13.1 Electrostatics, Magnetostatics & Maxwell's Equations

13.1 Electrostatics, Magnetostatics & Maxwell's Equations

Electromagnetic field theory forms the fundamental physical foundation for all electrical engineering disciplines. In the PRC Registered Electrical Engineer (REE) Licensure Examination, questions under Professional Electrical Engineering evaluate candidates on field vectors, force laws, electrostatic capacitance, magnetostatic inductance, electromagnetic boundary conditions, and Maxwell's unified equations.

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1. Electrostatics Foundations & Gauss's Law

Electrostatics deals with stationary electric charges and their resulting forces, electric field intensities, and energy storage.

Coulomb's Law & Electric Field Intensity

Coulomb's Law states that the electrostatic force $\mathbf{F}$ between two point charges $q_1$ and $q_2$ separated by a distance $r$ in a homogeneous dielectric medium is directly proportional to the product of the charges and inversely proportional to the square of the distance:

$\mathbf{F} = \frac{q_1 q_2}{4\pi \epsilon_0 \epsilon_r r^2} \mathbf{a}_r \quad (\text{N})$

where $\epsilon_0 \approx 8.854 \times 10^{-12}\ \text{F/m}$ is the absolute permittivity of free space, $\epsilon_r$ is the relative permittivity (dielectric constant) of the medium, and $\mathbf{a}_r$ is the unit vector directed from the source charge to the test charge.

The Electric Field Intensity $\mathbf{E}$ is defined as the force per unit positive test charge:

$\mathbf{E} = \frac{\mathbf{F}}{q} = \frac{q}{4\pi \epsilon r^2} \mathbf{a}_r \quad (\text{V/m})$

Electric Flux Density & Gauss's Law

The Electric Flux Density $\mathbf{D}$ represents the electric flux passing perpendicularly through a unit surface area, independent of the medium's dielectric properties:

$\mathbf{D} = \epsilon \mathbf{E} = \epsilon_0 \epsilon_r \mathbf{E} \quad (\text{C/m}^2)$

Gauss's Law states that the net outward electric flux passing through any closed surface (Gaussian surface) $S$ equals the total enclosed net free charge $Q_{\text{encl}}$:

$\text{Integral Form: } \oint_S \mathbf{D} \cdot d\mathbf{A} = Q_{\text{encl}} = \iiint_V \rho_v dV$

Applying the Divergence Theorem ($\oint_S \mathbf{D} \cdot d\mathbf{A} = \iiint_V (\nabla \cdot \mathbf{D}) dV$) yields Gauss's Law in point (differential) form:

$\text{Differential Form: } \nabla \cdot \mathbf{D} = \rho_v$

where $\rho_v$ is the volume charge density ($\text{C/m}^3$).

Electric Potential and Potential Gradient

The electrostatic potential $V$ at a point is the work done per unit positive charge in bringing it from infinity to that point against the electric field:

$V = -\int_{\infty}^{r} \mathbf{E} \cdot d\mathbf{L} \quad (\text{Volts})$

Because electrostatic fields are conservative ($\oint \mathbf{E} \cdot d\mathbf{L} = 0$), the electric field intensity equals the negative gradient of potential:

$\mathbf{E} = -\nabla V = -\left( \frac{\partial V}{\partial x} \mathbf{a}_x + \frac{\partial V}{\partial y} \mathbf{a}_y + \frac{\partial V}{\partial z} \mathbf{a}_z \right)$

Standard Capacitance Formulas

Capacitance $C$ measures electric charge storage per unit potential difference ($C = Q / V$):

• Parallel-Plate Capacitor: $C = \frac{\epsilon A}{d} = \frac{\epsilon_0 \epsilon_r A}{d} \quad (\text{F})$
• Coaxial Cylindrical Cable (length $L$, inner conductor radius $a$, outer sheath radius $b$): $C = \frac{2\pi \epsilon L}{\ln(b/a)} \quad (\text{F})$
• Concentric Spherical Capacitor (inner radius $a$, outer radius $b$): $C = \frac{4\pi \epsilon a b}{b - a} \quad (\text{F})$
• Electrostatic Stored Energy Density: $W_E = \frac{1}{2} C V^2 = \frac{1}{2} \iiint_V (\mathbf{D} \cdot \mathbf{E}) dV = \frac{1}{2} \iiint_V \epsilon |E|^2 dV \quad (\text{J})$

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2. Magnetostatics Foundations & Ampere's Law

Magnetostatics governs steady, non-time-varying magnetic fields produced by direct electric currents.

Biot-Savart Law & Ampere's Circuital Law

The Biot-Savart Law calculates the differential magnetic field intensity $d\mathbf{H}$ produced at a point by a differential current element $I d\mathbf{L}$:

$d\mathbf{H} = \frac{I d\mathbf{L} \times \mathbf{a}_r}{4\pi r^2} \quad (\text{A/m})$

$\mathbf{B} = \mu \mathbf{H} = \mu_0 \mu_r \mathbf{H} \quad (\text{Tesla or Wb/m}^2)$

where $\mu_0 = 4\pi \times 10^{-7}\ \text{H/m}$ is the magnetic permeability of free space, and $\mu_r$ is the relative permeability.

Ampere's Circuital Law states that the line integral of magnetic field intensity $\mathbf{H}$ around any closed path $C$ equals the net enclosed current $I_{\text{encl}}$ passing through the surface enclosed by $C$:

$\text{Integral Form: } \oint_C \mathbf{H} \cdot d\mathbf{L} = I_{\text{encl}} = \iint_S \mathbf{J} \cdot d\mathbf{A}$

Applying Stokes' Theorem ($\oint_C \mathbf{H} \cdot d\mathbf{L} = \iint_S (\nabla \times \mathbf{H}) \cdot d\mathbf{A}$) gives Ampere's Law in point form:

$\text{Differential Form: } \nabla \times \mathbf{H} = \mathbf{J}$

Lorentz Force Equation

The total force on a charge $q$ moving with velocity $\mathbf{v}$ through combined electric and magnetic fields is:

$\mathbf{F} = q (\mathbf{E} + \mathbf{v} \times \mathbf{B}) \quad (\text{N})$

For a straight conductor carrying current $I$ of length $L$ in a uniform magnetic flux density $\mathbf{B}$:

$\mathbf{F} = I (\mathbf{L} \times \mathbf{B}) \implies F = I L B \sin\theta$

Standard Inductance Formulas

Inductance $L$ measures magnetic flux linkage per unit current ($L = N\Phi / I$):

• Long Solenoid (length $l$, cross-sectional area $A$, $N$ turns): $L = \frac{\mu N^2 A}{l} \quad (\text{H})$
• Toroidal Coil (mean radius $r$, cross-sectional area $A$, $N$ turns): $L = \frac{\mu N^2 A}{2\pi r} \quad (\text{H})$
• Coaxial Cable Inductance (length $L$, inner radius $a$, outer radius $b$): $L = \frac{\mu L}{2\pi} \ln\left(\frac{b}{a}\right) \quad (\text{H})$
• Magnetostatic Stored Energy Density: $W_M = \frac{1}{2} L I^2 = \frac{1}{2} \iiint_V (\mathbf{B} \cdot \mathbf{H}) dV = \frac{1}{2} \iiint_V \mu |H|^2 dV \quad (\text{J})$

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3. Maxwell's Unified Equations & Displacement Current

Prior to James Clerk Maxwell, static electrical and magnetic laws were disconnected. Maxwell recognized an incompatibility between Ampere's static law ($\nabla \times \mathbf{H} = \mathbf{J}$) and the principle of conservation of charge, formulated by the continuity equation:

$\nabla \cdot \mathbf{J} = -\frac{\partial \rho_v}{\partial t}$

Taking the divergence of static Ampere's law gives $\nabla \cdot (\nabla \times \mathbf{H}) \equiv 0$, forcing $\nabla \cdot \mathbf{J} = 0$. This holds only for static conditions (steady DC), failing for time-varying AC circuits (such as charging a capacitor).

The Displacement Current Correction

To resolve this contradiction, Maxwell substituted Gauss's law ($\rho_v = \nabla \cdot \mathbf{D}$) into the continuity equation:

$\nabla \cdot \mathbf{J} = -\frac{\partial}{\partial t} (\nabla \cdot \mathbf{D}) = -\nabla \cdot \left( \frac{\partial \mathbf{D}}{\partial t} \right) \implies \nabla \cdot \left( \mathbf{J} + \frac{\partial \mathbf{D}}{\partial t} \right) = 0$

Maxwell defined the Displacement Current Density $\mathbf{J}_d$ as:

$\mathbf{J}_d = \frac{\partial \mathbf{D}}{\partial t} = \epsilon \frac{\partial \mathbf{E}}{\partial t} \quad (\text{A/m}^2)$

This modification completed the Ampere-Maxwell Law, proving that changing electric fields generate magnetic fields in the same manner as physical conduction currents.

Summary Table of Maxwell's Equations

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4. Electromagnetic Boundary Conditions

When electromagnetic fields cross the boundary interface between two different media (Medium 1 and Medium 2), boundary conditions determine the relationship between tangential and normal field components:

\text{Tangential Electric Field: } & E_{1t} = E_{2t} \quad (\text{Continuous}) \\[4pt] \text{Normal Electric Flux Density: } & D_{1n} - D_{2n} = \rho_s \quad (\text{Discontinuous by surface charge } \rho_s) \\[4pt] \text{Normal Magnetic Flux Density: } & B_{1n} = B_{2n} \quad (\text{Continuous}) \\[4pt] \text{Tangential Magnetic Field: } & H_{1t} - H_{2t} = K \quad (\text{Discontinuous by surface current } K) \end{aligned}$$ If the interface has no free surface charge ($\rho_s = 0$) and no surface current ($K = 0$), then $D_{1n} = D_{2n}$ and $H_{1t} = H_{2t}$. --- ## Solved Board Exam Numerical Problems ### Problem 1: Coaxial Power Cable Capacitance & Energy Calculation **Question:** A single-core coaxial underground power cable has an inner conductor diameter of $10\ \text{mm}$ ($a = 5\ \text{mm}$), an outer sheath inner diameter of $40\ \text{mm}$ ($b = 20\ \text{mm}$), a relative permittivity $\epsilon_r = 4.0$, and a total length $L = 500\ \text{m}$. The cable is energized at $13.8\ \text{kV}$ RMS line-to-ground. Calculate: (a) the total capacitance of the cable, and (b) the peak electrostatic stored energy in the dielectric insulation. **Solution:** 1. Calculate total cable capacitance $C$: $$C = \frac{2\pi \epsilon_0 \epsilon_r L}{\ln(b/a)}$$ $$C = \frac{2\pi \times (8.854 \times 10^{-12}\ \text{F/m}) \times 4.0 \times 500\ \text{m}}{\ln(20 / 5)}$$ $$C = \frac{1.1126 \times 10^{-7}}{1.3863} = 8.026 \times 10^{-8}\ \text{F} = 80.26\ \text{nF}$$ 2. Calculate peak voltage $V_{\text{peak}}$: $$V_{\text{peak}} = \sqrt{2} \times V_{\text{rms}} = \sqrt{2} \times 13,800\ \text{V} = 19,516.15\ \text{V}$$ 3. Calculate peak electrostatic stored energy $W_{\text{E, peak}}$: $$W_{\text{E, peak}} = \frac{1}{2} C V_{\text{peak}}^2 = \frac{1}{2} \times (8.026 \times 10^{-8}\ \text{F}) \times (19,516.15\ \text{V})^2$$ $$W_{\text{E, peak}} = 0.5 \times (8.026 \times 10^{-8}) \times (3.8088 \times 10^8) = 15.28\ \text{Joules}$$ --- ### Problem 2: Displacement Current in a Circular Parallel-Plate Capacitor **Question:** A circular parallel-plate capacitor with plate area $A = 0.05\ \text{m}^2$ and plate separation distance $d = 2.0\ \text{mm}$ is filled with an air dielectric ($\epsilon_r = 1.0$). A sinusoidal voltage source $v(t) = 150 \sin(120\pi t)\ \text{V}$ is connected across the capacitor plates. Calculate the displacement current $i_d(t)$ flowing through the capacitor. **Solution:** 1. Determine the electric field intensity $E(t)$ between the plates: $$E(t) = \frac{v(t)}{d} = \frac{150 \sin(120\pi t)}{0.002\ \text{m}} = 75,000 \sin(120\pi t)\ \text{V/m}$$ 2. Determine the electric flux density $D(t)$: $$D(t) = \epsilon_0 E(t) = (8.854 \times 10^{-12}) \times 75,000 \sin(120\pi t) = 6.6405 \times 10^{-7} \sin(120\pi t)\ \text{C/m}^2$$ 3. Calculate the displacement current density $J_d(t)$: $$J_d(t) = \frac{\partial D}{\partial t} = 6.6405 \times 10^{-7} \times (120\pi) \cos(120\pi t) = 2.5034 \times 10^{-4} \cos(120\pi t)\ \text{A/m}^2$$ 4. Calculate total displacement current $i_d(t)$: $$i_d(t) = J_d(t) \times A = (2.5034 \times 10^{-4}\ \text{A/m}^2) \times 0.05\ \text{m}^2$$ $$i_d(t) = 1.2517 \times 10^{-5} \cos(120\pi t)\ \text{A} = 12.52 \cos(120\pi t)\ \mu\text{A}$$