2105: Electromagnetic Field & Waves (Complete Study Notes) - Enhanced

ECE 2105: CLASS TEST STUDY GUIDE & STRATEGY

To ace Mashuk Sir’s Class Test, follow this high-impact, step-by-step revision strategy:

Table of Contents

1. Study Map: How and What to Study
2. High-Yield Revision Steps
3. Introduction & Course Materials
4. The Concept of Fields
5. Vector Calculus Postulates & Fundamentals
6. Coulomb’s Law & Gauss’s Law
7. Electric Potential ($V$)
8. Point Charges, Dipoles & Continuous Charge
9. Conductors and Dielectrics
10. Electric Flux Density ($\vec{D}$)
11. Electro-Static Boundary Conditions
12. Common Exam Pitfalls (Quick Diagnostic)
13. Quick-Reference Formula Table

1. Study Map: How and What to Study

• Vector Calculus Postulates (Topic 1 & Note 01): Focus on divergence/curl and Stokes’s/divergence theorems. You must know how to derive KVL from the curl postulate $\nabla \times \vec{E}=0$. Memorize and practice the vector identity proof $\nabla \times (\nabla V)\equiv 0$ (Cartesian coordinates).
• Gauss’s Law & Symmetric Applications (Topic 4 & Note 02): Focus on deriving Gauss’s Law from the divergence postulate. Practice applying Gauss’s Law to calculate the electric field of symmetric distributions: the infinite sheet of charge, the infinitely long line charge, and the uniformly charged spherical cloud (especially the case inside vs. outside the cloud showing linear growth and quadratic decay, maximizing at the surface).
• Electric Dipole (Topic 6 & Note 02): Derive the potential $V$ and field intensity $\vec{E}$ at a distant point. Pay special attention to the binomial/Taylor series expansion steps using $(1\pm x{)}^{-1/2}$ and $(1\pm x{)}^{-3/2}$ respectively. This is a very frequent exam question.
• Conductors & Dielectrics (Topic 7 & Note 02):
  • Memorize the 5 physical properties of conductors in a static field.
  • Study the detailed derivation of potential $V$ due to polarization, which defines surface and volume bound charge densities: ${\rho }_{ps}=\vec{P}\cdot {\hat{a}}_n$ and ${\rho }_p=-\nabla \cdot \vec{P}$.
  • Know how to derive $\vec{D}={ϵ}_0\vec{E}+\vec{P}$ starting from Gauss’s postulate.
• Boundary Conditions (Topic 9 & Note 02): Derive tangential continuity $E_{1t}=E_{2t}$ (using a tiny rectangular contour) and normal discontinuity $D_{1n}-D_{2n}={\rho }_s$ (using a cylindrical pillbox). Memorize the refraction law $\frac{\tan {\alpha }_1}{\tan {\alpha }_2}=\frac{{ϵ}_1}{{ϵ}_2}$.

2. High-Yield Revision Steps

1. Read Note 01 & 02: Focus on understanding the physical meaning of every equation, keeping the step-by-step mathematical proofs in mind.
2. Review the Numericals (Note 02, Section 11): Solve the 4 classic numericals (point charge symmetry, variable density spherical charge, work done in conservative field, electrohydrodynamic pump potential).
3. Use the Cheatsheet & Pitfall Guide (Note 06): Read the pitfalls (e.g., the $-{\rho }_v$ inward field lines trap) and take the 5-question interactive diagnostic quiz to test your exam-readiness.

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2105: Masuk sir upto week 3

Instructor: Mashuk Sir

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1. Introduction & Course Materials

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

• Books:

  • Field and Wave Electromagnetics - David K. Cheng {standard textbook focusing on field theory and electromagnetic wave propagation}

  • Elements of Electromagnetics - Matthew N.O. Sadiku {textbook with detailed vector analysis and step-by-step electrostatics/magnetostatics derivations}

• Action at a distance: {the physical phenomenon where two separated bodies exert forces on each other across space without direct mechanical contact} (Accompanied by a drawing of a bar magnet or metallic bar).

• Self-Study Reminders: Ex-3.5 | self (3.6, 3.7) {textbook exercise problems assigned for practice}

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2. The Concept of Fields

A field {a region of space where every point is assigned a vector or scalar value} represents the spatial distribution of a physical quantity.

• Static Electric Field: {an electric field created by stationary electric charges} The field of a static (stationary) electron ($e^{-}$). It doesn’t change with respect to time {i.e., $\frac{\partial \vec{E}}{\partial t}=0$}.

• Static Magnetic Field: {a magnetic field produced by steady, time-invariant electric currents} Produced by moving electrons at a steady / DC rate {constant velocity, zero acceleration}.

• Electromagnetic Waves: {self-propagating transverse oscillations of electric and magnetic fields carrying radiant energy} Produced when the movement of charges is not steady (i.e., acceleration or deacceleration) {time-varying currents radiate electromagnetic energy into space}.

Summary: Charge Dynamics & Field Types

Charge State	Field Produced	Time Variation	Physical Behavior
Stationary Charge ($v=0$)	Static Electric Field ($\vec{E}$)	Time-invariant	Electrostatic force (Coulomb’s Law)
Steady Motion ($v=\text{const}$)	Static Magnetic Field ($\vec{B}$)	Time-invariant	Magnetostatics (Biot-Savart / Ampere’s Law)
Accelerated Charge ($a\ne 0$)	Electromagnetic Waves ($\vec{E}(t),\vec{B}(t)$)	Time-varying	Coupled wave propagation at speed of light

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3. Vector Calculus Postulates & Fundamentals

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

Represents whether a field is pointing outward or inward from a source {measures net flux per unit volume exiting an infinitesimal volume}.

• Electric Field: $\vec{\nabla }\cdot \vec{E}=\frac{\rho }{{\epsilon }_0}\to$ Divergence of the electric field exists. (Where charge density $\rho =\frac{dQ}{dV}$ {volume charge density in Coulombs per cubic meter, ${\text{C/m}}^3$}).

• Magnetic Field: Divergence exists as 0 ($\vec{\nabla }\cdot \vec{B}=0$) {Gauss’s law for magnetism, indicating magnetic monopoles do not exist}.

Concept: Physical Meaning of Divergence

• $\vec{\nabla }\cdot \vec{E}>0$: Point acts as a source of electric field lines (positive charge $+Q$).
• $\vec{\nabla }\cdot \vec{E}<0$: Point acts as a sink of electric field lines (negative charge $-Q$).
• $\vec{\nabla }\cdot \vec{B}=0$: Magnetic field lines always form continuous closed loops; every field line entering a closed volume must also exit it.

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

Represents if a field is rotational {measures the circulation intensity or rotational vorticity of a vector field around a point}.

• Electric Field: $\vec{\nabla }\times \vec{E}=0\to$ Curl doesn’t exist. This fundamental postulate proves the electrostatic field is a conservative field {a field where work done along any path depends only on initial and final positions}.

  • 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) {stating the sum of electric potential differences around any closed circuit loop is zero}.

• Magnetic Field: No continuous curl exists (Curl exists) {Ampere’s law: $\vec{\nabla }\times \vec{B}={\mu }_0\vec{J}$, meaning magnetic fields curl around electric currents}.

Mechanism: Derivation of KVL from Electrostatic Curl Postulate

Since electrostatics dictates $\vec{\nabla }\times \vec{E}=0$, applying Stoke’s Theorem around a closed contour $C$ enclosing surface $S$: ${\oint }_C\vec{E}\cdot d\vec{l}={\int }_S(\vec{\nabla }\times \vec{E})\cdot d\vec{S}={\int }_{S}0\cdot d\vec{S}=0$ Physically, line-integrating the electric field $\vec{E}$ along a closed loop represents potential drops around a circuit path, directly yielding Kirchhoff’s Voltage Law ($\sum V=0$).

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Important Theorems

• Divergence Theorem: {Gauss-Ostrogradsky Theorem} ${\int }_v(\vec{\nabla }\cdot \vec{A})dv={\oint }_S\vec{A}\cdot d\vec{S}$ {converts a volume integral of divergence into a surface integral of flux}

• Stoke’s Theorem: ${\int }_S(\vec{\nabla }\times \vec{A})\cdot d\vec{S}={\oint }_C\vec{A}\cdot d\vec{l}$ {converts a surface integral of curl into a closed line integral of circulation}

• Vector Identity: $\vec{\nabla }\times (\vec{\nabla }V)=0$ {the curl of the gradient of any scalar field $V$ is identically zero}

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4. Coulomb’s Law & Gauss’s Law

Coulomb’s Force: {electrostatic attraction or repulsion between static point charges}

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

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

Limitation of Coulomb's Law

Coulomb’s Law directly calculates forces between discrete point charges in free space. For continuous distributions (lines, surfaces, or volumes of charge), we must integrate infinitesimal charge elements $dQ$ or utilize Gauss’s Law for symmetric systems.

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Gauss’s Law Derivation:

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

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

Applying the Divergence theorem ${\int }_v(\vec{\nabla }\cdot \vec{E})dv={\oint }_S\vec{E}\cdot d\vec{S}$ 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 ${\hat{a}}_n$).

Mechanism: Applying Gauss's Law

1. Select a closed Gaussian surface {an imaginary 3D surface matching the symmetry of the charge distribution}.
2. Ensure field magnitude $E$ is constant over the surface and directed strictly parallel or perpendicular to normal vector $d\vec{S}={\hat{a}}_{n}dS$.
3. Factor $E$ out of the integral: $E{\oint }_{S}dS=\frac{Q_{\text{enclosed}}}{{\epsilon }_0}$.

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5. Electric Potential ($V$)

Electric potential / voltage / EMF {scalar potential field representing potential energy per unit charge, measured in Volts}.

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: {work required to move a unit positive charge between two points $P_1$ and $P_2$}

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

Clarification: Negative Sign Convention in $\vec{E}=-\vec{\nabla }V$

The negative sign signifies that the electric field $\vec{E}$ points in the direction of maximum decreasing potential (from high voltage to low voltage). Positive work must be done against the field to move a positive charge to a higher potential.

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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 {unit radial vector points parallel to normal vector of spherical Gaussian surface}:

${\oint }_S{\hat{a}}_R{E}_R\cdot {\hat{a}}_{n}ds=\frac{q}{{\epsilon }_0}⟹E_R(4\pi R^2)=\frac{q}{{\epsilon }_0}$

$\vec{E}={\hat{a}}_R{E}_R={\hat{a}}_R\frac{1}{4\pi {\epsilon }_0}\frac{q}{R^2}\text{[Considering the charge is at origin]}$

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

$\vec{E}=\frac{\vec{R}-{\vec{R}}^{\prime }}{|\vec{R}-{\vec{R}}^{\prime }|}\frac{q}{4\pi {\epsilon }_0|\vec{R}-{\vec{R}}^{\prime }|{}^2}=\frac{q(\vec{R}-{\vec{R}}^{\prime })}{4\pi {\epsilon }_0|\vec{R}-{\vec{R}}^{\prime }|{}^3}\text{and}V=\frac{1}{4\pi {\epsilon }_0}{\sum }_{k=1}^n\frac{q_k}{|\vec{R}-{\vec{R}}_k^{\prime }|}$

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Field & Potential of a Dipole

Setup: $+q$ and $-q$ separated by distance $d$ {an electric dipole consisting of two equal and opposite charges}.

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 $\vec{P}=q\vec{d}$ {vector pointing from negative charge $-q$ to positive charge $+q$}:

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

• 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 $\vec{R}\cdot \vec{P}=RP\cos \theta$:

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

Resolution of Author Notes in Dipole Derivation

1. Binomial Expansion Formula: For arbitrary real exponent $n=-3/2$ (with $x\ll 1$), the generalized Taylor expansion is $(1+x{)}^n\approx 1+nx$. Here, ${\left(1\pm \frac{\vec{R}\cdot \vec{d}}{R^2}\right)}^{-3/2}\approx 1\mp \frac{3}{2}\frac{\vec{R}\cdot \vec{d}}{R^2}$. The discrete combinatorics formula $\left(\genfrac{}{}{0ex}{}{n}{k}\right)$ listed in original note applies to integer $n$, whereas Taylor expansion applies for negative exponents.
2. Unit Vector in Dipole Potential: The author note (Note: likely \hat{a}_R) is correct! $\vec{P}\cdot {\hat{a}}_R=P\cos \theta$, so $V=\frac{\vec{P}\cdot {\hat{a}}_R}{4\pi {\epsilon }_0{R}^2}$.

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Infinite Line Charge

Determine the electric field intensity of an infinitely long, straight line charge of uniform density ${\rho }_L$ {charge per unit length in Coulombs per meter, $\text{C/m}$} 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 $r$:

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

Mechanism: Integration Result for Infinite Line Charge

Due to z-axis symmetry, the $z$-components cancel (${\int }_{-\infty }^{\infty }{z}^{\prime }d{z}^{\prime }=0$), leaving only the radial component: $\vec{E}=\frac{{\rho }_L}{4\pi {\epsilon }_0}{\hat{a}}_r{\int }_{-\infty }^{\infty }\frac{rd{z}^{\prime }}{(r^2+z^{\prime 2}{)}^{3/2}}=\frac{{\rho }_L}{2\pi {\epsilon }_{0}r}{\hat{a}}_r$ This confirms that field intensity falls off as $1/r$ for a line charge, compared to $1/R^2$ for a point charge.

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7. Conductors and Dielectrics

Conductors

Inside a conductor {a material with abundant free electrons}, E field intensity $\vec{E}=0$, and charge density $\rho =0$.

Concept: Why $\vec{E}=0$ inside a Conductor

Any internal electric field exerts forces on free electrons, moving them to the conductor surface until the induced surface charges create an opposing field that completely cancels the internal field (${\vec{E}}_{\text{internal}}=0$). Thus, the interior of a static conductor is equipotential.

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Dielectrics (Polarization)

In an external $\vec{E}$ field, $e^{-}$ and nucleus reorient {dielectric materials have no free charges, but bound charges displace slightly under an applied field}.

Polarization vector: $\vec{P}={\lim }_{\Delta V\to 0}\frac{{\sum }_{k=1}^{n\Delta V}{\vec{p}}_k}{\Delta V}$ {dipole moment per unit volume, measured in ${\text{C/m}}^2$}

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 {bound charge density appearing on dielectric surface})

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

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

Applying Gauss’s Law with both charge types:

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

Substituting ${\rho }_p=-\vec{\nabla }\cdot \vec{P}$:

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

We define Electric flux density ($\vec{D}$) {also called displacement field, independent of medium polarization}:

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

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

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

Concept: Electric Flux Density ( $\vec{D}$) vs Electric Field ($\vec{E}$)

• $\vec{E}$ Field: Depends on all charges (both free charges $\rho$ and bound polarization charges ${\rho }_p$).
• $\vec{D}$ Field: Depends only on free charges $\rho$. This makes $\vec{D}$ convenient for analyzing boundary problems and dielectric interfaces without needing to measure internal polarization explicitly.

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Material Parameters:

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

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

Where ${\epsilon }_r=1+{\chi }_e=\frac{\epsilon }{{\epsilon }_0}$ {relative permittivity or dielectric constant}

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9. Electro-Static Boundary Conditions

For a closed loop $abcd$ across a boundary between Medium 1 (${\epsilon }_1$) and Medium 2 (${\epsilon }_2$):

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

${\oint }_{abcd}\vec{E}\cdot d\vec{l}={\vec{E}}_1\cdot \Delta w+{\vec{E}}_2\cdot (-\Delta w)=0$

${\vec{E}}_{1t}\Delta w-{\vec{E}}_{2t}\Delta w=0⟹{\vec{E}}_{1t}={\vec{E}}_{2t}$

(where $t\to$ tangential component).

Concept: Complete Boundary Conditions

Electrostatic boundary conditions govern field transitions across the interface between two media:

1. Tangential Electric Field: Continuous across boundary ${\vec{E}}_{1t}={\vec{E}}_{2t}⟹\frac{{\vec{D}}_{1t}}{{\epsilon }_1}=\frac{{\vec{D}}_{2t}}{{\epsilon }_2}$
2. Normal Electric Flux Density: Discontinuous by surface free charge density ${\rho }_s$ $D_{1n}-D_{2n}={\rho }_s⟹{\epsilon }_1{E}_{1n}-{\epsilon }_2{E}_{2n}={\rho }_s$ (If interface has no free surface charge ${\rho }_s=0$, then $D_{1n}=D_{2n}$).

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Conductor–Dielectric Boundary (Special Case)

Setting Medium 2 to be a perfect conductor forces ${\vec{E}}_2={\vec{D}}_2=0$ inside it. Substituting into the two general boundary conditions above collapses them to a single-medium result:

$E_t=0\text{and}{D}_n={\rho }_s⟹E_n=\frac{{\rho }_s}{\epsilon }$

Concept: Physical Meaning

A static field can never leave or enter a conductor’s surface at an angle — it must always be strictly normal (perpendicular), because any tangential component would drive free charge along the surface until it vanished. The entire surface charge shows up as a jump in the normal component of $\vec{D}$.

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Derivation: Law of Refraction (Bending) of Electric Field Lines

Setup: At a charge-free boundary (${\rho }_s=0$) between two dielectrics, ${\vec{E}}_1$ makes angle ${\alpha }_1$ with the normal in Medium 1, and ${\vec{E}}_2$ makes angle ${\alpha }_2$ with the normal in Medium 2.

1. Resolve each field into tangential and normal components using the angle from the normal: $E_{1t}=E_1\sin {\alpha }_1,E_{1n}=E_1\cos {\alpha }_1,E_{2t}=E_2\sin {\alpha }_2,E_{2n}=E_2\cos {\alpha }_2$
2. Apply the tangential continuity condition $E_{1t}=E_{2t}$: $E_1\sin {\alpha }_1=E_2\sin {\alpha }_2\text{...(i)}$
3. Apply the normal condition with ${\rho }_s=0$, i.e. $D_{1n}=D_{2n}⟹{\epsilon }_1{E}_{1n}={\epsilon }_2{E}_{2n}$: ${\epsilon }_1{E}_1\cos {\alpha }_1={\epsilon }_2{E}_2\cos {\alpha }_2\text{...(ii)}$
4. Divide (i) by (ii) to eliminate $E_1$ and $E_2$: $\frac{E_1\sin {\alpha }_1}{{\epsilon }_1{E}_1\cos {\alpha }_1}=\frac{E_2\sin {\alpha }_2}{{\epsilon }_2{E}_2\cos {\alpha }_2}⟹\frac{\tan {\alpha }_1}{{\epsilon }_1}=\frac{\tan {\alpha }_2}{{\epsilon }_2}$
5. Rearrange to the final form: $\boxed{\frac{\tan {\alpha }_1}{\tan {\alpha }_2}=\frac{{\epsilon }_1}{{\epsilon }_2}=\frac{{\epsilon }_{r1}}{{\epsilon }_{r2}}}$

Worked Numerical: Refraction Angle

Problem: Medium 1 (${\epsilon }_{r1}=2$) and Medium 2 (${\epsilon }_{r2}=8$) share a charge-free boundary. ${\vec{E}}_1$ makes ${\alpha }_1=30°$ with the normal. Find ${\alpha }_2$. $\tan {\alpha }_2=\tan {\alpha }_1\left(\frac{{\epsilon }_{r2}}{{\epsilon }_{r1}}\right)=\tan (30°)\times 4=0.5774\times 4=2.309$ ${\alpha }_2={\tan }^{-1}(2.309)\approx 66.6°$ The field bends away from the normal when entering the medium of higher permittivity — the field line becomes more tangential to the interface as ${\epsilon }_r$ increases.

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10. Common Exam Pitfalls (Quick Diagnostic)

Fast self-check before the CT — if any line below trips you up, go back and re-derive it on paper.

1. Negative charge cloud direction trap. For ${\rho }_v>0$, field lines point radially outward. For a negative cloud ($-{\rho }_v$), don’t just negate the magnitude — the field points radially inward, so sketch arrows toward the center: ${\vec{E}}_{in}=-{\hat{a}}_r\frac{|{\rho }_v|r}{3{\epsilon }_0}$, ${\vec{E}}_{out}=-{\hat{a}}_r\frac{|{\rho }_v|a^3}{3{\epsilon }_0{r}^2}$.
2. $\vec{E}$ vs $\vec{D}$ dependence. $\vec{E}$ responds to all charge (free + bound polarization). $\vec{D}$ responds only to free charge. If a problem gives you polarization $\vec{P}$ or a dielectric-filled region, check which one you were actually asked for before integrating.
3. Binomial exponent for the dipole. The expansion is not the integer combinatorics formula $\left(\genfrac{}{}{0ex}{}{n}{k}\right)$ — it’s the generalized Taylor series $(1+x{)}^n\approx 1+nx$ for non-integer $n$ (here $n=-1/2$ for potential, $n=-3/2$ for field). Mixing these up is the single most common CT dipole-derivation mistake.
4. Work-done sign convention. Work done by the field is $W=q\int \vec{E}\cdot d\vec{l}$ (no extra minus sign beyond what’s in $\vec{E}\cdot d\vec{l}$). Work done by an external agent against the field carries an explicit extra minus: $W_{ext}=-q{\int }_{P_1}^{P_2}\vec{E}\cdot d\vec{l}$. Forgetting this flips your final sign.
5. Tangential vs normal boundary conditions. Tangential $\vec{E}$ is always continuous ($E_{1t}=E_{2t}$) — this comes from the curl postulate. Normal $\vec{D}$ jumps by free surface charge ($D_{1n}-D_{2n}={\rho }_s$) — this comes from Gauss’s law. Don’t apply the wrong postulate to the wrong component.
6. Conductor boundary is a special case, not a new law. $E_t=0$ and $E_n={\rho }_s/\epsilon$ at a conductor surface fall directly out of the general two-medium boundary conditions by setting ${\vec{E}}_{inside}=0$ — you don’t need to re-derive from scratch, just substitute.
7. ${\hat{a}}_R$ direction in polarization potential. ${\nabla }^{\prime }(1/R)$ is a gradient taken with respect to the source coordinates $(x^{\prime },y^{\prime },z^{\prime })$, not the observation point — this sign convention is easy to flip under exam pressure.

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11. Quick-Reference Formula Table

A one-glance summary of every formula covered in this guide (Weeks 1–3 CT scope only).

Concept	Formula
Divergence postulate	$\vec{\nabla }\cdot \vec{E}=\rho /{\epsilon }_0$
Curl postulate	$\vec{\nabla }\times \vec{E}=0⟹{\oint }_C\vec{E}\cdot d\vec{l}=0$ (KVL)
Coulomb’s Law	${\vec{F}}_{12}=\frac{1}{4\pi {\epsilon }_0}\frac{Q_1{Q}_2}{R_{12}^2}{\hat{a}}_{R_{12}}$
Gauss’s Law (integral)	${\oint }_S\vec{E}\cdot d\vec{S}=Q/{\epsilon }_0$
Infinite sheet of charge	$\vec{E}={\hat{a}}_n\frac{{\rho }_s}{2{\epsilon }_0}$
Infinite line charge	$\vec{E}={\hat{a}}_r\frac{{\rho }_L}{2\pi {\epsilon }_{0}r}$
Uniform sphere, inside ($R\le a$)	${\vec{E}}_{in}={\hat{a}}_R\frac{{\rho }_{0}R}{3{\epsilon }_0}$ (linear)
Uniform sphere, outside ($R>a$)	${\vec{E}}_{out}={\hat{a}}_R\frac{{\rho }_0{a}^3}{3{\epsilon }_0{R}^2}$ (inverse-square)
Potential of point charge	$V=\frac{Q}{4\pi {\epsilon }_{0}R}$
Field–potential relation	$\vec{E}=-\vec{\nabla }V$
Dipole potential	$V=\frac{\vec{P}\cdot {\hat{a}}_R}{4\pi {\epsilon }_0{R}^2}$
Dipole field (spherical)	$\vec{E}=\frac{P}{4\pi {\epsilon }_0{R}^3}(2\cos \theta {\hat{a}}_R+\sin \theta {\hat{a}}_{\theta })$
Polarization surface bound charge	${\rho }_{ps}=\vec{P}\cdot {\hat{a}}_n$
Polarization volume bound charge	${\rho }_p=-\vec{\nabla }\cdot \vec{P}$
Electric flux density	$\vec{D}={\epsilon }_0\vec{E}+\vec{P}=\epsilon \vec{E}$
Susceptibility relation	$\vec{D}={\epsilon }_0(1+{\chi }_e)\vec{E}$, ${\epsilon }_r=1+{\chi }_e$
Tangential BC	$E_{1t}=E_{2t}$
Normal BC	$D_{1n}-D_{2n}={\rho }_s$
Conductor surface BC	$E_t=0,E_n={\rho }_s/\epsilon$
Refraction law	$\frac{\tan {\alpha }_1}{\tan {\alpha }_2}=\frac{{\epsilon }_1}{{\epsilon }_2}$

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Final Pre-CT Checklist

• Can derive KVL and Gauss’s Law from the two fundamental postulates without looking.
• Can derive $\vec{E}$ for infinite sheet, infinite line, and both inside/outside a uniform sphere.
• Can derive dipole $V$ and $\vec{E}$ including the binomial expansion step.
• Can derive ${\rho }_{ps}$, ${\rho }_p$, and $\vec{D}={\epsilon }_0\vec{E}+\vec{P}$ from the polarization potential integral.
• Can derive tangential BC, normal BC, the conductor special case, and the refraction law — all four, back to back.
• Can sketch field/equipotential lines for a $+{\rho }_v$ sphere, a $-{\rho }_v$ sphere, and a dipole without hesitating on arrow direction.