineering), but the symbols hide a lot of structure — what is a point versus a line versus a sheet source, when a field falls off as

1/r^2

versus

1/r

, and which permittivity or permeability to use. This concept makes you fluent in all of it. Expect 1 to 2 of your 4 to 6 Electromagnetics questions to live right here, and they are usually fast points if you keep the geometry and units straight.

Coulomb's law and the electric field

Two point charges exert a force along the line joining them: like charges repel, opposite charges attract. The electric field

\mathbf{E}

is that force per unit positive test charge, so

\mathbf{F}=Q\mathbf{E}

. For a single point charge

Q

the field points radially outward (for

Q>0

) and falls off as the inverse square of distance. In free space or air use

\varepsilon_0=8.85\times10^{-12}\,\text{F/m}

; in a dielectric use

\varepsilon=\varepsilon_r\varepsilon_0

. Work in vectors when directions matter — superpose the contributions of each charge as vectors, not magnitudes.

\mathbf{E}=\dfrac{Q}{4\pi\varepsilon r^{2}}\,\mathbf{a}_r,\qquad \mathbf{F}=Q\mathbf{E}

Line and sheet sources: the geometry sets the falloff

Symmetry changes how the field decays. A point charge gives

E\propto 1/r^2

; an infinite line charge of density

\rho_L

(C/m) gives a radial field that decays only as

1/r

; an infinite sheet of charge

\rho_s

(C/m

^2

) gives a uniform field that does not decay with distance at all. Recognizing the source geometry is half the battle on these problems — a question that mentions "per meter of length" is a line source, and "per unit area" is a sheet.

E_{\text{line}}=\dfrac{\rho_L}{2\pi\varepsilon r},\qquad E_{\text{sheet}}=\dfrac{\rho_s}{2\varepsilon}

Gauss's law

Gauss's law is the fast path whenever the geometry is symmetric. It says the flux of the electric flux density

\mathbf{D}=\varepsilon\mathbf{E}

through any closed surface equals the charge enclosed — fields outside the surface contribute nothing. Pick a Gaussian surface that matches the symmetry (sphere for a point charge, cylinder for a line, pillbox for a sheet) so that

\mathbf{D}

is constant and either parallel or perpendicular to the surface, then

D

comes straight out of the integral.

\oint_S \mathbf{D}\cdot d\mathbf{S}=Q_{\text{enc}},\qquad \mathbf{D}=\varepsilon\mathbf{E}

Potential, voltage, and stored energy

Potential difference is work per unit charge moved against the field,

V=-\int_{p_1}^{p_2}\mathbf{E}\cdot d\mathbf{r}

. The most-tested special case is the parallel-plate field: a uniform

E=V/d

pointing from the

+

plate to the

-

plate. Energy can be tracked either as a field integral

\tfrac{1}{2}\varepsilon E^2

over the volume or, for a capacitor, with the lumped forms below — they agree.

E_{\text{plates}}=\dfrac{V}{d},\qquad W_E=\int_V \tfrac{1}{2}\varepsilon E^{2}\,dV

Capacitance and stored charge

Capacitance is the charge stored per volt,

C=q/v

, set purely by geometry and the dielectric. For parallel plates of area

A

separated by

d

with a dielectric of permittivity

\varepsilon=\varepsilon_r\varepsilon_0

, capacitance scales with area and inversely with spacing. The energy stored has three equivalent forms; pick whichever matches the quantities you already have.

C=\dfrac{\varepsilon A}{d},\qquad W=\tfrac{1}{2}Cv^{2}=\dfrac{q^{2}}{2C}=\tfrac{1}{2}qv

Magnetostatics: Biot-Savart and Ampere

Steady currents make magnetic fields. The handbook gives the result for a long straight wire on the

z

-axis: the magnetic field strength

\mathbf{H}

circles the wire (right-hand rule) and decays as

1/r

, with flux density

\mathbf{B}=\mu\mathbf{H}

. This is exactly what Ampere's law

\oint \mathbf{H}\cdot d\mathbf{l}=I_{\text{enc}}

delivers when you take a circular Amperian loop of radius

r

. Use

\mu_0=4\pi\times10^{-7}\,\text{H/m}

in air,

\mu=\mu_r\mu_0

in magnetic material. The force on a current-carrying conductor in a field is

\mathbf{F}=I\mathbf{L}\times\mathbf{B}

H=\dfrac{I}{2\pi r},\qquad \mathbf{B}=\mu\mathbf{H},\qquad \oint \mathbf{H}\cdot d\mathbf{l}=I_{\text{enc}}

Inductance and magnetic energy

Inductance relates flux linkage to current,

N\phi=Li

. For a coil of

N

turns on a core of cross-section

A

, mean magnetic path length

\ell

, and permeability

\mu

, inductance grows as

N^2

and falls with path length. The magnetic-circuit view writes

L=N^2/\mathcal{R}

with reluctance

\mathcal{R}=\ell/(\mu A)

— the magnetic analog of resistance. Magnetic energy parallels electric energy: a field integral

\tfrac{1}{2}\mu H^2

over the volume, or the lumped

\tfrac{1}{2}Li^2

L=\dfrac{N^{2}\mu A}{\ell}=\dfrac{N^{2}}{\mathcal{R}},\qquad W_H=\tfrac{1}{2}Li^{2}

Exam strategy

First classify the source — point, line, sheet, or current — because that fixes the falloff and the right formula before you touch numbers. Use Gauss's law (electric) or Ampere's law (magnetic) whenever the geometry is symmetric; they are far faster than integrating Coulomb or Biot-Savart contributions. Watch the constants:

\varepsilon_0=8.85\times10^{-12}\,\text{F/m}

and

\mu_0=4\pi\times10^{-7}\,\text{H/m}

are on the formula sheet, but

1/(4\pi\varepsilon_0)\approx 8.99\times10^{9}

is not — derive it. Keep

\mathbf{E}

(V/m),

\mathbf{D}

(C/m

^2

\mathbf{H}

(A/m), and

\mathbf{B}

(T) distinct, and remember the energy forms come in matched electric/magnetic pairs:

\tfrac{1}{2}Cv^2

mirrors

\tfrac{1}{2}Li^2

Key equations

Coulomb force

\mathbf{F}_2=\dfrac{Q_1 Q_2}{4\pi\varepsilon r^{2}}\,\mathbf{a}_{r12}

Force on charge 2 from charge 1; $r$ in m, $Q$ in C, $\varepsilon=\varepsilon_r\varepsilon_0$ (F/m). Positive product = repulsive (along $\mathbf{a}_{r12}$ ).

Point-charge E field

\mathbf{E}=\dfrac{Q}{4\pi\varepsilon r^{2}}\,\mathbf{a}_r,\quad \mathbf{F}=Q\mathbf{E}

$E$ in V/m, radial, $\propto 1/r^2$ . Multiply by a test charge to get force.

Line and sheet fields

E_{\text{line}}=\dfrac{\rho_L}{2\pi\varepsilon r},\quad E_{\text{sheet}}=\dfrac{\rho_s}{2\varepsilon}

Infinite line charge $\rho_L$ (C/m) falls as $1/r$ ; infinite sheet $\rho_s$ (C/m $^2$ ) is uniform.

Gauss's law

\oint_S \mathbf{D}\cdot d\mathbf{S}=Q_{\text{enc}},\quad \mathbf{D}=\varepsilon\mathbf{E}

Flux of $\mathbf{D}$ (C/m $^2$ ) through a closed surface equals enclosed charge (C). Use with symmetric Gaussian surfaces.

Parallel-plate E field

E=\dfrac{V}{d}

Uniform field between plates, directed from $+$ to $-$ ; $V$ in volts, $d$ in m, $E$ in V/m.

Parallel-plate capacitance

C=\dfrac{\varepsilon A}{d}=\dfrac{\varepsilon_r\varepsilon_0 A}{d}

$A$ = plate area (m $^2$ ), $d$ = spacing (m), $\varepsilon_r$ = dielectric constant; $C$ in farads.

Capacitor stored energy

W=\tfrac{1}{2}Cv^{2}=\dfrac{q^{2}}{2C}=\tfrac{1}{2}qv

Three equivalent forms in joules with $q=Cv$ .

Electric / magnetic field energy

W_E=\int_V \tfrac{1}{2}\varepsilon E^{2}\,dV,\quad W_H=\int_V \tfrac{1}{2}\mu H^{2}\,dV

Energy as field integrals over a volume; energy densities $\tfrac12\varepsilon E^2$ (J/m $^3$ ) and $\tfrac12\mu H^2$ .

Long-wire magnetic field

H=\dfrac{I}{2\pi r},\quad \mathbf{B}=\mu\mathbf{H}

$H$ in A/m circles the wire (right-hand rule); $B=\mu H$ in tesla; $r$ from the wire axis (m).

Ampere's law

\oint \mathbf{H}\cdot d\mathbf{l}=I_{\text{enc}}

Circulation of $\mathbf{H}$ around a closed loop equals enclosed current; the fast path for symmetric geometries.

Force on a conductor

\mathbf{F}=I\,\mathbf{L}\times\mathbf{B}

Force on length $\mathbf{L}$ of conductor carrying current $I$ in flux density $\mathbf{B}$ ; magnitude $BIL\sin\theta$ in N.

Inductance

L=\dfrac{N^{2}\mu A}{\ell}=\dfrac{N^{2}}{\mathcal{R}},\quad \mathcal{R}=\dfrac{\ell}{\mu A}

$N$ turns, core area $A$ (m $^2$ ), mean path $\ell$ (m), permeability $\mu=\mu_r\mu_0$ ; reluctance $\mathcal{R}$ in H $^{-1}$ .

Inductor stored energy

W=\tfrac{1}{2}Li^{2}

Magnetic energy in joules; the dual of $\tfrac{1}{2}Cv^2$ .

Worked examples

Coulomb force and field between two charges

Problem. Charge $Q_1=+20\,\text{nC}$ and $Q_2=-15\,\text{nC}$ sit $5.0\,\text{cm}$ apart in air. Find the magnitude and nature of the force between them, and the magnitude of the field $Q_1$ alone produces at $Q_2$ 's location.

Solution. Use

F=\dfrac{|Q_1 Q_2|}{4\pi\varepsilon_0 r^2}

with

r=0.05\,\text{m}

and

1/(4\pi\varepsilon_0)=8.99\times10^{9}

F=8.99\times10^{9}\cdot\dfrac{(20\times10^{-9})(15\times10^{-9})}{(0.05)^2}=8.99\times10^{9}\cdot\dfrac{3.0\times10^{-16}}{2.5\times10^{-3}}

F=1.08\times10^{-3}\,\text{N}=1.08\,\text{mN}

, attractive (opposite signs). Field from

Q_1

at that point:

E=\dfrac{|Q_1|}{4\pi\varepsilon_0 r^2}=8.99\times10^{9}\cdot\dfrac{20\times10^{-9}}{2.5\times10^{-3}}=7.19\times10^{4}\,\text{V/m}

. Sanity check:

F=|Q_2|E=15\times10^{-9}\cdot7.19\times10^{4}=1.08\times10^{-3}\,\text{N}

— matches.

F=\dfrac{|Q_1 Q_2|}{4\pi\varepsilon_0 r^{2}}\approx 1.08\,\text{mN}\ (\text{attractive})

Parallel-plate capacitor: C, charge, energy, field

Problem. A parallel-plate capacitor has plates of area $A=0.020\,\text{m}^2$ separated by $d=0.50\,\text{mm}$ of dielectric with $\varepsilon_r=4.0$ . It is charged to $V=12\,\text{V}$ . Find $C$ , the stored charge, the stored energy, and the field strength between the plates.

Solution.

C=\dfrac{\varepsilon_r\varepsilon_0 A}{d}=\dfrac{4.0\,(8.85\times10^{-12})(0.020)}{0.50\times10^{-3}}=1.42\times10^{-9}\,\text{F}=1.42\,\text{nF}

. Charge:

q=CV=1.416\times10^{-9}\cdot12=1.70\times10^{-8}\,\text{C}=17.0\,\text{nC}

. Energy:

W=\tfrac{1}{2}CV^2=\tfrac{1}{2}(1.416\times10^{-9})(12)^2=1.02\times10^{-7}\,\text{J}=102\,\text{nJ}

. Field:

E=V/d=12/(0.50\times10^{-3})=2.40\times10^{4}\,\text{V/m}=24.0\,\text{kV/m}

. Sanity check: energy density

\tfrac12\varepsilon E^2\cdot(Ad)=\tfrac12(4)(8.85\text{e-}12)(2.4\text{e}4)^2(0.02\cdot5\text{e-}4)=1.02\times10^{-7}\,\text{J}

— agrees.

C=\dfrac{\varepsilon_r\varepsilon_0 A}{d}=1.42\,\text{nF},\quad W=\tfrac12 CV^2=102\,\text{nJ}

Solenoid: inductance, B-field, and stored energy

Problem. A 200-turn air-core solenoid has length $\ell=0.30\,\text{m}$ and cross-sectional area $A=4.0\,\text{cm}^2$ , carrying $I=3.0\,\text{A}$ . Find the inductance, the axial flux density $B$ inside, and the stored magnetic energy.

Solution. Inductance:

L=\dfrac{\mu_0 N^2 A}{\ell}=\dfrac{(4\pi\times10^{-7})(200)^2(4.0\times10^{-4})}{0.30}

. Numerator

=(1.2566\times10^{-6})(40000)(4\times10^{-4})=2.011\times10^{-5}

; divide by

0.30

L=6.70\times10^{-5}\,\text{H}=67.0\,\mu\text{H}

. Field (ideal solenoid

B=\mu_0 N I/\ell

B=\dfrac{(4\pi\times10^{-7})(200)(3.0)}{0.30}=2.51\times10^{-3}\,\text{T}=2.51\,\text{mT}

. Energy:

W=\tfrac12 L I^2=\tfrac12(6.70\times10^{-5})(3.0)^2=3.02\times10^{-4}\,\text{J}=302\,\mu\text{J}

. Sanity check: field-energy density

\tfrac{1}{2}B^2/\mu_0=\tfrac12(2.51\text{e-}3)^2/(1.2566\text{e-}6)=2.51\,\text{J/m}^3

, times volume

A\ell=1.2\times10^{-4}\,\text{m}^3

gives

3.02\times10^{-4}\,\text{J}

— agrees.

L=\dfrac{\mu_0 N^{2} A}{\ell}=67.0\,\mu\text{H},\quad W=\tfrac12 LI^2=302\,\mu\text{J}

Common pitfalls

•Forgetting the $1/(4\pi\varepsilon_0)\approx 8.99\times10^{9}$ prefactor: the handbook lists $\varepsilon_0$ , not the combined Coulomb constant. Compute $1/(4\pi\varepsilon_0)$ yourself or you will be off by $\sim10^{10}$ .
•Using the point-charge $1/r^2$ falloff for a line or sheet source. A line charge field goes as $1/r$ and an infinite sheet field is uniform — classify the source first.
•Omitting $\varepsilon_r$ (or $\mu_r$ ): air uses $\varepsilon_0$ / $\mu_0$ , but a dielectric or magnetic core multiplies by the relative value. A dielectric of $\varepsilon_r=4$ quadruples capacitance.
•Adding field magnitudes instead of vectors when multiple charges are present. Superpose $\mathbf{E}$ as vectors (components), then take the magnitude.
•Mixing up $\mathbf{H}$ (A/m) and $\mathbf{B}$ (T): the wire formula gives $H=I/(2\pi r)$ ; multiply by $\mu$ to get $B$ . Reporting $H$ as if it were $B$ drops a factor of $\mu$ .
•Treating inductance as linear in turns. $L\propto N^2$ — doubling turns quadruples $L$ , not doubles it.
•Unit slips on area and length: $\text{cm}^2=10^{-4}\,\text{m}^2$ and $\text{mm}=10^{-3}\,\text{m}$ . A single misplaced power of ten is the most common wrong multiple-choice answer.

References

NCEES FE Reference Handbook — Electrical and Computer Engineering: Electrostatics / Electrostatic Fields — Coulomb's law, point/line/sheet fields, Gauss's law, voltage, parallel-plate E.
NCEES FE Reference Handbook — Electrical and Computer Engineering: Magnetic Fields and Capacitors/Inductors — Long-wire H/B, force on a conductor, capacitance, inductance, and stored-energy forms.
Hayt & Buck, Engineering Electromagnetics — Standard derivations of Gauss's law and Ampere's law with symmetric surfaces/loops.

B. Electrodynamics

Electrodynamics and Maxwell's Equations

Faraday's law and displacement current close Maxwell's four equations, which then predict plane waves, the Poynting power flow, and every wave parameter you will compute.

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C. Transmission lines (high frequency)

High-Frequency Transmission Lines

Characteristic impedance and propagation, the load reflection coefficient and VSWR, the input-impedance formula, and quarter-wave matching for high-frequency lines.

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