masuk sir ct topics

Instructor: Mashuk Sir

1. Introduction & Course Materials

• Subject: Electromagnetic fields & waves (Electronic is crossed out)

• Books:

  • Field and Wave Electromagnetics - David K. Cheng

  • Elements of Electromagnetics - Matthew N.O. Sadiku

2. The Concept of Fields

A field represents the spatial distribution of a physical quantity.

• Static Electric Field: The field of a static (stationary) electron ($e^{-}$). It doesn’t change with respect to time.

• Static Magnetic Field: Produced by moving electrons at a steady / DC rate.

• Electromagnetic Waves: Produced when the movement of charges is not steady (i.e., acceleration or deacceleration).

3. Vector Calculus Postulates & Fundamentals

Divergence ($\vec{\nabla }\cdot \vec{A}$)

Represents whether a field is pointing outward or inward from a source.

• Electric Field: $\vec{\nabla }\cdot \vec{E}=\frac{\rho }{{\epsilon }_0}\to$ Divergence of the electric field exists. (Where charge density $\rho =\frac{dQ}{dV}$).

• Magnetic Field: Divergence exists as 0 ($\vec{\nabla }\cdot \vec{B}=0$).

Curl ($\vec{\nabla }\times \vec{A}$)

Represents if a field is rotational.

• Electric Field: $\vec{\nabla }\times \vec{E}=0\to$ Curl doesn’t exist. This fundamental postulate proves the electrostatic field is a conservative field.

  • Work done is the same in all paths for a certain 2 points.

  • For circuits, ${\oint }_C\vec{E}\cdot d\vec{l}=0$ becomes KVL (Kirchhoff’s Voltage Law).

• Magnetic Field: No continuous curl exists (Curl exists).

Important Theorems

• Divergence Theorem: ${\int }_v(\vec{\nabla }\cdot \vec{A})dv={\oint }_S\vec{A}\cdot d\vec{S}$

• Stoke’s Theorem: ${\int }_S(\vec{\nabla }\times \vec{A})\cdot d\vec{S}={\oint }_C\vec{A}\cdot d\vec{l}$

• Vector Identity: $\vec{\nabla }\times (\vec{\nabla }V)=0$

4. Coulomb’s Law & Gauss’s Law

Coulomb’s Force:

${\vec{F}}_{12}=\frac{1}{4\pi {\epsilon }_0}\frac{Q_1{Q}_2}{R_{12}^2}{\hat{a}}_{R_{12}}$

(Note: This works for point charges, doesn’t work for continuous fields).

Gauss’s Law Derivation:

Starting with $\vec{\nabla }\cdot \vec{E}=\frac{\rho }{{\epsilon }_0}$, integrate both sides:

${\int }_v(\vec{\nabla }\cdot \vec{E})dv=\frac{1}{{\epsilon }_0}{\int }_v\rho dv=\frac{Q}{{\epsilon }_0}$

Applying the Divergence theorem to the left side yields Gauss’s Law:

${\oint }_S\vec{E}\cdot d\vec{S}=\frac{Q}{{\epsilon }_0}$

(Note: Surface/area is a vector with direction to its normal).

5. Electric Potential ($V$)

Electric potential / voltage / EMF.

Comparing the vector identity $\vec{\nabla }\times (\vec{\nabla }V)=0$ with the postulate $\vec{\nabla }\times \vec{E}=0$:

$\vec{E}=-\vec{\nabla }V$

(The negative sign is a direction convention and has no mathematical significance).

(Equipotential lines form concentric circles around a charge).

Work done to bring a charge from infinity to distance $R$:

$V=-{\int }_{\infty }^R\left({\hat{a}}_R\frac{Q}{4\pi {\epsilon }_0{R}^2}\right)\cdot {\hat{a}}_{R}dR=\frac{Q}{4\pi {\epsilon }_{0}R}$

Potential Difference:

$V_{21}=V_{P_1}-V_{P_2}=\frac{Q}{4\pi {\epsilon }_0}\left(\frac{1}{R_1}-\frac{1}{R_2}\right)$

6. Point Charges, Dipoles & Continuous Charge

Field & Potential of a Point Charge

By applying Gauss’s Law with a radially outward field where ${\hat{a}}_R$ and ${\hat{a}}_n$ are the same:

\oint_S \hat{a}_R E_R \cdot \hat{a}_n ds = \frac{q}{\varepsilon_0} \implies E_R(4\pi R^2) = \frac{q}{\varepsilon_0}$$$$\vec{E} = \hat{a}_R E_R = \hat{a}_R \frac{1}{4\pi\varepsilon_0} \frac{q}{R^2} \quad \text{[Considering the charge is at origin]}

If the charge is not in the center (source at ${\vec{R}}^{\prime }$, observation at $\vec{R}$):

$\vec{E}=\frac{\vec{R}-{\vec{R}}^{\prime }}{|\vec{R}-{\vec{R}}^{\prime }|}\frac{q}{4\pi {\epsilon }_0|\vec{R}-{\vec{R}}^{\prime }|{}^2}\text{and}V=\frac{1}{4\pi {\epsilon }_0}{\sum }_{k=1}^n\frac{q_k}{|\vec{R}-{\vec{R}}_k^{\prime }|}$

Field & Potential of a Dipole

Setup: $+q$ and $-q$ separated by distance $d$.

Potential $V=\frac{q}{4\pi {\epsilon }_0}\left(\frac{1}{R^{+}}-\frac{1}{R^{-}}\right)$

Electric Field $\vec{E}=\frac{q}{4\pi {\epsilon }_0}\left\{\frac{\vec{R}-\vec{d}/2}{|\vec{R}-\vec{d}/2|{}^3}-\frac{\vec{R}+\vec{d}/2}{|\vec{R}+\vec{d}/2|{}^3}\right\}$

Binomial Expansion Approximation (if $d\ll R$):

Formula: $(1+x{)}^n={\sum }_{k=0}^n{}^n{C}_k{x}^k{\text{where }}^n{C}_k=(\genfrac{}{}{0ex}{}{n}{k})=\frac{n!}{k!(n-k)!}\text{ ? (Note: Author questions this formula)}$

Expanding the denominators (ignoring $d^2/4$ as it is close to zero) yields:

$|\vec{R}\pm \vec{d}/2|{}^{-3}\cong R^{-3}\left[1\mp \frac{3}{2}\frac{\vec{R}\cdot \vec{d}}{R^2}\right]$

Applying this gives the potential and field for a dipole moment $\vec{P}=q\vec{d}$:

• Potential: $V=\frac{qd\cos \theta }{4\pi {\epsilon }_0{R}^2}=\frac{\vec{P}\cdot {\hat{a}}_n}{4\pi {\epsilon }_0{R}^2}\text{[For a special case] (Note: likely }{\hat{a}}_R\text{)}$

• Field (Cartesian prep): $\vec{E}=\frac{q}{4\pi {\epsilon }_0{R}^3}\left[\frac{3\vec{R}\cdot \vec{d}}{R^2}\vec{R}-\vec{d}\right]$

• Field (Spherical form): Using $\vec{P}=P({\hat{a}}_R\cos \theta -{\hat{a}}_{\theta }\sin \theta )$ and $\vec{R}\cdot \vec{P}=RP\cos \theta$:

  $\vec{E}=\frac{P}{4\pi {\epsilon }_0{R}^3}({\hat{a}}_{R}2\cos \theta +{\hat{a}}_{\theta }\sin \theta )$

Infinite Line Charge

Determine the electric field intensity of an infinitely long, straight line charge of uniform density ${\rho }_L$ in air.

$d\vec{E}={\hat{a}}_R\frac{\rho d{v}^{\prime }}{4\pi {\epsilon }_0{R}^2}⟹\vec{E}=\frac{1}{4\pi {\epsilon }_0}{\int }_{v^{\prime }}{\hat{a}}_R\frac{\rho }{R^2}d{v}^{\prime }$

Setting up the integral with source at $z^{\prime }$ and observation at $r$:

$d\vec{E}=\frac{\underset{Q}{\underbrace{{\rho }_{L}d{z}^{\prime }}}({\hat{a}}_{r}r-{\hat{a}}_z{z}^{\prime })}{4\pi {\epsilon }_0(r^2+z^{\prime 2}{)}^{3/2}}$

7. Conductors and Dielectrics

Conductors

Inside a conductor, E field intensity $\vec{E}=0$, and charge density $\rho =0$.

Dielectrics (Polarization)

In an external $\vec{E}$ field, $e^{-}$ and nucleus reorient.

Polarization vector: $\vec{P}={\lim }_{\Delta V\to 0}\frac{{\sum }_{k=1}^{n\Delta V}{\vec{p}}_k}{\Delta V}$

Calculating potential $V$ from polarization:

$d{\vec{p}}_k=\vec{P}d{v}^{\prime }⟹dV=\frac{\vec{P}\cdot {\hat{a}}_R}{4\pi {\epsilon }_0{R}^2}d{v}^{\prime }$

$V=\frac{1}{4\pi {\epsilon }_0}{\int }_{v^{\prime }}\frac{\vec{P}\cdot {\hat{a}}_R}{R^2}d{v}^{\prime }$

Using vector identities ${\vec{\nabla }}^{\prime }(1/R)={\hat{a}}_R/R^2$ and $\vec{\nabla }\cdot (b\vec{C})=b\vec{\nabla }\cdot \vec{C}+\vec{C}\cdot \vec{\nabla }b$:

$V=\frac{1}{4\pi {\epsilon }_0}\left[{\int }_{v^{\prime }}{\vec{\nabla }}^{\prime }\cdot \left(\frac{\vec{P}}{R}\right)d{v}^{\prime }-{\int }_{v^{\prime }}\frac{{\vec{\nabla }}^{\prime }\cdot \vec{P}}{R}d{v}^{\prime }\right]$

Apply divergence theorem to the first integral:

$V=\frac{1}{4\pi {\epsilon }_0}{\oint }_{S^{\prime }}\frac{\vec{P}\cdot {\hat{a}}_n}{R}d{S}^{\prime }+\frac{1}{4\pi {\epsilon }_0}{\int }_{v^{\prime }}\frac{-({\vec{\nabla }}^{\prime }\cdot \vec{P})}{R}d{v}^{\prime }$

This defines two charge densities:

1. ${\rho }_{ps}=\vec{P}\cdot {\hat{a}}_n$ (Surface polarization charge density)

2. ${\rho }_p=-\vec{\nabla }\cdot \vec{P}$ (Volume polarization charge density)

8. Electric Flux Density ($\vec{D}$)

Applying Gauss’s Law with both charge types:

$\vec{\nabla }\cdot \vec{E}=\frac{\rho \text{ (due to E)}+{\rho }_p\text{ (due to effects E on dielectric)}}{{\epsilon }_0}$

Substituting ${\rho }_p$:

$\vec{\nabla }\cdot ({\epsilon }_0\vec{E}+\vec{P})=\rho$

We define Electric flux density ($\vec{D}$) (Note arrows: depends on?, def, vs field density):

$\vec{D}={\epsilon }_0\vec{E}+\vec{P}$

• Differential form: $\vec{\nabla }\cdot \vec{D}=\rho$

• Integral form: ${\oint }_S\vec{D}\cdot d\vec{S}=Q$

Material Parameters:

$\vec{P}={\epsilon }_0{\chi }_e\vec{E}({\chi }_e\to \text{electric susceptibility, a material parameter})$

$\vec{D}={\epsilon }_0(1+{\chi }_e)\vec{E}={\epsilon }_0{\epsilon }_r\vec{E}=\epsilon \vec{E}$

Where ${\epsilon }_r=1+{\chi }_e=\frac{\epsilon }{{\epsilon }_0}$

9. Electro-Static Boundary Conditions

For a closed loop $abcd$ across a boundary between Medium 1 and Medium 2:

If height $\Delta h\to 0$, the total line integral of electric field in closed loop is zero:

\oint_{abcd} \vec{E} \cdot d\vec{l} = \vec{E}_1 \cdot \Delta w + \vec{E}_2 \cdot (-\Delta w) = 0$$$$\vec{E}_{1t} \Delta w - \vec{E}_{2t} \Delta w = 0 \implies \vec{E}_{1t} = \vec{E}_{2t}

(where $t\to$ tangential component).