Related Concepts: 2.02 Electric Potential, Equipotential Contours & Dipole Derivations | 2.04 Electrostatic Boundary Conditions, Refraction Law & Poisson-Laplace Equations | 2.01 Fundamental Postulates of Electrostatics & Gauss_s Law Applications | 1.04 Media Properties & Conductor-Insulator Behaviour

2.03 Conductors, Dielectrics & Polarization Charge Densities

Core Idea

Materials in a static electric field split into two families. Conductors have free electrons that move until the internal field is exactly cancelled. Dielectrics have no free electrons, but their bound charges shift microscopically to form aligned dipoles — a polarization $\vec{P}$. Accounting for that polarization is what forces us to invent $\vec{D}={ϵ}_0\vec{E}+\vec{P}$, a result asked in 2016, 2019, 2021.

graph TD
    A["Material in a static E field"] --> B{"Free electrons available?"}
    B -->|"Yes"| C["CONDUCTOR"]
    B -->|"No"| D["DIELECTRIC"]
    C --> C1["Charges migrate to the surface"]
    C1 --> C2["E inside = 0, rho_v inside = 0<br/>equipotential body, E normal at surface"]
    D --> D1["Bound charges displace slightly"]
    D1 --> D2["Microscopic dipoles align<br/>Polarization vector P"]
    D2 --> D3["Bound charges rho_ps and rho_p appear"]
    D3 --> D4["D = eps0 E + P"]

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1. Conductors in Static Electric Fields [PYQ: 2016, 2020]

The mechanism, step by step

A conductor contains a sea of electrons free to move. Apply an external field ${\vec{E}}_{\text{ext}}$ and those electrons drift against it (they are negative), piling up on one face and leaving positive ions exposed on the opposite face. This separated surface charge creates an internal field ${\vec{E}}_{\text{ind}}$ opposing ${\vec{E}}_{\text{ext}}$. Migration continues — in about $10^{-19}$ s for copper — until the two exactly cancel. At that point no force acts on the remaining free charges, so motion stops. This is electrostatic equilibrium.

1.1 The Five Properties of a Conductor in Equilibrium

Property 1 — Zero internal field

${\vec{E}}_{\text{inside}}=0$ Reason: if it were non-zero, free charges would still feel a force and keep moving — contradicting equilibrium.

Property 2 — Zero internal volume charge density

${\rho }_v=0\text{everywhere inside}$ Proof: from Gauss’s postulate, ${\rho }_v=\nabla \cdot \vec{D}=\nabla \cdot (ϵ\vec{E})=ϵ\nabla \cdot (0)=0$. Consequence: all excess charge must reside on the outer surface of the conductor. This is the basis of electrostatic shielding (the Faraday cage).

Property 3 — The conductor is an equipotential body

$V_b-V_a=-{\int }_a^b\vec{E}\cdot d\vec{l}=-{\int }_a^{b}0\cdot d\vec{l}=0⟹V_a=V_b$ Every point in and on the conductor sits at the same potential. Its surface is therefore an equipotential surface.

Property 4 — The surface field is purely normal

$E_t=0\text{at the surface}$ Reason: any tangential component would drive surface charges sideways along the surface — again contradicting equilibrium. Static field lines must therefore meet a conductor surface perpendicularly, which is consistent with Property 3 (field lines are always perpendicular to equipotentials).

Property 5 — The normal surface field magnitude

${\vec{E}}_n={\hat{a}}_n\frac{{\rho }_s}{{ϵ}_0},D_n={\rho }_s$ Obtained from a Gaussian pillbox straddling the surface, with one face inside the conductor (where $\vec{D}=0$) and one face just outside.

1.2 Conductor–Free-Space Boundary [PYQ: 2016, 2020]

PYQ — 2016, 2020 (09 marks)

Determine the normal and tangential components of electric field intensity $\vec{E}$, and electric flux density $\vec{D}$, at the boundary of a conductor and free space.

Derivation

Let medium 1 be free space and medium 2 the conductor, with ${\vec{E}}_2={\vec{D}}_2=0$ inside.

Tangential component — small rectangular loop. Apply ${\oint }_C\vec{E}\cdot d\vec{l}=0$ to a loop of width $\Delta w$ and height $\Delta h\to 0$ straddling the surface. The two short sides contribute nothing as $\Delta h\to 0$: $E_{1t}\Delta w-E_{2t}\Delta w=0⟹E_{1t}=E_{2t}=0$ $\boxed{E_t=0,D_t={ϵ}_0{E}_t=0}$

Normal component — Gaussian pillbox. Apply ${\oint }_S\vec{D}\cdot d\vec{s}=Q_{\text{enc}}$ to a pillbox of face area $\Delta s$ and height $\Delta h\to 0$. Only the outer face contributes since ${\vec{D}}_2=0$: $D_{1n}\Delta s={\rho }_s\Delta s$ $\boxed{D_n={\rho }_s,E_n=\frac{{\rho }_s}{{ϵ}_0}}$

Summary: $\vec{E}={\hat{a}}_n\frac{{\rho }_s}{{ϵ}_0},\vec{D}={\hat{a}}_n{\rho }_s\text{just outside; both zero inside.}$

[FIGURE: Conductor–free-space interface with (a) a rectangular loop $abcd$ of width $\Delta w$ and height $\Delta h\to 0$ used for the tangential condition, and (b) a cylindrical pillbox of face area $\Delta s$ and height $\Delta h\to 0$ used for the normal condition; surface charge ${\rho }_s$ shown on the conductor face — source: David K. Cheng, Ch. 3, Fig. 3-19/3-23]

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2. Dielectric Polarization [PYQ: 2016, 2019, 2021]

A dielectric {an insulator — a material with no free electrons available for conduction} cannot pass current, but it is far from inert in a field.

2.1 The Two Polarization Mechanisms

Mechanism	What happens	Typical materials
Non-polar (induced) polarization	Molecules have no intrinsic dipole moment. The applied field pulls the electron cloud one way and the nucleus the other, inducing a dipole moment aligned with $\vec{E}$.	Hydrogen, oxygen, most plastics
Polar (orientational) polarization	Molecules already possess a permanent dipole moment but are randomly oriented, giving zero net effect. The applied field exerts a torque $\vec{\tau }=\vec{p}\times \vec{E}$ that partially aligns them.	Water, ammonia

Either way the result is the same: a large number of microscopic dipoles all pointing more or less along $\vec{E}$.

Definition — Polarization Vector $\vec{P}$

The polarization vector is the net electric dipole moment per unit volume: $\vec{P}={\lim }_{\Delta v\to 0}\frac{{\sum }_{i=1}^{n\Delta v}{\vec{p}}_i}{\Delta v}{\text{(C/m}}^2\text{)}$ where ${\vec{p}}_i=q_i{\vec{d}}_i$ is an individual microscopic dipole moment and $n$ is the dipole number density.

Why $\vec{P}$ has the same units as $\vec{D}$

Dipole moment per volume $=$ (C·m)/m³ $=$ C/m². That is not a coincidence — $\vec{P}$ genuinely represents a bound-charge surface density, as the next derivation shows.

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3. Master Derivation: Bound Charge Densities [PYQ: 2016, 2019, 2021]

PYQ — 2016 (11 marks), 2019, 2021 (07/08 marks)

Show that the total electric flux density in a dielectric material is $\vec{D}={ϵ}_0\vec{E}+\vec{P}$, where the symbols have their usual meanings.

This requires first deriving the bound charge densities ${\rho }_{ps}$ and ${\rho }_p$.

Part A — Deriving ${\rho }_{ps}$ and ${\rho }_p$

Step 1 — Set up the potential integral. An infinitesimal volume $d{v}^{\prime }$ of polarized dielectric behaves as a dipole of moment $d\vec{p}=\vec{P}d{v}^{\prime }$. Using the dipole potential from 2.02 Electric Potential, Equipotential Contours & Dipole Derivations and integrating over the dielectric volume $V^{\prime }$: $V=\frac{1}{4\pi {ϵ}_0}{\int }_{V^{\prime }}\frac{\vec{P}\cdot {\hat{a}}_R}{R^2}d{v}^{\prime }$

Step 2 — Substitute the gradient identity. Since $\frac{{\hat{a}}_R}{R^2}={\nabla }^{\prime }\left(\frac{1}{R}\right)$, where ${\nabla }^{\prime }$ differentiates with respect to the source coordinates $(x^{\prime },y^{\prime },z^{\prime })$: $V=\frac{1}{4\pi {ϵ}_0}{\int }_{V^{\prime }}\vec{P}\cdot {\nabla }^{\prime }\left(\frac{1}{R}\right)d{v}^{\prime }$

Step 3 — Apply the vector product rule. Using ${\nabla }^{\prime }\cdot \left(\frac{\vec{P}}{R}\right)=\frac{{\nabla }^{\prime }\cdot \vec{P}}{R}+\vec{P}\cdot {\nabla }^{\prime }\left(\frac{1}{R}\right)$, rearrange and substitute: $V=\frac{1}{4\pi {ϵ}_0}\left[{\int }_{V^{\prime }}{\nabla }^{\prime }\cdot \left(\frac{\vec{P}}{R}\right)d{v}^{\prime }-{\int }_{V^{\prime }}\frac{{\nabla }^{\prime }\cdot \vec{P}}{R}d{v}^{\prime }\right]$

Step 4 — Apply the Divergence Theorem to the first integral, converting it to a closed surface integral over the dielectric’s boundary $S^{\prime }$ with outward normal ${\hat{a}}_n$: $V=\frac{1}{4\pi {ϵ}_0}{\oint }_{S^{\prime }}\frac{\vec{P}\cdot {\hat{a}}_n}{R}d{s}^{\prime }+\frac{1}{4\pi {ϵ}_0}{\int }_{V^{\prime }}\frac{-({\nabla }^{\prime }\cdot \vec{P})}{R}d{v}^{\prime }$

Step 5 — Compare with the standard potential formulas. The potential produced by a surface charge density ${\rho }_{ps}$ plus a volume charge density ${\rho }_p$ is: $V=\frac{1}{4\pi {ϵ}_0}{\oint }_{S^{\prime }}\frac{{\rho }_{ps}}{R}d{s}^{\prime }+\frac{1}{4\pi {ϵ}_0}{\int }_{V^{\prime }}\frac{{\rho }_p}{R}d{v}^{\prime }$ Matching the two expressions term by term: $\boxed{{\rho }_{ps}=\vec{P}\cdot {\hat{a}}_n{\text{(C/m}}^2\text{)}}\boxed{{\rho }_p=-\nabla \cdot \vec{P}{\text{(C/m}}^3\text{)}}■$

Physical meaning of the two bound charges

• ${\rho }_{ps}=\vec{P}\cdot {\hat{a}}_n$ — on any surface where the dipole chains are cut, the uncancelled ends of the dipoles appear as a surface bound charge. It is largest where $\vec{P}$ is perpendicular to the surface, and zero where $\vec{P}$ lies along it.
• ${\rho }_p=-\nabla \cdot \vec{P}$ — inside the material, adjacent dipole ends cancel provided $\vec{P}$ is uniform. Only where $\vec{P}$ diverges (is non-uniform) does an uncancelled volume bound charge appear. The minus sign is because a positive divergence of $\vec{P}$ carries positive charge away, leaving a net negative charge behind.

Conservation check worth a mark

Total bound charge must be zero — polarization only rearranges charge, it never creates it: ${\oint }_{S^{\prime }}{\rho }_{ps}d{s}^{\prime }+{\int }_{V^{\prime }}{\rho }_{p}d{v}^{\prime }={\oint }_{S^{\prime }}\vec{P}\cdot {\hat{a}}_{n}d{s}^{\prime }-{\int }_{V^{\prime }}\nabla \cdot \vec{P}d{v}^{\prime }=0$ by the Divergence Theorem. ✓

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4. Deriving $\vec{D}={ϵ}_0\vec{E}+\vec{P}$ [PYQ: 2016, 2019, 2021]

Part B — The main result

Step 1. In a dielectric, Gauss’s postulate in free-space form must account for both free charge ${\rho }_v$ and bound charge ${\rho }_p$, since $\vec{E}$ responds to all charge: $\nabla \cdot ({ϵ}_0\vec{E})={\rho }_{\text{total}}={\rho }_v+{\rho }_p$

Step 2. Substitute ${\rho }_p=-\nabla \cdot \vec{P}$ from Part A: $\nabla \cdot ({ϵ}_0\vec{E})={\rho }_v-\nabla \cdot \vec{P}$

Step 3. Bring the polarization term to the left: $\nabla \cdot ({ϵ}_0\vec{E})+\nabla \cdot \vec{P}={\rho }_v⟹\nabla \cdot ({ϵ}_0\vec{E}+\vec{P})={\rho }_v$

Step 4. Define the bracketed quantity as the electric flux density: $\boxed{\vec{D}={ϵ}_0\vec{E}+\vec{P}{\text{(C/m}}^2\text{)}}■$

Step 5 — The payoff. With this definition, Gauss’s law recovers its clean form involving free charge only: $\nabla \cdot \vec{D}={\rho }_v\text{and}{\oint }_S\vec{D}\cdot d\vec{s}=Q_{\text{free}}$

Say this sentence in the exam

“Defining $\vec{D}={ϵ}_0\vec{E}+\vec{P}$ absorbs the unknown bound polarization charge into the definition of the flux density, so that Gauss’s law needs only the free charge — which is the quantity we can actually control and measure.”

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5. Susceptibility, Permittivity and the Constitutive Chain

For a linear, isotropic dielectric, $\vec{P}$ is proportional to $\vec{E}$: $\vec{P}={ϵ}_0{\chi }_e\vec{E}$

where ${\chi }_e$ {electric susceptibility — a dimensionless number measuring how easily the material polarizes} is a constant of the material.

Substituting into $\vec{D}={ϵ}_0\vec{E}+\vec{P}$: $\vec{D}={ϵ}_0\vec{E}+{ϵ}_0{\chi }_e\vec{E}={ϵ}_0(1+{\chi }_e)\vec{E}={ϵ}_0{ϵ}_r\vec{E}=ϵ\vec{E}$

Quantity	Symbol	Relation	Notes
Electric susceptibility	${\chi }_e$	$\vec{P}={ϵ}_0{\chi }_e\vec{E}$	Dimensionless; zero for vacuum
Relative permittivity (dielectric constant)	${ϵ}_r$	${ϵ}_r=1+{\chi }_e=ϵ/{ϵ}_0$	Dimensionless; $\ge 1$ always
Absolute permittivity	$ϵ$	$ϵ={ϵ}_0{ϵ}_r$	F/m

Why a dielectric weakens the field

For a fixed free charge, $\vec{D}$ is unchanged (Gauss’s law sees only free charge), so $\vec{E}=\vec{D}/ϵ=\vec{D}/({ϵ}_0{ϵ}_r)$ is reduced by the factor ${ϵ}_r$. The induced bound charges partially cancel the applied field. This is exactly why inserting a dielectric increases capacitance by ${ϵ}_r$ — see 2.04 Electrostatic Boundary Conditions, Refraction Law & Poisson-Laplace Equations.

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6. Conductors vs. Dielectrics — Comparison

Basis	Conductor	Dielectric
Free charges	Abundant free electrons	Essentially none; all charges bound
Response to $\vec{E}$	Free charges migrate to the surface	Bound charges displace slightly in place
Internal field at equilibrium	Exactly zero	Reduced by factor ${ϵ}_r$, but non-zero
Charge produced	Real free surface charge ${\rho }_s$	Bound charges ${\rho }_{ps}=\vec{P}\cdot {\hat{a}}_n$, ${\rho }_p=-\nabla \cdot \vec{P}$
Conductivity $\sigma$	Very high ($\sim 10^7$ S/m for copper)	Very low ($\sim 10^{-12}$ S/m)
Potential	Constant throughout (equipotential body)	Varies from point to point
Field at the surface	Purely normal, $E_n={\rho }_s/{ϵ}_0$	Both components generally non-zero
Examples	Copper, aluminium, silver	Glass, mica, Teflon, distilled water

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7. PYQ Coverage for This Note

Question (verbatim, condensed)	Marks	Year(s)
Show that the total electric flux density in a dielectric is $\vec{D}={ϵ}_0\vec{E}+\vec{P}$.	07/08/11	2016, 2019, 2021
Determine the normal and tangential components of $\vec{E}$ and $\vec{D}$ at the boundary of a conductor and free space.	09	2016, 2020
…What happens when one of the media is a conductor? (boundary-condition question)	07	2015 → see 2.04 Electrostatic Boundary Conditions, Refraction Law & Poisson-Laplace Equations

Syllabus items in this note without a direct PYQ

The five conductor properties and the polarization mechanisms (polar vs non-polar) are syllabus content (Week 6, Instructor 1) that appear as supporting theory inside the boundary-condition and $\vec{D}={ϵ}_0\vec{E}+\vec{P}$ questions rather than as standalone questions. Learn them as the “why” behind those answers.

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8. Exam Hacks & Traps

Key Exam Checkpoints

1. The $\vec{D}={ϵ}_0\vec{E}+\vec{P}$ derivation is a two-parter. You must first derive ${\rho }_p=-\nabla \cdot \vec{P}$; you cannot quote it. Budget time accordingly.
2. Name the tools: the gradient identity ${\hat{a}}_R/R^2={\nabla }^{\prime }(1/R)$, the product rule, and the Divergence Theorem. Method marks live there.
3. The primes matter. ${\nabla }^{\prime }$ acts on source coordinates, $\nabla$ on field coordinates. Mention it once, then drop primes in the final answer.
4. Bound charge is not free charge. Gauss’s law in terms of $\vec{D}$ counts only free charge; in terms of ${ϵ}_0\vec{E}$ it counts all charge. Confusing these two is the most common error here.
5. The minus sign in ${\rho }_p=-\nabla \cdot \vec{P}$ is physical, not cosmetic. Explain it: outward-diverging polarization carries positive charge away, leaving negative behind.
6. “Conductor” in a boundary question means set ${\vec{E}}_2={\vec{D}}_2=0$. Everything then follows in two lines.
7. ${ϵ}_r\ge 1$ always. An answer with ${ϵ}_r<1$ is wrong.

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9. Self-Check

1. List the five properties of a conductor in electrostatic equilibrium, with a one-line reason each.
2. Prove ${\rho }_v=0$ inside a conductor.
3. Derive ${\rho }_{ps}$ and ${\rho }_p$ from the polarization potential integral.
4. Derive $\vec{D}={ϵ}_0\vec{E}+\vec{P}$ and state why the definition is useful.
5. Why does inserting a dielectric reduce $\vec{E}$ but leave $\vec{D}$ unchanged?
6. What are $E_t$ and $D_n$ at a conductor–free-space boundary?

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Next: 2.04 Electrostatic Boundary Conditions, Refraction Law & Poisson-Laplace Equations