Electromagnetism Concepts and Principles

The Quantities

• Sources
  • Charge: $Q$ (C)
  • Line charge density: $\rho_l$ (C/m)
  • Surface charge density: $\rho_s$ (C/m²)
  • Volume charge density: $\rho_v$ (C/m³)
• Current
  • I : Current (A)
  • J : Conduction current density (A/m²)
  • $I = \frac{dQ}{dt}$

Fields

• Electric
  • $E$ : Electric field strength (V/m)
  • $D$ : Electric flux density (C/m²)
• Magnetic
  • $H$ : Magnetic field strength (A/m)
  • $B$ : Magnetic flux density (Wb/m²)

Material Properties

• $\epsilon$ : Permittivity of a material (F/m)
• $\mu$ : Permeability of a material (H/m)
• $\sigma$ : Conductivity of a material (S/m)
• $D = \epsilon E$
• $B = \mu H$

Electromagnetism History

• ~1820 – 1831 – 1835
  • Ampere quantified the relationship between magnetic field and current.
  • Faraday discovered that changing magnetic flux induces EMF (electromotive force).
  • Gauss showed that the source of electric fields were electric charges (no magnetic charge).

Maxwell's Equations

• Gauss’s law: $\nabla \cdot D = \rho_v$
• Gauss’s law for magnetic fields: $\nabla \cdot B = 0$
• Faraday’s law: $\nabla \times E = -\frac{\partial B}{\partial t}$
• Ampere’s law: $\nabla \times H = J$

Maxwell's Correction to Ampere's Law

• Maxwell noticed that the original equations were incomplete.
• The divergence of Ampere’s equation leads to $\nabla \cdot (\nabla \times H) = 0$, which implies $\nabla \cdot J = 0$.
• This implication suggests that the source of electric current is zero, conflicting with the continuity equation: $\nabla \cdot J = -\frac{\partial \rho_v}{\partial t}$.
• Maxwell added a displacement current term to satisfy the continuity equation and Gauss's law: $\nabla \cdot D = \rho_v$
• The complete Ampere’s law is then expressed as: $\nabla \times H = J + \frac{\partial D}{\partial t}$

Maxwell’s Equations - General Form

• Integral form of Maxwell's equations:
  • Gauss's law: $\oint<em>S D \cdot ds = \int</em>v \rho_v dv = Q$
  • Gauss's law for magnetism: $\oint_S B \cdot ds = 0$
  • Faraday's law: $\oint<em>C E \cdot dl = -\frac{d}{dt} \oint</em>S B \cdot ds$
  • Ampere-Maxwell's law: $\oint<em>C H \cdot dl = \oint</em>S J \cdot ds + \frac{d}{dt} \oint_S D \cdot ds$

Maxwell’s Equations - Static Fields

• Electrostatics
  • $\nabla \times E = 0 \Rightarrow \oint_C E \cdot dl = 0$
  • $\nabla \cdot D = \rho \Rightarrow \oint_S D \cdot ds = Q$
• Magnetostatics
  • $\nabla \times H = J \Rightarrow \oint<em>C H \cdot dl = \int</em>S J \cdot ds$
  • $\nabla \cdot B = 0 \Rightarrow \oint_S B \cdot ds = 0$

Charge Distributions

• Volume charge density: $\rho<em>v = \lim</em>{\Delta V \to 0} \frac{\Delta q}{\Delta V} = \frac{dq}{dV}$ (C/m³)
• Total charge in a volume: $Q = \int<em>V \rho</em>v dv$ (C)
• Total charge on a surface: $Q = \int<em>S \rho</em>s dS$
• Total charge along a line: $Q = \int<em>l \rho</em>l dl$

Current Density

• Current flowing in a tube: $\Delta I = J \cdot \Delta s$
• Current density: $J = \rho_v u$ (A/m²), where $u$ is the velocity of the charge.
• Total current: $I = \int_S J \cdot ds$ (A)

Convection Current

• The amount of charge that crosses the tube's cross-sectional surface $\Delta s$ in time $\Delta t$ is $\Delta q = \rho<em>v \Delta V = \rho</em>v u \Delta s' \Delta t$
• For a surface with any orientation: $J = \rho_v u$ (A/m²)
• $I = \int_S J \cdot ds$ (A)

Conduction Current

• When current is due to movement of charged particles relative to their host material, $J$ is called conduction current density.
• Conduction current obeys Ohm's law, whereas convection current does not.

Coulomb's Law

• Electric field at point P due to single charge.
• Electric force on a test charge placed at P.

Electric Flux Density

• $D = \epsilon E$
• $\epsilon = \epsilon<em>r \epsilon</em>0$
• $\epsilon_0 = 8.85 \times 10^{-12} \approx (1/36\pi) \times 10^{-9}$ (F/m)
• If $\epsilon$ is independent of the magnitude of $E$, the material is linear.
• If $\epsilon$ is independent of the direction of $E$, the material is isotropic.

Electric Field Due to Multiple Charges

• Electric field due to charge 1: $E<em>1 = \frac{q</em>1 (R - R<em>1)}{4\pi\epsilon |R - R</em>1|^3}$
• Electric field due to charge 2: $E<em>2 = \frac{q</em>2 (R - R<em>2)}{4\pi\epsilon |R - R</em>2|^3}$
• The electric field obeys the principle of linear superposition: $E = E<em>1 + E</em>2$
• $E = \frac{1}{4\pi\epsilon} \sum<em>{i=1}^N \frac{q</em>i (R - R<em>i)}{|R - R</em>i|^3}$

Electric Field Due to Charge Distributions

• Field due to a line charge: $dE = \frac{\rho_l dl' \hat{R}}{4\pi\epsilon R^2}$
• Field due to a surface charge: $dE = \frac{\rho_s dS' \hat{R}}{4\pi\epsilon R^2}$
• Field due to a volume charge: $dE = \frac{\rho_v dV' \hat{R}}{4\pi\epsilon R^2}$
• Where: $\hat{R} = \frac{R - R'}{|R - R'|}$

Example: Electric Field Due to a Finite Line Charge

• Problem: Find the electric field, $E(P)$, due to a constant line charge $\rho_l$ between $z = a$ and $z = b$.
• Geometry: The problem is independent of $\phi$, so use cylindrical coordinates. $\hat{R} = \frac{R - R'}{|R - R'|}$
• $dE(R) = \frac{\rho_l dz' \hat{R}}{4 \pi \epsilon R^2}$
• $\vec{R'} = z' \hat{z}$
• $\vec{R} = r \hat{r} + z \hat{z}$
• $\vec{R} = \vec{R} - \vec{R'} = r \hat{r} + (z - z') \hat{z}$
• $|R| = \sqrt{r^2 + (z - z')^2}$
• $E(R) = \int<em>a^b dE = \int</em>a^b \frac{\rho_l (r \hat{r} + (z - z') \hat{z}) dz'}{4 \pi \epsilon (r^2 + (z - z')^2)^{3/2}}$
• Using the substitution $u = z' - z$:
  • $E(R) = \frac{\rho<em>l}{4 \pi \epsilon} \left[ \hat{r} \int</em>{u(a)}^{u(b)} \frac{r du}{(r^2 + u^2)^{3/2}} + \hat{z} \int_{u(a)}^{u(b)} \frac{-u du}{(r^2 + u^2)^{3/2}} \right]$
• $E(R) = \frac{\rho<em>l}{4 \pi \epsilon} \left[ \hat{r} \frac{u}{r \sqrt{r^2 + u^2}} \Big|</em>{z-b}^{z-a} + \hat{z} \frac{1}{\sqrt{r^2 + u^2}} \Big|_{z-b}^{z-a} \right]$
• $E(R) = \frac{\rho<em>l}{4 \pi \epsilon} \left[ \frac{1}{r \sqrt{r^2 + (z'-z)^2}} \Big|</em>{a}^{b} \hat{r} + \frac{1}{\sqrt{r^2 + (z'-z)^2}} \Big|_{a}^{b}\hat{z} \right]$
• $E(R) = \frac{\rho<em>l}{4 \pi \epsilon} \left[ \left(\frac{z-z'}{\sqrt{r^2 + (z - z')^2}}\right)</em>{z=a}^{z=b}\hat{z} + \left(\frac{r}{\sqrt{r^2 + (z-z')^2}}\right)_{z=a}^{z=b}\hat{r} \right]$

Example 4-4: Electric Field of a Ring of Charge

• Problem: Determine the electric field intensity $E$ at a point $P = (0, 0, h)$ along the axis of a ring of charge with radius $b$ and uniform line charge density $\rho_l$.
• Solution: Consider a differential ring segment with cylindrical coordinates $(b, \phi, 0)$. The segment has length $dl = b d\phi$ and charge $dq = \rho<em>l dl = \rho</em>l b d\phi$.
• The distance vector from segment 1 to point $P = (0, 0, h)$ is $\vec{R_1} = -b \hat{r} + h \hat{z}$.
• $|R_1| = \sqrt{b^2 + h^2}$
• $\hat{R_1} = \frac{-b \hat{r} + h \hat{z}}{\sqrt{b^2 + h^2}}$
• The electric field at $P = (0, 0, h)$ due to the charge in segment 1 is:
  • $dE<em>1 = \frac{\rho</em>l dl \hat{R<em>1}}{4 \pi \epsilon</em>0 R<em>1^2} = \frac{\rho</em>l b (-b \hat{r} + h \hat{z}) d\phi}{4 \pi \epsilon_0 (b^2 + h^2)^{3/2}}$
• Symmetry considerations: For every ring segment, there is a diametrically opposite segment. The $\hat{r}$-components cancel, and the $\hat{z}$-components add.
• $dE = dE<em>1 + dE</em>2 = \hat{z} \frac{\rho<em>l b h d\phi}{2 \pi \epsilon</em>0 (b^2 + h^2)^{3/2}}$
• Integrating over a semicircle ($0 \leq \phi \leq \pi$) to account for the symmetry:
  • $E = 2 \int<em>0^{\pi} \frac{\rho</em>l b h}{4 \pi \epsilon_0 (b^2 + h^2)^{3/2}} d\phi$
• $E = 2 \frac{\rho<em>l b h}{4 \epsilon</em>0 (b^2 + h^2)^{3/2}} = \hat{z} \frac{Q}{4 \pi \epsilon<em>0 (b^2 + h^2)^{3/2}}$, where $Q = 2 \pi b \rho</em>l$ is the total charge on the ring.

Example 4-5: Electric Field of a Circular Disk of Charge

• Problem: Find the electric field at point $P = (0, 0, h)$ due to a circular disk of radius $a$ and uniform charge density $\rho_s$.
• Treating the disk as a set of concentric rings, a ring of radius $r$ and width $dr$ has area $ds = 2 \pi r dr$ and charge $dq = \rho<em>s ds = 2 \pi \rho</em>s r dr$.
• Using the result from Example 4-4 for the on-axis electric field due to a ring:
  • $dE = \hat{z} \frac{h (2 \pi \rho<em>s r dr)}{4 \pi \epsilon</em>0 (r^2 + h^2)^{3/2}}$
• Integrating over the limits $r = 0$ to $r = a$:
  • $E = \int<em>0^a \frac{\rho</em>s h r dr}{2 \epsilon_0 (r^2 + h^2)^{3/2}} \hat{z}$
    Solved:
• $E = \frac{\rho<em>s}{2 \epsilon</em>0} \left( \frac{|h|}{\sqrt{a^2 + h^2})} - 1\right) = \pm \frac{\rho<em>s}{2 \epsilon</em>0} \left(1 - \frac{|h|}{\sqrt{a^2 + h^2})} \right)$
  • The plus sign for h > 0 (P above the disk) and the minus sign when h < 0 (P below the disk).
• For an infinite sheet of charge with $a = \infty$:
  • $E = \pm \frac{\rho<em>s}{2 \epsilon</em>0}$
  • $E$ is the same at all points above the x-y plane and at all points below the x-y plane.