Chapter 2 - Electrostatics & Dielectric Boundary Conditions

---

📄 Section: 02 Chapter Map - Electrostatics & Boundary Conditions

02 Chapter Map — Electrostatics & Boundary Conditions

What this chapter covers

The two fundamental postulates, Coulomb’s law and continuous charge distributions, Gauss’s law and its symmetric applications, electric potential and the dipole derivations, conductors and dielectric polarization, boundary conditions and the refraction law, Poisson’s and Laplace’s equations with all capacitance derivations, and electrostatic energy. This is the heaviest-weighted chapter in the paper.

---

📚 Study Notes Index (read in this order)

#	Note	What it gives you	PYQ weight
1	2.01 Fundamental Postulates of Electrostatics & Gauss_s Law Applications	Two postulates + derivations, Coulomb’s law, Gauss’s law applications (sheet, line, charge cloud)	Very high — 2015–2025, nearly every year
2	2.02 Electric Potential, Equipotential Contours & Dipole Derivations	$V$, $\vec{E}=-\nabla V$, $\vec{E}$ vs $\vec{D}$, equipotential sketches, full dipole derivations	Very high — 2016, 2017, 2019, 2021, 2022, 2023, 2024, 2025
3	2.03 Conductors, Dielectrics & Polarization Charge Densities	Five conductor properties, polarization, bound charges, $\vec{D}={ϵ}_0\vec{E}+\vec{P}$	High — 2016, 2019, 2020, 2021
4	2.04 Electrostatic Boundary Conditions, Refraction Law & Poisson-Laplace Equations	Boundary conditions, refraction law + numerical, Poisson/Laplace, parallel-plate & cylindrical capacitance, EHD pump	Very high — 2015–2025
5	2.05 Electrostatic Energy, Work Done & Solved PYQ Numericals	Energy to assemble $n$ charges, energy density, and every Ch. 2 numerical solved	Very high — 2016–2025
✅	00 Chapter 2 Active-Recall Diagnostic Quiz	Test yourself before and after	—

---

🎯 Highest-Yield Items in This Chapter

Concept	Years asked	Where
Fundamental postulates (both forms + significance)	2015, 2016, 2017, 2020, 2022, 2025	2.01 Fundamental Postulates of Electrostatics & Gauss_s Law Applications
Parallel-plate capacitance / surface charge density	2016, 2018, 2019, 2021, 2022, 2023, 2024, 2025	2.04 Electrostatic Boundary Conditions, Refraction Law & Poisson-Laplace Equations
Electrostatic energy to assemble $n$ charges	2018, 2022, 2023, 2024, 2025	2.05 Electrostatic Energy, Work Done & Solved PYQ Numericals
Electric dipole potential / field derivation	2016, 2017, 2019, 2021, 2023, 2024	2.02 Electric Potential, Equipotential Contours & Dipole Derivations
Poisson’s equation derivation (+ its solution)	2015, 2016, 2017, 2022, 2023, 2025	2.04 Electrostatic Boundary Conditions, Refraction Law & Poisson-Laplace Equations
Dielectric boundary conditions + refraction numerical	2015, 2017, 2019, 2022, 2025	2.04 Electrostatic Boundary Conditions, Refraction Law & Poisson-Laplace Equations
Uniformly charged cloud (zero at centre / linear / inverse / max at surface)	2018, 2019, 2020, 2022	2.01 Fundamental Postulates of Electrostatics & Gauss_s Law Applications
$\vec{D}={ϵ}_0\vec{E}+\vec{P}$	2016, 2019, 2021	2.03 Conductors, Dielectrics & Polarization Charge Densities
Gauss’s law + sheet / line charge	2015, 2016, 2023	2.01 Fundamental Postulates of Electrostatics & Gauss_s Law Applications
Energy stored in 4 point charges (numerical)	2019, 2020, 2023	2.05 Electrostatic Energy, Work Done & Solved PYQ Numericals
$\vec{E}$ vs $\vec{D}$ distinction	2022, 2025	2.02 Electric Potential, Equipotential Contours & Dipole Derivations
Conductor–free-space boundary components	2016, 2020	2.03 Conductors, Dielectrics & Polarization Charge Densities
Cylindrical capacitor	2015, 2017	2.04 Electrostatic Boundary Conditions, Refraction Law & Poisson-Laplace Equations

---

🔗 Logical Flow of the Chapter

graph TD
    P["Two postulates: ∇·D = ρv and ∇×E = 0"] --> G["Gauss's law applications"]
    P --> V["Scalar potential V exists, E = −∇V"]
    V --> DIP["Electric dipole V and E"]
    DIP --> POL["Dielectric polarization P"]
    POL --> D["D = ε0E + P"]
    D --> BC["Boundary conditions + refraction law"]
    V --> PL["Poisson / Laplace equations"]
    PL --> CAP["Capacitance: parallel plate, cylindrical, EHD pump"]
    CAP --> EN["Electrostatic energy and energy density"]

---

🔗 Related Resources

• Course teaching plan: ECE 2105 - Electromagnetic Fields and Waves
• Full checklist with PYQ years: field checklist
• Verbatim question bank: ece 2105 field pyq
• Previous chapter: 01 Chapter Map - Vector Calculus & Fundamentals
• Next chapter: 03 Chapter Map - Magnetostatics & Magnetic Materials

---

📄 Section: 2.01 Fundamental Postulates of Electrostatics & Gauss_s Law Applications

Related Concepts: 1.03 Integral Theorems (Divergence Theorem & Stokes_s Theorem) & Identity Proofs | 2.02 Electric Potential, Equipotential Contours & Dipole Derivations | 2.03 Conductors, Dielectrics & Polarization Charge Densities | 2.04 Electrostatic Boundary Conditions, Refraction Law & Poisson-Laplace Equations

2.01 Fundamental Postulates of Electrostatics & Gauss’s Law Applications

Core Idea

The whole of electrostatics rests on exactly two postulates: $\nabla \cdot \vec{D}={\rho }_v$ (electric flux has sources — charges) and $\nabla \times \vec{E}=0$ (the field does no net work around a loop). Everything else in Chapter 2 — Coulomb’s law, potential, capacitance, boundary conditions — is a consequence. This is the single most-repeated question in the paper: 2015, 2016, 2017, 2020, 2022, 2025.

---

1. The Two Fundamental Postulates [PYQ: 2015, 2016, 2017, 2020, 2022, 2025]

Heavily tested — six of the last eleven papers

The question is always some combination of: write the differential form, write the integral form, derive one from the other, state their physical significance. Prepare all four parts.

Postulate	Differential (point) form	Integral form	Physical law
Divergence postulate	$\nabla \cdot \vec{D}={\rho }_v$	${\oint }_S\vec{D}\cdot d\vec{s}=Q_{\text{enc}}$	Gauss’s Law
Curl postulate	$\nabla \times \vec{E}=0$	${\oint }_C\vec{E}\cdot d\vec{l}=0$	Conservative field / KVL

Symbols:

Symbol	Meaning	Unit
$\vec{E}$	Electric field intensity	V/m
$\vec{D}={ϵ}_0\vec{E}$	Electric flux density (free space)	C/m²
${\rho }_v=dQ/dv$	Volume charge density	C/m³
$Q_{\text{enc}}$	Total free charge enclosed by $S$	C
${ϵ}_0$	Permittivity of free space $=8.854\times 10^{-12}$	F/m

1.1 Physical Significance [PYQ: 2015, 2016, 2017, 2020, 2022, 2025]

Postulate 1 — $\nabla \cdot \vec{D}={\rho }_v$

The static electric field is divergent: its flux lines have genuine sources and sinks, and those sources are electric charges. Positive charge acts as a source (flux diverges outward), negative charge acts as a sink (flux converges inward). Quantitatively, the net outward electric flux through any closed surface equals exactly the free charge enclosed — no more, no less, and independent of how that charge is arranged inside.

Postulate 2 — $\nabla \times \vec{E}=0$

The static electric field is irrotational (curl-free): it never forms closed loops on its own. Physically, the work done in carrying a unit charge around any closed path in an electrostatic field is exactly zero — the field is conservative. This is the field-theory statement of Kirchhoff’s Voltage Law, and it is precisely what allows a single-valued scalar potential $V$ to exist with $\vec{E}=-\nabla V$ (see 1.03 Integral Theorems (Divergence Theorem & Stokes_s Theorem) & Identity Proofs, Identity I).

graph TD
    P1["∇·D = ρv<br/>(divergence postulate)"] -->|"Divergence Theorem"| I1["∮S D · ds = Q enclosed<br/>GAUSS'S LAW"]
    P2["∇×E = 0<br/>(curl postulate)"] -->|"Stokes's Theorem"| I2["∮C E · dl = 0<br/>CONSERVATIVE FIELD / KVL"]
    I1 --> A["Field of symmetric<br/>charge distributions"]
    I2 --> B["Scalar potential V exists<br/>E = −∇V"]

---

1.2 Derivation: Differential $\to$ Integral Form [PYQ: 2020, 2022]

PYQ — 2020, 2022 (10 marks)

Write down the differential form of fundamental postulates of electrostatics in free space. Then derive the integral form of them. Also state their physical significance.

Derivation 1 — Gauss's Law from $\nabla \cdot \vec{D}={\rho }_v$

Step 1. Start from the differential postulate: $\nabla \cdot \vec{D}={\rho }_v$

Step 2. Integrate both sides over an arbitrary volume $V$: ${\int }_V(\nabla \cdot \vec{D})dv={\int }_V{\rho }_{v}dv$

Step 3. Apply the Divergence Theorem to the left-hand side, converting the volume integral of a divergence into a closed surface flux integral over the bounding surface $S$: ${\int }_V(\nabla \cdot \vec{D})dv={\oint }_S\vec{D}\cdot d\vec{s}$

Step 4. Recognise that the volume integral of charge density is the total enclosed charge: ${\int }_V{\rho }_{v}dv=Q_{\text{enc}}$

Step 5. Equating gives the integral form: $\boxed{{\oint }_S\vec{D}\cdot d\vec{s}=Q_{\text{enc}}}■$

Derivation 2 — Conservative Law from $\nabla \times \vec{E}=0$

Step 1. Start from the differential postulate: $\nabla \times \vec{E}=0$

Step 2. Integrate over an arbitrary open surface $S$ bounded by the closed contour $C$: ${\int }_S(\nabla \times \vec{E})\cdot d\vec{s}=0$

Step 3. Apply Stokes’s Theorem to the left-hand side: ${\int }_S(\nabla \times \vec{E})\cdot d\vec{s}={\oint }_C\vec{E}\cdot d\vec{l}$

Step 4. Therefore: $\boxed{{\oint }_C\vec{E}\cdot d\vec{l}=0}■$

Method marks live in the theorem names

Write “applying the Divergence Theorem” and “applying Stokes’s Theorem” explicitly. Students who jump straight to the boxed result lose 2–3 marks even with the correct answer.

---

2. Coulomb’s Law & Continuous Charge Distributions [PYQ: 2018, 2021]

PYQ — 2018, 2021 (10 marks)

State Coulomb’s law. Determine the electric field intensity due to a continuous distribution of charge with (i) surface charge density and (ii) line charge density.

2.1 Coulomb’s Law (point charges)

Statement — Coulomb's Law

The force between two stationary point charges is directly proportional to the product of the charges, inversely proportional to the square of the distance between them, and directed along the line joining them — repulsive for like charges, attractive for unlike.

${\vec{F}}_{12}={\hat{a}}_{R_{12}}\frac{q_1{q}_2}{4\pi {ϵ}_0{R}_{12}^2}\text{(N)}$

where ${\vec{R}}_{12}={\vec{R}}_2-{\vec{R}}_1$ is the vector from $q_1$ to $q_2$, $R_{12}=|{\vec{R}}_{12}|$, and ${\hat{a}}_{R_{12}}={\vec{R}}_{12}/R_{12}$.

Dividing by the test charge gives the field of a single point charge: $\vec{E}={\hat{a}}_R\frac{q}{4\pi {ϵ}_0{R}^2}\text{(V/m)}$

2.2 Continuous Distributions

Replace the point charge $q$ by a differential element $d{q}^{\prime }$ and integrate over the source geometry:

$\vec{E}=\frac{1}{4\pi {ϵ}_0}\int \frac{{\hat{a}}_{R}d{q}^{\prime }}{R^2}$

Distribution	Density (unit)	Element	Field integral
Line	${\rho }_l$ (C/m)	$d{q}^{\prime }={\rho }_{l}d{l}^{\prime }$	$\vec{E}=\frac{1}{4\pi {ϵ}_0}{\int }_{L^{\prime }}{\hat{a}}_R\frac{{\rho }_{l}d{l}^{\prime }}{R^2}$
Surface	${\rho }_s$ (C/m²)	$d{q}^{\prime }={\rho }_{s}d{s}^{\prime }$	$\vec{E}=\frac{1}{4\pi {ϵ}_0}{\int }_{S^{\prime }}{\hat{a}}_R\frac{{\rho }_{s}d{s}^{\prime }}{R^2}$
Volume	${\rho }_v$ (C/m³)	$d{q}^{\prime }={\rho }_{v}d{v}^{\prime }$	$\vec{E}=\frac{1}{4\pi {ϵ}_0}{\int }_{V^{\prime }}{\hat{a}}_R\frac{{\rho }_{v}d{v}^{\prime }}{R^2}$

[FIGURE: Line (${\rho }_{l}d{l}^{\prime }$), surface (${\rho }_{s}d{s}^{\prime }$) and volume (${\rho }_{v}d{v}^{\prime }$) source elements each contributing $d\vec{E}$ at a common observation point P — source: David K. Cheng, Ch. 3, Fig. 3-4]

Example

Problem: find $\vec{E}$ at perpendicular distance $r$ from an infinitely long straight line of uniform density ${\rho }_l$ lying along the $z$-axis.

Step 1 — Geometry (write this out).

• Source element $d{q}^{\prime }={\rho }_{l}d{z}^{\prime }$ at $(0,0,z^{\prime })$; observation point $P(r,0,0)$.
• Distance vector: $\vec{R}={\hat{a}}_{r}r-{\hat{a}}_z{z}^{\prime }$, magnitude $R=\sqrt{r^2+z^{\prime 2}}$.
• Unit vector: ${\hat{a}}_R=\frac{{\hat{a}}_{r}r-{\hat{a}}_z{z}^{\prime }}{\sqrt{r^2+z^{\prime 2}}}$.

Step 2 — Set up the integral. $d\vec{E}=\frac{{\rho }_l({\hat{a}}_{r}r-{\hat{a}}_z{z}^{\prime })d{z}^{\prime }}{4\pi {ϵ}_0(r^2+z^{\prime 2}{)}^{3/2}}⟹\vec{E}=\frac{{\rho }_l}{4\pi {ϵ}_0}{\int }_{-\infty }^{\infty }\frac{{\hat{a}}_{r}r-{\hat{a}}_z{z}^{\prime }}{(r^2+z^{\prime 2}{)}^{3/2}}d{z}^{\prime }$

Step 3 — Symmetry cancellation. The integrand of the $z$-component, $z^{\prime }/(r^2+z^{\prime 2}{)}^{3/2}$, is an odd function, so integrating over the symmetric limits $(-\infty ,\infty )$ gives exactly zero: $E_z=-\frac{{\rho }_l}{4\pi {ϵ}_0}{\int }_{-\infty }^{\infty }\frac{z^{\prime }d{z}^{\prime }}{(r^2+z^{\prime 2}{)}^{3/2}}=0$ Only the radial component survives.

Step 4 — Evaluate the radial integral. Substitute $z^{\prime }=r\tan \theta$, so $d{z}^{\prime }=r{\sec }^2\theta d\theta$ and $(r^2+z^{\prime 2}{)}^{3/2}=r^3{\sec }^3\theta$, with limits $\theta :-\pi /2\to \pi /2$: ${\int }_{-\infty }^{\infty }\frac{d{z}^{\prime }}{(r^2+z^{\prime 2}{)}^{3/2}}=\frac{1}{r^2}{\int }_{-\pi /2}^{\pi /2}\cos \theta d\theta =\frac{1}{r^2}[\sin \theta {]}_{-\pi /2}^{\pi /2}=\frac{2}{r^2}$

Step 5 — Result. $E_r=\frac{{\rho }_{l}r}{4\pi {ϵ}_0}\cdot \frac{2}{r^2}=\frac{{\rho }_l}{2\pi {ϵ}_{0}r}⟹\boxed{\vec{E}={\hat{a}}_r\frac{{\rho }_l}{2\pi {ϵ}_{0}r}}\text{(V/m)}■$

Physical reading: the field falls off as $1/r$ — the first power — not $1/R^2$ like a point charge, because the source extends infinitely in one dimension so the flux spreads cylindrically instead of spherically.

---

3. Gauss’s Law and Its Applications [PYQ: 2015, 2016, 2019, 2020, 2023]

Abstract

The total outward electric flux through any closed surface equals the total free charge enclosed by that surface. ${\oint }_S\vec{D}\cdot d\vec{s}=Q_{\text{enc}}={\int }_V{\rho }_{v}dv$ It is independent of the shape of the surface and of how the enclosed charge is distributed inside it. Charges outside the surface contribute zero net flux — their lines enter and leave again.

3.1 Applications of Gauss’s Law [PYQ: 2019, 2020]

The 2019 and 2020 papers explicitly asked to “write down some applications”. List these:

1. Finding $\vec{E}$ of an infinitely long line charge (cylindrical symmetry).
2. Finding $\vec{E}$ of an infinite sheet of charge (planar symmetry).
3. Finding $\vec{E}$ of a uniformly charged sphere / charge cloud, inside and outside.
4. Finding $\vec{E}$ of a coaxial cable and hence its capacitance.
5. Proving ${\rho }_v=0$ inside a conductor and that all excess charge resides on its surface.
6. Deriving the normal boundary condition $D_{1n}-D_{2n}={\rho }_s$ using a pillbox.
7. Deriving Poisson’s and Laplace’s equations (combined with $\vec{E}=-\nabla V$).
8. Explaining electrostatic shielding (the Faraday cage).

The prerequisite for using Gauss's law at all

Gauss’s law is always true but only useful when symmetry lets you pull $D$ out of the integral. Always write the justification sentence: “We choose a Gaussian surface $S$ on which $\vec{D}$ is everywhere either normal to $S$ with constant magnitude, or tangential to $S$ (zero flux).” This one line earns setup marks in every application question.

---

3.2 Application 1 — Infinite Sheet of Charge [PYQ: 2015, 2023]

PYQ — 2015 (11 marks), 2023 (10 marks)

State and explain Gauss’s law. Using this law, determine the electric field intensity of an infinite sheet of charge.

[FIGURE: Infinite sheet of charge in the $z=0$ plane; Gaussian pillbox of end-cap area $A$ extending symmetrically to $z=\pm h$ — source: David K. Cheng, Ch. 3, Fig. 3-7]

Derivation

Step 1 — Gaussian surface. A cylinder (pillbox) of cross-sectional area $A$ with its axis normal to the sheet, extending symmetrically to $z=\pm h$.

Step 2 — Symmetry argument. By symmetry $\vec{D}$ can only point along $\pm {\hat{a}}_z$ and can only depend on $|z|$. Therefore no flux crosses the curved side wall ($\vec{D}\perp d\vec{s}$ there).

Step 3 — Evaluate the flux. Both end caps contribute equally and outwardly: ${\oint }_S\vec{D}\cdot d\vec{s}=D_{z}A+D_{z}A=2D_{z}A$

Step 4 — Enclosed charge. $Q_{\text{enc}}={\rho }_{s}A$

Step 5 — Equate and solve. $2D_{z}A={\rho }_{s}A⟹D_z=\frac{{\rho }_s}{2}$ $\boxed{\vec{E}={\hat{a}}_n\frac{{\rho }_s}{2{ϵ}_0}}\text{(V/m)}■$

Physical reading: the field is uniform — independent of distance from the sheet. An infinite plane looks the same from any distance, so the field cannot decay. ${\hat{a}}_n$ points away from the sheet on both sides for positive ${\rho }_s$.

---

3.3 Application 2 — Infinitely Long Line Charge [PYQ: 2016]

PYQ — 2016 (12 marks)

State and explain Gauss’s law. Using this law determine electric field intensity and electric potential of an infinitely long straight line charge of uniform density ${\rho }_l$ in air.

Derivation

Step 1 — Gaussian surface. A coaxial cylinder of radius $r$ and length $L$ around the line.

Step 2 — Symmetry. $\vec{D}=D_r{\hat{a}}_r$ only (radial), constant on the curved surface. The flat end caps carry no flux since $\vec{D}\perp {\hat{a}}_z$ there.

Step 3 — Flux. ${\oint }_S\vec{D}\cdot d\vec{s}=D_r(2\pi rL)$

Step 4 — Enclosed charge. $Q_{\text{enc}}={\rho }_{l}L$.

Step 5 — Field. $D_r(2\pi rL)={\rho }_{l}L⟹D_r=\frac{{\rho }_l}{2\pi r}⟹\boxed{\vec{E}={\hat{a}}_r\frac{{\rho }_l}{2\pi {ϵ}_{0}r}}$ Note this matches the direct-integration result of §2.2 — but in five lines instead of five steps of calculus. That is the whole point of Gauss’s law.

Step 6 — Potential. Because the field extends to infinity, we cannot take $V(\infty )=0$; instead choose an arbitrary reference radius $r_0$: $V(r)-V(r_0)=-{\int }_{r_0}^r\vec{E}\cdot d\vec{l}=-{\int }_{r_0}^r\frac{{\rho }_l}{2\pi {ϵ}_0{r}^{\prime }}d{r}^{\prime }=-\frac{{\rho }_l}{2\pi {ϵ}_0}\ln \frac{r}{r_0}$ $\boxed{V(r)=\frac{{\rho }_l}{2\pi {ϵ}_0}\ln \left(\frac{r_0}{r}\right)}\text{(V)}$

Why you must mention the reference point

Setting $V(\infty )=0$ for an infinite line charge gives a divergent (infinite) potential, because the source itself is infinite. Stating “a finite reference radius $r_0$ is chosen since the charge distribution extends to infinity” is a marked point in this question.

---

3.4 Application 3 — Uniformly Charged Spherical Cloud [PYQ: 2018, 2019, 2020, 2022]

The most-repeated derivation in Chapter 2

Asked in 2018, 2019, 2020, 2022 with different phrasings — “zero at centre”, “varies linearly up to the surface”, “varies inversely outside”, “maximum at the surface”. Learn one derivation; it answers all four.

[FIGURE: Uniformly charged spherical cloud of radius $a$ with concentric Gaussian spheres drawn for $R<a$ and $R>a$ — source: David K. Cheng, Ch. 3, Fig. 3-8]

Master Derivation — cloud of radius $a$, uniform density ${\rho }_0$

Case 1 — Inside the cloud ($R\le a$)

Gaussian surface: concentric sphere of radius $R<a$. ${\oint }_S\vec{D}\cdot d\vec{s}=D_R(4\pi R^2)$ Enclosed charge — only the charge within radius $R$: $Q_{\text{enc}}={\int }_0^R{\rho }_{0}d{v}^{\prime }={\rho }_0\left(\frac{4}{3}\pi R^3\right)$ Equate: $D_R(4\pi R^2)={\rho }_0\left(\frac{4}{3}\pi R^3\right)⟹D_R=\frac{{\rho }_{0}R}{3}$ $\boxed{{\vec{E}}_{\text{in}}={\hat{a}}_R\frac{{\rho }_{0}R}{3{ϵ}_0}}\text{— grows LINEARLY with }R$

At the centre, $R=0$: $\vec{E}=0$. (This is the “zero at its centre” part of the 2018/2022 question — one line.)

Case 2 — Outside the cloud ($R>a$)

Gaussian surface: concentric sphere of radius $R>a$. $Q_{\text{enc}}=Q_{\text{total}}={\rho }_0\left(\frac{4}{3}\pi a^3\right)$ $D_R(4\pi R^2)={\rho }_0\left(\frac{4}{3}\pi a^3\right)⟹D_R=\frac{{\rho }_0{a}^3}{3R^2}$ $\boxed{{\vec{E}}_{\text{out}}={\hat{a}}_R\frac{{\rho }_0{a}^3}{3{ϵ}_0{R}^2}}\text{— decays INVERSELY with }{R}^2$

Outside, the cloud behaves exactly like a point charge $Q_{\text{total}}$ at the centre.

Case 3 — Maximum is at the surface [PYQ: 2019]

Evaluate both expressions at $R=a$: $E_{\text{in}}(a)=\frac{{\rho }_{0}a}{3{ϵ}_0},E_{\text{out}}(a)=\frac{{\rho }_0{a}^3}{3{ϵ}_0{a}^2}=\frac{{\rho }_{0}a}{3{ϵ}_0}$ They agree — the field is continuous at the surface (no surface charge exists there). Since $E$ increases monotonically inside and decreases monotonically outside, the maximum must occur exactly at $R=a$: $\boxed{E_{\max }=\frac{{\rho }_{0}a}{3{ϵ}_0}}\text{at }R=a■$

[GRAPH: $E$ versus $R$ for the uniform charge cloud — a straight line $E={\rho }_{0}R/3{ϵ}_0$ from $(0,0)$ rising to the peak ${\rho }_{0}a/3{ϵ}_0$ at $R=a$, then a $1/R^2$ decay $E={\rho }_0{a}^3/3{ϵ}_0{R}^2$ for $R>a$. Governing equations as given. Source: David K. Cheng, Ch. 3, Fig. 3-9]

Answering all four phrasings from one derivation

• “zero at its centre” → set $R=0$ in $E_{\text{in}}$
• “varies linearly up to the surface” → $E_{\text{in}}\propto R$
• “varies inversely outside” → $E_{\text{out}}\propto 1/R^2$
• “maximum at the surface” → both expressions equal ${\rho }_{0}a/3{ϵ}_0$ at $R=a$, and the function is monotonic on either side

Always draw the $E$–$R$ graph. It is worth marks on its own and takes ten seconds.

---

3.5 Variant — Non-Uniform Density ${\rho }_v=2r$ [PYQ: 2021]

Fully solved in 2.05 Electrostatic Energy, Work Done & Solved PYQ Numericals. The method is identical; only $Q_{\text{enc}}$ changes because ${\rho }_v$ must stay inside the integral: $Q_{\text{enc}}={\int }_0^R{\rho }_v(r^{\prime })4\pi r^{\prime 2}d{r}^{\prime }$

The trap in the 2021 question

${\rho }_v=2r$ is not uniform, despite the question saying “uniform charge distribution”. You cannot pull it out of the integral. Students who write $Q={\rho }_v\cdot \frac{4}{3}\pi R^3$ lose the whole question.

---

4. PYQ Coverage for This Note

Question (verbatim, condensed)	Marks	Year(s)
Fundamental postulates in differential + integral form + physical significance	08–10	2015, 2016, 2017, 2020, 2022, 2025
…Then derive the integral form of them.	10	2020, 2022
Identify each postulate with its proper experimental law	07/08	2016, 2017 (also in Ch. 4 Maxwell context)
State Coulomb’s law. Determine $\vec{E}$ due to (i) surface (ii) line charge density.	10	2018, 2021
State and explain Gauss’s law… determine $\vec{E}$ of an infinite sheet of charge.	10/11	2015, 2023
State and explain Gauss’s law… determine $\vec{E}$ and $V$ of an infinitely long line charge ${\rho }_l$ in air.	12	2016
State Gauss’s law and also write some applications of it.	05	2019
Write down some applications of Gauss’s law. Show that $\vec{E}$ inside a uniformly charged cloud varies linearly up to the surface and varies inversely outside.	20	2020
Show that the strength of $\vec{E}$ due to a charge cloud is maximum at the surface.	12	2019
Show that $\vec{E}$ inside a uniformly charged cloud is zero at its centre and varies linearly up to the surface.	09/10	2018, 2022
Numerical: spherical distribution ${\rho }_v=2r$ nC/m³, find $\vec{E}$ at $r=2$ m and $r=12$ m	10	2021 → see 2.05 Electrostatic Energy, Work Done & Solved PYQ Numericals

---

5. Exam Hacks & Traps

Key Exam Checkpoints

1. Name the theorem in the postulate derivation. Divergence Theorem for Gauss, Stokes’s Theorem for the curl postulate. This is where the method marks are.
2. Always justify the Gaussian surface. One sentence about $\vec{D}$ being normal-and-constant or tangential-and-zero. Never just assert $\oint \vec{D}\cdot d\vec{s}=D(4\pi R^2)$.
3. Enclosed charge means enclosed, not total. Inside a charge cloud, $Q_{\text{enc}}\propto R^3$, not $a^3$. This is the single biggest error in the cloud derivation.
4. Check the exponent of the answer. Point charge $\to 1/R^2$; infinite line $\to 1/r$; infinite sheet $\to$ constant. If your infinite-sheet answer contains a distance, you have made an error.
5. $\vec{D}$ versus $\vec{E}$. Gauss’s law is cleanest in terms of $\vec{D}$ because it involves only free charge. Convert to $\vec{E}$ at the end via $\vec{E}=\vec{D}/ϵ$.
6. For the line-charge potential, state the finite reference radius. $V(\infty )=0$ fails for infinite sources.
7. Sketch the $E$–$R$ graph for the cloud even if not explicitly asked. It demonstrates you understand continuity at $R=a$.
8. Units, every time: $\vec{E}$ in V/m, $\vec{D}$ in C/m², ${\rho }_l$ in C/m, ${\rho }_s$ in C/m², ${\rho }_v$ in C/m³.

---

6. Self-Check

1. State both postulates in both forms, and name the theorem linking each pair.
2. Why does an infinite sheet produce a distance-independent field?
3. In the charge-cloud derivation, what is $Q_{\text{enc}}$ for $R<a$, and why is it not ${\rho }_0\frac{4}{3}\pi a^3$?
4. Prove the field is maximum at $R=a$ in two lines.
5. What justification sentence must precede every use of Gauss’s law?
6. List six applications of Gauss’s law.

---

Next: 2.02 Electric Potential, Equipotential Contours & Dipole Derivations

---

📄 Section: 2.02 Electric Potential, Equipotential Contours & Dipole Derivations

Related Concepts: 2.01 Fundamental Postulates of Electrostatics & Gauss_s Law Applications | 2.03 Conductors, Dielectrics & Polarization Charge Densities | 1.02 Vector Operators (Gradient, Divergence, Curl & Laplacian) | 2.05 Electrostatic Energy, Work Done & Solved PYQ Numericals

2.02 Electric Potential, Equipotential Contours & Dipole Derivations

Core Idea

Electric potential $V$ is a scalar field — potential energy per unit charge. Working with $V$ instead of $\vec{E}$ is a massive simplification: you add scalars instead of vectors, and then recover the field by one differentiation, $\vec{E}=-\nabla V$. The dipole potential derivation built on this idea is one of the two or three most frequently examined derivations in the whole course (2016, 2017, 2019, 2021, 2023, 2024).

---

1. Electric Potential and the Relation $\vec{E}=-\nabla V$ [PYQ: 2016, 2017, 2021]

Abstract

The electric potential $V$ at a point is the work done by an external agent, per unit positive charge, in bringing a test charge from a reference point (usually infinity, where $V=0$) to that point, without acceleration. $V=-{\int }_{\infty }^P\vec{E}\cdot d\vec{l}\text{(V or J/C)}$

Abstract

The electric field intensity $\vec{E}$ at a point is the force per unit positive test charge placed at that point: $\vec{E}={\lim }_{q\to 0}\frac{\vec{F}}{q}\text{(V/m or N/C)}$ The limit $q\to 0$ ensures the test charge does not disturb the source distribution it is measuring.

Potential difference between two points: $V_{AB}=V_B-V_A=-{\int }_A^B\vec{E}\cdot d\vec{l}$

1.1 Proof: Work Done Moving a Unit Charge = Potential Difference [PYQ: 2017]

PYQ — 2017 (08 marks)

Define electric potential. Show that in an electric field, work done in moving a unit charge from one point to another is equal to the electric potential difference between those two points.

Proof

Step 1. A charge $q$ in a field $\vec{E}$ experiences a force ${\vec{F}}_E=q\vec{E}$.

Step 2. To move it without acceleration, an external agent must apply an equal and opposite force: ${\vec{F}}_{\text{ext}}=-q\vec{E}$

Step 3. The differential work done by the external agent over a displacement $d\vec{l}$ is: $dW={\vec{F}}_{\text{ext}}\cdot d\vec{l}=-q\vec{E}\cdot d\vec{l}$

Step 4. Integrate from $A$ to $B$: $W_{A\to B}=-q{\int }_A^B\vec{E}\cdot d\vec{l}$

Step 5. For a unit positive charge ($q=1$ C): $W_{A\to B}=-{\int }_A^B\vec{E}\cdot d\vec{l}=V_B-V_A=V_{AB}■$

Step 6 — Path independence. Because $\nabla \times \vec{E}=0$ (second postulate), ${\oint }_C\vec{E}\cdot d\vec{l}=0$, so the integral depends only on the endpoints, never the path. This is what makes $V$ a single-valued function of position and hence a legitimate scalar field.

1.2 Deriving $\vec{E}=-\nabla V$ for a Point Charge [PYQ: 2016, 2021]

PYQ — 2016 (10 marks), 2021 (06 marks)

Define electric field intensity and electric potential. Also derive the relation between them when caused by a point charge.

Derivation

Step 1. The field of a point charge $q$ at distance $R$ is $\vec{E}={\hat{a}}_R\frac{q}{4\pi {ϵ}_0{R}^2}$.

Step 2. Bring a unit charge from $\infty$ to $R$ along a radial path, so $d\vec{l}={\hat{a}}_{R}d{R}^{\prime }$: $V=-{\int }_{\infty }^R\frac{q}{4\pi {ϵ}_0{R}^{\prime 2}}d{R}^{\prime }=-\frac{q}{4\pi {ϵ}_0}{\left[-\frac{1}{R^{\prime }}\right]}_{\infty }^R=\frac{q}{4\pi {ϵ}_{0}R}$

Step 3. Now take the gradient. Since $V$ depends only on $R$, in spherical coordinates only the radial term survives: $\nabla V={\hat{a}}_R\frac{\partial V}{\partial R}={\hat{a}}_R\frac{\partial }{\partial R}\left(\frac{q}{4\pi {ϵ}_{0}R}\right)=-{\hat{a}}_R\frac{q}{4\pi {ϵ}_0{R}^2}$

Step 4. The bracketed quantity is exactly $\vec{E}$: $\nabla V=-\vec{E}⟹\boxed{\vec{E}=-\nabla V}■$

Why the minus sign is physically necessary

$\nabla V$ points toward increasing potential. But a positive charge released in a field naturally accelerates toward lower potential energy. The field must therefore point “downhill” in $V$, which is exactly what the minus sign encodes.

The identity that licenses all of this

$\vec{E}$ can be written as a gradient only because $\nabla \times \vec{E}=0$ and $\nabla \times (\nabla V)\equiv 0$ — see 1.03 Integral Theorems (Divergence Theorem & Stokes_s Theorem) & Identity Proofs, Identity I. Cite this in a “derive the relation” answer.

---

2. Electric Field Intensity $\vec{E}$ vs. Electric Flux Density $\vec{D}$ [PYQ: 2022, 2025]

PYQ — 2022 (05 marks), 2025 (07 marks)

2022: Differentiate between electric field intensity and electric flux density, emphasizing their physical significance. 2025: Distinguish between $\vec{E}$ and $\vec{D}$ with respect to definition, unit, governing relation, and physical significance.

The 2025 phrasing names the exact four rows the examiner wants. Reproduce this table:

Basis	Electric Field Intensity $\vec{E}$	Electric Flux Density $\vec{D}$
Definition	Force experienced per unit positive test charge at a point	Electric flux passing per unit area normal to the flux, arising from free charge only
Unit	volt per metre (V/m), equivalently N/C	coulomb per square metre (C/m²)
Governing relation	$\vec{E}=-\nabla V$; $\vec{E}=\vec{F}/q$; $\nabla \times \vec{E}=0$	$\vec{D}=ϵ\vec{E}={ϵ}_0\vec{E}+\vec{P}$; $\nabla \cdot \vec{D}={\rho }_v$
Physical significance	Describes the force effect of the field on charges; the quantity that appears in energy and work calculations	Describes the source–flux relationship; measures the flux due to free charge and is what Gauss’s law counts
Medium dependence	Depends on the medium — for the same free charge, $\vec{E}$ falls by a factor ${ϵ}_r$ inside a dielectric	Independent of the medium — determined solely by the free charge distribution
Behaviour at a dielectric boundary	Its tangential component is continuous: $E_{1t}=E_{2t}$	Its normal component is continuous when ${\rho }_s=0$: $D_{1n}=D_{2n}$
Related to polarization?	No — it is the total field acting on a charge	Yes — explicitly contains $\vec{P}$

The single sentence that earns the "physical significance" mark

”$\vec{D}$ is a source-related quantity determined only by free charge and unaffected by the medium; $\vec{E}$ is a force-related quantity that does depend on the medium, since the dielectric’s polarization partially cancels the applied field.”

Why we need two vectors at all

Inside a dielectric the bound polarization charges create their own field opposing the applied one. If we only had $\vec{E}$, Gauss’s law would have to count both free and bound charge, which we usually cannot measure. Defining $\vec{D}={ϵ}_0\vec{E}+\vec{P}$ absorbs the bound charge into the definition, so $\nabla \cdot \vec{D}={\rho }_v$ involves free charge only. See 2.03 Conductors, Dielectrics & Polarization Charge Densities.

---

3. Equipotential Lines & Field-Line Sketches [PYQ: 2024, 2025]

Abstract

An equipotential line (in 2D) or equipotential surface (in 3D) is the locus of all points having the same electric potential. No work is done in moving a charge along it, since $dW=-qdV=0$.

3.1 Three properties always worth stating

1. Orthogonality. Field lines and equipotentials meet at exactly $90^{\circ }$ everywhere. Proof: along an equipotential $dV=\nabla V\cdot d\vec{l}=0$, so $\nabla V\perp d\vec{l}$, and $\vec{E}=-\nabla V$ is therefore perpendicular to the equipotential.
2. Direction. Field lines point from high $V$ to low $V$ — the direction of maximum decrease of potential.
3. Zero work. Moving a charge anywhere along an equipotential requires no work. Equipotentials never cross each other (a point cannot have two potentials).

3.2 Sketching the Two Required Cases [PYQ: 2024, 2025]

PYQ — 2024 (07 marks), 2025 (10 marks)

2024: Define equipotential line. Draw the electric field lines and the equipotential lines of a uniform charge sphere. 2025: Make a two-dimensional sketch of the electric field lines and the equipotential lines of a uniform charge sphere and a dipole. Ensure the lines are distinguishable.

[FIGURE: Two-panel 2D sketch. Left — uniformly charged sphere: straight radial field lines with arrowheads pointing outward from the sphere; concentric dashed circles as equipotentials, spaced increasingly far apart with radius. Right — electric dipole: curved solid field lines emerging from $+q$ and terminating on $-q$, symmetric about the dipole axis; dashed equipotential contours forming closed ovals around each charge, with the perpendicular bisector plane being the $V=0$ equipotential (a straight dashed line). — source: David K. Cheng, Ch. 3, Fig. 3-12]

Uniform charge sphere (radius $a$, total charge $Q$):

Feature	Field lines $\vec{E}$	Equipotentials $V$
Shape	Straight, radial, pointing outward (for $+Q$)	Concentric circles (spheres in 3D)
Outside ($R>a$)	$E=\frac{Q}{4\pi {ϵ}_0{R}^2}$	$V=\frac{Q}{4\pi {ϵ}_{0}R}$
Spacing	Lines spread apart as $R$ grows (field weakens)	Circles get further apart as $R$ grows

Electric dipole:

Feature	Field lines $\vec{E}$	Equipotentials $V$
Shape	Curved loops leaving $+q$, entering $-q$	Closed ovals around each charge
Symmetry plane	Field is purely perpendicular to the bisecting plane ($\theta =90^{\circ }$)	The bisecting plane is the $V=0$ equipotential — a straight line in the 2D sketch
Far field	$E\propto 1/R^3$	$V\propto 1/R^2$

Marks the examiner is looking for in the sketch

1. Arrowheads on the field lines (direction from $+$ to $-$).
2. Different line styles — the 2025 paper explicitly says “ensure the lines are distinguishable”: use solid for $\vec{E}$, dashed for equipotentials, and add a legend.
3. Visible perpendicularity at every crossing.
4. The $V=0$ line through the dipole’s midplane — this is the detail most students omit.
5. Field lines start and end on charges; equipotentials are closed and never cross.

---

4. Master Derivation: The Electric Dipole [PYQ: 2016, 2017, 2019, 2021, 2023, 2024]

The most-repeated derivation in Chapter 2

Asked in six separate years. The potential derivation ($V$) is asked in 2017, 2019, 2021, 2024; the field derivation ($\vec{E}$ from the dipole moment) in 2016, 2023.

Abstract

An electric dipole is a pair of equal and opposite point charges, $+q$ and $-q$, separated by a distance $d$ that is small compared with the distance to the observation point.

The electric dipole moment is the vector $\vec{p}=q\vec{d}\text{(C}\cdot \text{m)}$ whose magnitude is $qd$ and whose direction is from the negative charge toward the positive charge.

[FIGURE: Electric dipole with $+q$ at $(0,0,d/2)$ and $-q$ at $(0,0,-d/2)$; observation point $P$ at distance $R$ from the origin making angle $\theta$ with the $z$-axis; distances $R_{+}$ and $R_{-}$ marked, with the far-field approximation showing $R_{-}-R_{+}\approx d\cos \theta$ — source: David K. Cheng, Ch. 3, Fig. 3-10]

4.1 Derivation of the Dipole Potential $V(R,\theta )$ [PYQ: 2017, 2019, 2021, 2024]

Step-by-step derivation

Step 1 — Exact expression. Place $+q$ at $z=+d/2$ and $-q$ at $z=-d/2$. Potential is a scalar, so simply add: $V=\frac{q}{4\pi {ϵ}_0}\left(\frac{1}{R_{+}}-\frac{1}{R_{-}}\right)=\frac{q}{4\pi {ϵ}_0}\left(\frac{R_{-}-R_{+}}{R_{+}{R}_{-}}\right)$

Step 2 — Far-field approximation ($R\gg d$). By the law of cosines, $R_{\pm }^2=R^2+{\left(\frac{d}{2}\right)}^2\mp Rd\cos \theta \approx R^2\left(1\mp \frac{d}{R}\cos \theta \right)$ neglecting $(d/2R{)}^2$. Taking the inverse square root and applying the binomial expansion $(1+x{)}^n\approx 1+nx$ with $n=-1/2$: $\frac{1}{R_{\pm }}\approx \frac{1}{R}\left(1\pm \frac{d}{2R}\cos \theta \right)$

Step 3 — Subtract. $\frac{1}{R_{+}}-\frac{1}{R_{-}}\approx \frac{1}{R}\left[\left(1+\frac{d\cos \theta }{2R}\right)-\left(1-\frac{d\cos \theta }{2R}\right)\right]=\frac{d\cos \theta }{R^2}$

Equivalently: $R_{-}-R_{+}\approx d\cos \theta$ and $R_{+}{R}_{-}\approx R^2$.

Step 4 — Substitute back. $V=\frac{qd\cos \theta }{4\pi {ϵ}_0{R}^2}$

Step 5 — Express via the dipole moment. With $\vec{p}={\hat{a}}_{z}qd$ and $\vec{p}\cdot {\hat{a}}_R=p\cos \theta$: $\boxed{V=\frac{p\cos \theta }{4\pi {ϵ}_0{R}^2}=\frac{\vec{p}\cdot {\hat{a}}_R}{4\pi {ϵ}_0{R}^2}}\text{(V)}■$

4.2 How $V$ Varies with Distance and Angle [PYQ: 2019]

The 2019 paper explicitly asked you to “explain how electric potential varies with distance and angle of position”:

Dependence	Behaviour	Physical reason
With distance $R$	$V\propto 1/R^2$ — decays faster than a point charge’s $1/R$	At large $R$ the two opposite charges nearly cancel; only their small separation leaves a residual, so the potential falls off one power faster
With angle $\theta$	$V\propto \cos \theta$	Maximum positive at $\theta =0^{\circ }$ (on the $+q$ side of the axis); maximum negative at $\theta =180^{\circ }$; exactly zero at $\theta =90^{\circ }$
At $\theta =90^{\circ }$	$V=0$ everywhere on the bisecting plane	Every point there is equidistant from $+q$ and $-q$, so their contributions cancel exactly

[GRAPH: $V$ versus $\theta$ at fixed $R$ — a cosine curve peaking at $\theta =0$, crossing zero at $\theta =90^{\circ }$, minimum at $\theta =180^{\circ }$. Governing equation $V=p\cos \theta /(4\pi {ϵ}_0{R}^2)$.]

4.3 Derivation of the Dipole Field $\vec{E}(R,\theta )$ [PYQ: 2016, 2023]

Step-by-step derivation

Step 1. Apply $\vec{E}=-\nabla V$ in spherical coordinates. Since $V$ is independent of $\varphi$, the ${\hat{a}}_{\varphi }$ term vanishes: $\vec{E}=-\left[{\hat{a}}_R\frac{\partial V}{\partial R}+{\hat{a}}_{\theta }\frac{1}{R}\frac{\partial V}{\partial \theta }\right]$

Step 2 — Radial component. $E_R=-\frac{\partial }{\partial R}\left(\frac{p\cos \theta }{4\pi {ϵ}_0{R}^2}\right)=-\left(\frac{-2p\cos \theta }{4\pi {ϵ}_0{R}^3}\right)=\frac{2p\cos \theta }{4\pi {ϵ}_0{R}^3}$

Step 3 — Polar component. $E_{\theta }=-\frac{1}{R}\frac{\partial }{\partial \theta }\left(\frac{p\cos \theta }{4\pi {ϵ}_0{R}^2}\right)=-\frac{1}{R}\left(\frac{-p\sin \theta }{4\pi {ϵ}_0{R}^2}\right)=\frac{p\sin \theta }{4\pi {ϵ}_0{R}^3}$

Step 4 — Combine. $\boxed{\vec{E}=\frac{p}{4\pi {ϵ}_0{R}^3}\left(2\cos \theta {\hat{a}}_R+\sin \theta {\hat{a}}_{\theta }\right)}\text{(V/m)}■$

Coordinate-free form (occasionally useful): $\vec{E}=\frac{1}{4\pi {ϵ}_0{R}^3}\left[\frac{3(\vec{p}\cdot \vec{R})}{R^2}\vec{R}-\vec{p}\right]$

4.4 Special Cases Worth Quoting

Position	$\theta$	$\vec{E}$	Note
On the axis	$0^{\circ }$	${\hat{a}}_R\frac{2p}{4\pi {ϵ}_0{R}^3}$	Twice as strong as at the same distance broadside
Broadside (bisecting plane)	$90^{\circ }$	${\hat{a}}_{\theta }\frac{p}{4\pi {ϵ}_0{R}^3}$	Points antiparallel to $\vec{p}$; note $V=0$ here even though $\vec{E}\ne 0$

The classic conceptual trap

On the bisecting plane $V=0$ but $\vec{E}\ne 0$. Zero potential does not mean zero field — $\vec{E}$ depends on the rate of change of $V$ across the surface, not on its value. Examiners like this one.

4.5 Dipole Scaling Summary

Quantity	Point charge	Dipole
Potential $V$	$\propto 1/R$	$\propto 1/R^2$
Field $\vec{E}$	$\propto 1/R^2$	$\propto 1/R^3$
Angular dependence	None (isotropic)	$\cos \theta$ for $V$; $2\cos \theta {\hat{a}}_R+\sin \theta {\hat{a}}_{\theta }$ for $\vec{E}$

---

5. PYQ Coverage for This Note

Question (verbatim, condensed)	Marks	Year(s)
Define electric field intensity and electric potential. Also derive the relation between them when caused by a point charge.	10	2016
Define electric potential. Show that work done in moving a unit charge from one point to another equals the potential difference.	08	2017
Define electric potential and state the relation between electric potential and electric field intensity.	06	2021
Differentiate between $\vec{E}$ and $\vec{D}$, emphasizing their physical significance.	05	2022
Distinguish between $\vec{E}$ and $\vec{D}$ with respect to definition, unit, governing relation, and physical significance.	07	2025
Define equipotential line. Draw the electric field lines and equipotential lines of a uniform charge sphere.	07	2024
2D sketch of field lines and equipotential lines of a uniform charge sphere and a dipole.	10	2025
Define electric dipole moment. Deduce $\vec{E}$ of an electric dipole in terms of its dipole moment.	10/13	2016, 2023
Define electric dipole moment. Estimate $V$ at any point P due to an electric dipole ($R\gg d$).	12	2021
Define electric dipole. Derive $V$ due to a dipole and explain how $V$ varies with distance and angle of position.	13	2019
Two equal opposite charges separated by $l$ form a dipole. Derive $V$ at an arbitrary point P.	12	2017, 2024

---

6. Exam Hacks & Traps

Key Exam Checkpoints

1. State the approximation explicitly. Write “Since $R\gg d$, we apply the binomial expansion $(1+x{)}^n\approx 1+nx$“. Students who silently jump to $d\cos \theta$ lose presentation marks.
2. Never compute exact $R_{+}$ and $R_{-}$. The whole point is the far-field expansion.
3. $\vec{p}$ points from $-q$ to $+q$. The opposite of the field direction between the charges. Getting this backwards flips the sign of $V$.
4. $V=0$ but $\vec{E}\ne 0$ on the bisecting plane. Know why.
5. For the $\vec{E}$ derivation, use spherical $\nabla$ with the $1/R$ factor on the $\theta$ term. Forgetting the $1/R$ gives $E_{\theta }$ with the wrong power of $R$.
6. In the 2025 sketch question, use two distinguishable line styles and a legend. The question says so directly.
7. For the $\vec{E}$ vs $\vec{D}$ table, use the four headings named in 2025 — definition, unit, governing relation, physical significance. Add medium-dependence as a bonus row.
8. Units: $V$ in volts, $\vec{E}$ in V/m, $\vec{D}$ in C/m², $\vec{p}$ in C·m.

---

7. Self-Check

1. Define $V$ and prove that the work per unit charge equals $V_B-V_A$.
2. Why can $\vec{E}$ be written as $-\nabla V$? Which identity licenses it?
3. Give four differences between $\vec{E}$ and $\vec{D}$.
4. Derive $V$ for a dipole and state the approximation used.
5. Why is $V\propto 1/R^2$ for a dipole but $1/R$ for a point charge?
6. Where is $V=0$ for a dipole, and is $\vec{E}$ zero there?

---

Next: 2.03 Conductors, Dielectrics & Polarization Charge Densities

---

📄 Section: 2.03 Conductors, Dielectrics & Polarization Charge Densities

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"]

---

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]

---

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.

---

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. ✓

---

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.”

---

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.

---

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

---

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.

---

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.

---

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?

---

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

---

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

Related Concepts: 2.03 Conductors, Dielectrics & Polarization Charge Densities | 2.05 Electrostatic Energy, Work Done & Solved PYQ Numericals | 3.04 Magnetization, Magnetic Materials, Boundary Conditions & Hall Effect | 1.02 Vector Operators (Gradient, Divergence, Curl & Laplacian)

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

Core Idea

Two closely-related high-yield blocks live here. Boundary conditions tell you how $\vec{E}$ and $\vec{D}$ behave when crossing an interface — derived by shrinking a loop (for $\vec{E}$) and a pillbox (for $\vec{D}$) onto the surface. Poisson’s and Laplace’s equations convert the postulates into a single differential equation for $V$, which is then solved to get every capacitance formula in the syllabus. Between them these topics appear in nine of the eleven papers.

---

1. Electrostatic Boundary Conditions [PYQ: 2015, 2017, 2022, 2025]

Two media with permittivities ${ϵ}_1$ and ${ϵ}_2$ share a plane boundary. Take the fields close to the interface and decompose each into a tangential component (parallel to the surface) and a normal component (perpendicular to it).

[FIGURE: Interface between media 1 (${ϵ}_1$) and 2 (${ϵ}_2$) showing (a) a rectangular closed loop $abcd$ of width $\Delta w$ and height $\Delta h\to 0$ straddling the boundary, used for the tangential condition; (b) a cylindrical Gaussian pillbox of face area $\Delta s$ and height $\Delta h\to 0$, used for the normal condition. Field vectors ${\vec{E}}_1$, ${\vec{E}}_2$ shown at angles ${\alpha }_1$, ${\alpha }_2$ to the normal — source: David K. Cheng, Ch. 3, Fig. 3-23]

1.1 Tangential Component: $E_{1t}=E_{2t}$

Derivation — shrinking loop

Step 1. Apply the conservative postulate ${\oint }_C\vec{E}\cdot d\vec{l}=0$ to a small rectangular loop $abcd$ of width $\Delta w$ lying along the interface and height $\Delta h$ straddling it.

Step 2. Let $\Delta h\to 0$. The two short sides (of length $\Delta h$) contribute nothing to the line integral in this limit.

Step 3. Only the two long sides survive, traversed in opposite directions: ${\oint }_C\vec{E}\cdot d\vec{l}=E_{1t}\Delta w-E_{2t}\Delta w=0$

Step 4. Divide by $\Delta w$: $\boxed{E_{1t}=E_{2t}}\text{equivalently}\frac{D_{1t}}{{ϵ}_1}=\frac{D_{2t}}{{ϵ}_2}$

Vector form: ${\hat{a}}_{n2}\times ({\vec{E}}_1-{\vec{E}}_2)=0$

Statement: the tangential component of $\vec{E}$ is always continuous across any interface, with no exceptions.

1.2 Normal Component: $D_{1n}-D_{2n}={\rho }_s$

Derivation — shrinking pillbox

Step 1. Apply Gauss’s law ${\oint }_S\vec{D}\cdot d\vec{s}=Q_{\text{enc}}$ to a small cylindrical pillbox of face area $\Delta s$ and height $\Delta h$ straddling the interface.

Step 2. Let $\Delta h\to 0$. The flux through the curved side wall vanishes.

Step 3. Only the two flat faces contribute, with outward normals in opposite directions: ${\oint }_S\vec{D}\cdot d\vec{s}=D_{1n}\Delta s-D_{2n}\Delta s$

Step 4. The enclosed charge is the free surface charge on the interface: $Q_{\text{enc}}={\rho }_s\Delta s$

Step 5. Equate and divide by $\Delta s$: $\boxed{D_{1n}-D_{2n}={\rho }_s}\text{equivalently}{ϵ}_1{E}_{1n}-{ϵ}_2{E}_{2n}={\rho }_s$

Vector form: ${\hat{a}}_{n2}\cdot ({\vec{D}}_1-{\vec{D}}_2)={\rho }_s$

Charge-free boundary (${\rho }_s=0$, the usual dielectric–dielectric case): $D_{1n}=D_{2n}⟹{ϵ}_1{E}_{1n}={ϵ}_2{E}_{2n}$

1.3 Summary Table

Component	Condition	Charge-free case	Physical statement
Tangential $\vec{E}$	$E_{1t}=E_{2t}$	Same	${\vec{E}}_t$ is always continuous
Normal $\vec{D}$	$D_{1n}-D_{2n}={\rho }_s$	$D_{1n}=D_{2n}$	${\vec{D}}_n$ jumps by the free surface charge
Tangential $\vec{D}$	$D_{1t}/{ϵ}_1=D_{2t}/{ϵ}_2$	Same	${\vec{D}}_t$ is discontinuous unless ${ϵ}_1={ϵ}_2$
Normal $\vec{E}$	${ϵ}_1{E}_{1n}-{ϵ}_2{E}_{2n}={\rho }_s$	${ϵ}_1{E}_{1n}={ϵ}_2{E}_{2n}$	${\vec{E}}_n$ is discontinuous

Memory hook

“E is tangentially continuous, D is normally continuous.” Tangential $\to$ $\vec{E}$; normal $\to$ $\vec{D}$.

1.4 What Happens if One Medium is a Conductor? [PYQ: 2015]

PYQ — 2015 (07 marks)

Consider a plane boundary between two dielectric media (with zero conductivities) and establish a relationship between the tangential and normal components of the electric field on both sides. What happens when one of the media is a conductor?

Set medium 2 to be a perfect conductor, so ${\vec{E}}_2={\vec{D}}_2=0$ (see 2.03 Conductors, Dielectrics & Polarization Charge Densities):

General dielectric–dielectric	Becomes, with medium 2 a conductor
$E_{1t}=E_{2t}$	$E_{1t}=0$ — the field has no tangential component; it must meet the conductor perpendicularly
$D_{1n}-D_{2n}={\rho }_s$	$D_{1n}={\rho }_s$, so $E_{1n}={\rho }_s/{ϵ}_1$ — a real free surface charge appears on the conductor

In words: at a dielectric–conductor boundary the field lines become strictly normal to the conducting surface, the conductor surface becomes an equipotential, and the entire field terminates on induced free surface charge. The refraction law also degenerates — ${\alpha }_1\to 0$, meaning the field is refracted to the normal regardless of the incident angle.

---

2. The Law of Refraction for Electric Field Lines [PYQ: 2015, 2017, 2019, 2022, 2025]

Derivation

Let ${\alpha }_1$, ${\alpha }_2$ be the angles that ${\vec{E}}_1$, ${\vec{E}}_2$ make with the normal to the interface. Assume a charge-free boundary (${\rho }_s=0$).

Step 1 — Resolve into components: $E_{1t}=E_1\sin {\alpha }_1,E_{1n}=E_1\cos {\alpha }_1,E_{2t}=E_2\sin {\alpha }_2,E_{2n}=E_2\cos {\alpha }_2$

Step 2 — Apply the tangential condition: E_1\sin\alpha_1 = E_2\sin\alpha_2 \tag{i}

Step 3 — Apply the normal condition: \epsilon_1 E_1\cos\alpha_1 = \epsilon_2 E_2\cos\alpha_2 \tag{ii}

Step 4 — Divide (i) by (ii): $\frac{\tan {\alpha }_1}{{ϵ}_1}=\frac{\tan {\alpha }_2}{{ϵ}_2}$

$\boxed{\frac{\tan {\alpha }_1}{\tan {\alpha }_2}=\frac{{ϵ}_1}{{ϵ}_2}=\frac{{ϵ}_{r1}}{{ϵ}_{r2}}}■$

Physical reading: field lines bend away from the normal when entering a medium of higher permittivity. If ${ϵ}_2>{ϵ}_1$, then $\tan {\alpha }_2>\tan {\alpha }_1$, so ${\alpha }_2>{\alpha }_1$.

2.1 The Standard Numerical [PYQ: 2017, 2019, 2022, 2025]

PYQ — 2017, 2019, 2022 (08/09 marks); 2025 (08 marks)

Two dielectric media with permittivities ${ϵ}_1$ and ${ϵ}_2$ are separated by a charge-free boundary. The electric field intensity in medium 1 at point $P_1$ has magnitude $E_1$ and makes an angle ${\alpha }_1$ with the normal. Determine the magnitude and direction of the electric field intensity at point $P_2$ in medium 2.

Standard solution method — memorise these four steps

Step 1 — Direction (the angle). From the refraction law: ${\alpha }_2={\tan }^{-1}\left(\frac{{ϵ}_2}{{ϵ}_1}\tan {\alpha }_1\right)$

Step 2 — Tangential component carries over unchanged: $E_{2t}=E_{1t}=E_1\sin {\alpha }_1$

Step 3 — Normal component scales by the permittivity ratio: $E_{2n}=\frac{{ϵ}_1}{{ϵ}_2}{E}_{1n}=\frac{{ϵ}_1}{{ϵ}_2}{E}_1\cos {\alpha }_1$

Step 4 — Magnitude by Pythagoras: $E_2=\sqrt{E_{2t}^2+E_{2n}^2}=E_1\sqrt{{\sin }^2{\alpha }_1+{\left(\frac{{ϵ}_1}{{ϵ}_2}\right)}^2{\cos }^2{\alpha }_1}$

Answer format: ”${\vec{E}}_2$ has magnitude $E_2$ and makes an angle ${\alpha }_2$ with the normal to the interface.”

The trap in this numerical

The question says “magnitude and direction” — two separate marks. Many students compute ${\alpha }_2$ and stop, or compute $E_2$ and stop. Always give both, and always state whether your angle is measured from the normal or from the interface — the 2017/2023 magnetic version of this question deliberately switches between the two.

---

3. Poisson’s and Laplace’s Equations [PYQ: 2015, 2016, 2017, 2022, 2023, 2025]

Heavily tested — six papers

2015 (deduce both), 2016 (write both + use Laplace for capacitance), 2017 & 2023 (derive Poisson + its solution), 2022 & 2025 (derive Poisson).

Derivation

Step 1. Start from Gauss’s postulate in a material medium: $\nabla \cdot \vec{D}={\rho }_v$

Step 2. Substitute the constitutive relation $\vec{D}=ϵ\vec{E}$: $\nabla \cdot (ϵ\vec{E})={\rho }_v$

Step 3. For a homogeneous medium, $ϵ$ is constant and comes outside the divergence: $ϵ(\nabla \cdot \vec{E})={\rho }_v⟹\nabla \cdot \vec{E}=\frac{{\rho }_v}{ϵ}$

Step 4. Substitute $\vec{E}=-\nabla V$: $\nabla \cdot (-\nabla V)=\frac{{\rho }_v}{ϵ}⟹-{\nabla }^{2}V=\frac{{\rho }_v}{ϵ}$

$\boxed{{\nabla }^{2}V=-\frac{{\rho }_v}{ϵ}}\text{Poisson’s Equation}■$

Step 5. In a charge-free region (${\rho }_v=0$): $\boxed{{\nabla }^{2}V=0}\text{Laplace’s Equation}■$

Where the "homogeneous medium" assumption is used

Step 3 requires $ϵ$ to be constant so it can be pulled out of the divergence. If $ϵ=f(x,y,z)$ (an inhomogeneous medium — see 1.04 Media Properties & Conductor-Insulator Behaviour), you must keep $\nabla \cdot (ϵ\nabla V)$ intact and the simple Poisson form fails. State this assumption — it is a marked point.

3.1 The Solution of Poisson’s Equation [PYQ: 2017, 2023]

PYQ — 2017, 2023 (08/11 marks)

Derive Poisson’s equation with respect to an electric potential. What will be the solution of it?

The general (particular) solution, obtained by superposing the point-charge potential over the whole source distribution, is:

$\boxed{V=\frac{1}{4\pi ϵ}{\int }_{V^{\prime }}\frac{{\rho }_v}{R}d{v}^{\prime }}$

with the corresponding forms for surface and line sources: $V=\frac{1}{4\pi ϵ}{\int }_{S^{\prime }}\frac{{\rho }_s}{R}d{s}^{\prime },V=\frac{1}{4\pi ϵ}{\int }_{L^{\prime }}\frac{{\rho }_l}{R}d{l}^{\prime }$

where $R$ is the distance from the source element to the observation point. To this particular solution one adds any solution of the homogeneous equation ${\nabla }^{2}V=0$ needed to satisfy the boundary conditions.

Why Laplace's equation matters more in practice

Most exam problems place all the charge on conductor surfaces, leaving the region between them charge-free. There you solve ${\nabla }^{2}V=0$ subject to the electrode potentials — a boundary-value problem. Every capacitance derivation below is exactly this.

3.2 The Standard Solution Procedure

graph TD
    A["Identify the symmetry:<br/>V varies with only ONE coordinate"] --> B["Reduce ∇²V = 0 to an ODE"]
    B --> C["Integrate twice → two constants C1, C2"]
    C --> D["Apply the two electrode boundary conditions"]
    D --> E["Get V as a function of position"]
    E --> F["E = −∇V"]
    F --> G["rho_s = D_n = ε E_n at the plate"]
    G --> H["Q = rho_s × A, then C = Q / V0"]

---

4. Application 1: Parallel Plate Capacitor [PYQ: 2016, 2018, 2019, 2021, 2022, 2023, 2024, 2025]

The single most-repeated numerical derivation in Chapter 2 — eight years

Asked as “determine the capacitance” (2018, 2019, 2022, 2023, 2024), “using Laplace’s equation find the capacitance” (2016), “estimate the potential and surface charge density” (2021), and “determine the surface charge density on each plate” (2022, 2025).

Master Derivation

Setup. Two large parallel plates separated by distance $d$, filled with a dielectric of permittivity $ϵ$, plate area $A$. Lower plate at $y=0$ held at $V=0$; upper plate at $y=d$ held at $V=V_0$. Fringing at the edges is neglected.

Step 1 — Reduce Laplace’s equation. By symmetry $V$ varies only with $y$: ${\nabla }^{2}V=\frac{d^{2}V}{d{y}^2}=0$

Step 2 — Integrate twice: $\frac{dV}{dy}=C_1⟹V(y)=C_{1}y+C_2$

Step 3 — Apply boundary conditions:

• At $y=0$: $V(0)=0⟹C_2=0$
• At $y=d$: $V(d)=V_0⟹C_{1}d=V_0⟹C_1=V_0/d$

Step 4 — Potential distribution (this alone answers the 2021 part (i)): $\boxed{V(y)=\frac{V_0}{d}y\text{(V)}}$ The potential varies linearly between the plates.

Step 5 — Electric field: $\vec{E}=-\nabla V=-{\hat{a}}_y\frac{dV}{dy}=-{\hat{a}}_y\frac{V_0}{d}\text{(V/m)}$ Uniform, and directed from the high-potential plate to the low-potential plate.

Step 6 — Surface charge density (this answers 2021 part (ii), 2022 and 2025). On the upper plate at $y=d$, the outward normal into the dielectric is $-{\hat{a}}_y$: $\boxed{{\rho }_s=D_n=ϵE_n=\frac{ϵV_0}{d}{\text{(C/m}}^2\text{)}}$ On the lower plate the density is $-ϵV_0/d$ — equal and opposite.

Step 7 — Total charge: $Q={\rho }_{s}A=\frac{ϵA{V}_0}{d}\text{(C)}$

Step 8 — Capacitance: $\boxed{C=\frac{Q}{V_0}=\frac{ϵA}{d}\text{(F)}}■$

Reading the result

$C\propto A$ (bigger plates hold more charge at the same voltage), $C\propto 1/d$ (closer plates produce a stronger field for the same voltage), and $C\propto ϵ$ (the dielectric polarizes and partially cancels the field, letting more charge sit at the same voltage — see 2.03 Conductors, Dielectrics & Polarization Charge Densities).

---

5. Application 2: Cylindrical (Coaxial) Capacitor [PYQ: 2015, 2017]

PYQ — 2015 (10 marks), 2017 (13 marks)

A cylindrical capacitor consists of an inner conductor of radius $a$ and an outer conductor of inner radius $b$. The space between is filled with a dielectric of permittivity $ϵ$, and the length is $L$. Determine the capacitance.

Master Derivation

Setup. Inner conductor radius $a$ at potential $V_0$; outer conductor inner radius $b$ at potential $0$; length $L\gg b$ so fringing is negligible.

Step 1 — Laplace’s equation in cylindrical coordinates. $V$ varies only with $r$: ${\nabla }^{2}V=\frac{1}{r}\frac{d}{dr}\left(r\frac{dV}{dr}\right)=0$

Step 2 — Integrate once: $r\frac{dV}{dr}=C_1⟹\frac{dV}{dr}=\frac{C_1}{r}$

Step 3 — Integrate again: $V(r)=C_1\ln r+C_2$

Step 4 — Apply boundary conditions:

• At $r=b$: $V(b)=0⟹C_2=-C_1\ln b$, so $V(r)=C_1\ln (r/b)$
• At $r=a$: $V(a)=V_0⟹V_0=C_1\ln (a/b)=-C_1\ln (b/a)$ $⟹C_1=-\frac{V_0}{\ln (b/a)}$

Step 5 — Potential distribution: $V(r)=V_0\frac{\ln (b/r)}{\ln (b/a)}$ Note this is logarithmic, not linear — the cylindrical geometry concentrates the field near the inner conductor.

Step 6 — Electric field: $\vec{E}=-{\hat{a}}_r\frac{dV}{dr}={\hat{a}}_r\frac{V_0}{r\ln (b/a)}\text{(V/m)}$

Step 7 — Surface charge density on the inner conductor ($r=a$, outward normal $+{\hat{a}}_r$): ${\rho }_s=ϵE_r(a)=\frac{ϵV_0}{a\ln (b/a)}$

Step 8 — Total charge on the inner cylinder of area $A=2\pi aL$: $Q={\rho }_{s}A=\frac{ϵV_0}{a\ln (b/a)}(2\pi aL)=\frac{2\pi ϵL{V}_0}{\ln (b/a)}$

Step 9 — Capacitance: $\boxed{C=\frac{Q}{V_0}=\frac{2\pi ϵL}{\ln (b/a)}\text{(F)}}■$

Per unit length: $C^{\prime }=\frac{2\pi ϵ}{\ln (b/a)}$ (F/m) — the standard coaxial cable result.

Common slip

$\ln (b/a)$, not $\ln (a/b)$. Since $b>a$, $\ln (b/a)>0$ and the capacitance comes out positive. A negative capacitance means you inverted the ratio.

---

6. Application 3: Electrohydrodynamic Pump (Poisson with ${\rho }_v\ne 0$) [PYQ: 2024]

PYQ — 2024 (13 marks)

In an electrohydrodynamic pump, the region between two electrodes is filled with a uniform charge density ${\rho }_0$. If the left electrode has potential $V_0$ and the right electrode has potential 0 V, determine the expressions for electric potential and electric field intensity at any point between the electrodes.

Solution — this one needs Poisson, not Laplace

Setup. Left electrode at $x=0$ with $V=V_0$; right electrode at $x=d$ with $V=0$; the region between carries uniform ${\rho }_v={\rho }_0$.

Step 1 — Poisson’s equation in 1D: $\frac{d^{2}V}{d{x}^2}=-\frac{{\rho }_0}{ϵ}$

Step 2 — Integrate once: $\frac{dV}{dx}=-\frac{{\rho }_0}{ϵ}x+C_1$

Step 3 — Integrate again: $V(x)=-\frac{{\rho }_0}{2ϵ}{x}^2+C_{1}x+C_2$

Step 4 — Boundary conditions:

• At $x=0$: $V(0)=V_0⟹C_2=V_0$
• At $x=d$: $-\frac{{\rho }_0{d}^2}{2ϵ}+C_{1}d+V_0=0⟹C_1=\frac{{\rho }_{0}d}{2ϵ}-\frac{V_0}{d}$

Step 5 — Potential distribution: $V(x)=-\frac{{\rho }_0}{2ϵ}{x}^2+\left(\frac{{\rho }_{0}d}{2ϵ}-\frac{V_0}{d}\right)x+V_0$ $\boxed{V(x)=V_0\left(1-\frac{x}{d}\right)+\frac{{\rho }_0}{2ϵ}x(d-x)}$

Step 6 — Electric field: $E_x=-\frac{dV}{dx}=-\left[-\frac{{\rho }_0}{ϵ}x+\frac{{\rho }_{0}d}{2ϵ}-\frac{V_0}{d}\right]$ $\boxed{\vec{E}(x)={\hat{a}}_x\left[\frac{V_0}{d}+\frac{{\rho }_0}{ϵ}\left(x-\frac{d}{2}\right)\right]\text{(V/m)}}$

Sanity checks worth writing down

• Setting ${\rho }_0=0$ recovers the empty parallel-plate result: $V=V_0(1-x/d)$, $\vec{E}={\hat{a}}_x{V}_0/d$. ✓
• The second term $\frac{{\rho }_0}{2ϵ}x(d-x)$ is a parabola vanishing at both electrodes — the space charge bows the potential without violating the fixed electrode values. ✓
• The field is now non-uniform, varying linearly with $x$ — that is exactly what drives the pumping action.

Laplace vs Poisson — pick the right one

If the region between the electrodes is charge-free, use ${\nabla }^{2}V=0$ (Laplace). If it contains a volume charge density, you must use ${\nabla }^{2}V=-{\rho }_v/ϵ$ (Poisson). The 2024 question is deliberately the Poisson case; students who default to Laplace lose the entire question.

---

7. PYQ Coverage for This Note

Question (verbatim, condensed)	Marks	Year(s)
Plane boundary between two dielectrics — relate tangential and normal components. What happens when one medium is a conductor?	07	2015
Two dielectric media ${ϵ}_1$, ${ϵ}_2$, charge-free boundary, $E_1$ at angle ${\alpha }_1$ — determine magnitude and direction of ${\vec{E}}_2$.	08/09	2017, 2019, 2022
…Derive the boundary conditions and determine the magnitude and direction of the electric field in medium 2.	08	2025
Deduce the equations of Poisson’s and Laplace’s expressing the space rate of variation of electric field component.	08	2015
Write down the Laplace and Poisson’s equation. Using Laplace equation find out the capacitance of a parallel plate capacitor.	10	2016
Derive Poisson’s equation with respect to an electric potential. What will be the solution of it?	08/11	2017, 2023
Derive Poisson’s equation for electrostatics.	05/08	2022, 2025
Parallel plate capacitor, separation $d$, dielectric $ϵ$, area $A$ — determine the capacitance.	12/13	2018, 2019, 2022, 2023, 2024
Plates at 0 and $V_0$ — estimate (i) potential at any point between the plates, (ii) surface charge density on the plates.	11	2021
Fixed voltage $V_0$ across a parallel plate capacitor — determine the surface charge density on each plate.	11/12	2022, 2025
Cylindrical capacitor, inner radius $a$, outer inner radius $b$, dielectric $ϵ$, length $L$ — determine the capacitance.	10/13	2015, 2017
Electrohydrodynamic pump with uniform ${\rho }_0$ — determine $V$ and $\vec{E}$ at any point between the electrodes.	13	2024

---

8. Exam Hacks & Traps

Key Exam Checkpoints

1. Boundary conditions are derived, not quoted. Draw the loop for the tangential condition and the pillbox for the normal condition, and say $\Delta h\to 0$ explicitly.
2. “E tangentially continuous, D normally continuous.” Never write $E_{1n}=E_{2n}$ — that is only true if ${ϵ}_1={ϵ}_2$.
3. State whether your angle is from the normal or from the interface. Papers switch between the two conventions deliberately; a correct number measured from the wrong reference scores zero.
4. Give both magnitude AND direction when the refraction numerical asks for it. Two separate marks.
5. State the “homogeneous medium” assumption when pulling $ϵ$ out of the divergence in the Poisson derivation.
6. Charge-free region $\to$ Laplace. Charge present $\to$ Poisson. Reading the question for this distinction is worth 13 marks in 2024.
7. Follow the fixed capacitance chain: ${\nabla }^{2}V=0\to V(x)\to \vec{E}=-\nabla V\to {\rho }_s=ϵE_n\to Q={\rho }_{s}A\to C=Q/V_0$. Do not shortcut to $C=ϵA/d$ when the question says “using Laplace’s equation”.
8. $\ln (b/a)$ with $b>a$ for the coaxial result.
9. Always state “fringing effects are neglected” — the question usually says so, and repeating it earns the assumption mark.

---

9. Self-Check

1. Derive both electrostatic boundary conditions from the postulates.
2. What happens to each condition when medium 2 becomes a perfect conductor?
3. Derive $\tan {\alpha }_1/\tan {\alpha }_2={ϵ}_1/{ϵ}_2$.
4. Given $E_1$, ${\alpha }_1$, ${ϵ}_1$, ${ϵ}_2$, write the four steps to find $E_2$ and ${\alpha }_2$.
5. Derive Poisson’s equation and name the assumption used.
6. Derive $C=ϵA/d$ starting from Laplace’s equation.
7. Derive $C=2\pi ϵL/\ln (b/a)$.
8. Why does the electrohydrodynamic pump need Poisson rather than Laplace?

---

Next: 2.05 Electrostatic Energy, Work Done & Solved PYQ Numericals

---

📄 Section: 2.05 Electrostatic Energy, Work Done & Solved PYQ Numericals

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

2.05 Electrostatic Energy, Work Done & Solved PYQ Numericals

Core Idea

Assembling a set of charges takes work, and that work is stored in the field as electrostatic energy. The derivation of $W_e=\frac{1}{2}\sum q_i{V}_i$ has been asked in 2018, 2022, 2023, 2024 and 2025 — five consecutive-era papers — and its numerical counterpart in 2019, 2020, 2023. This note also collects every remaining Chapter 2 numerical in fully worked form.

Why this note exists

Electrostatic energy is one of the highest-frequency topics in the paper (7 exam years) but had no coverage anywhere in the Chapter 2 atomic notes, and several repeated numericals had no worked solution. Both gaps are closed here.

---

1. Electrostatic Energy [PYQ: 2018, 2019, 2020, 2022, 2023, 2024, 2025]

Abstract

The electrostatic energy $W_e$ of a system of charges is the total work done by an external agent in assembling that configuration by bringing each charge, one at a time and without acceleration, from infinity to its final position. Since electrostatic forces are conservative, this work is fully recoverable and is regarded as stored in the electric field.

1.1 Master Derivation: Assembling $n$ Point Charges [PYQ: 2018, 2022, 2023, 2024, 2025]

PYQ — 2018, 2022 (12), 2023 (12), 2024 (12), 2025 (13 marks)

Define electrostatic potential energy. Derive the expression for the work done in assembling $n$ (or $k$) point charges one after another from infinity.

Step-by-step derivation

Start with empty space and bring in the charges one at a time.

Step 1 — Bring in $q_1$. Empty space exerts no force, so: $W_1=0$

Step 2 — Bring in $q_2$. It must be moved through the potential $V_{21}$ established at position 2 by $q_1$: $W_2=q_2{V}_{21}=q_2\frac{q_1}{4\pi {ϵ}_0{R}_{21}}$

Step 3 — Bring in $q_3$. It moves through the combined potential of $q_1$ and $q_2$: $W_3=q_3(V_{31}+V_{32})=q_3\left(\frac{q_1}{4\pi {ϵ}_0{R}_{31}}+\frac{q_2}{4\pi {ϵ}_0{R}_{32}}\right)$

Step 4 — Continue to $q_n$, which moves through the potential of all $n-1$ charges already placed. The total work is: $W_e={\sum }_{i=1}^n{\sum }_{j<i}\frac{q_i{q}_j}{4\pi {ϵ}_0{R}_{ij}}$

Step 5 — Symmetrise. Note that assembling the charges in the reverse order gives the same total energy (the final configuration is identical). Writing $W_e$ as the average of the forward and reverse sums, every pair $(i,j)$ is now counted twice, so we divide by 2: $W_e=\frac{1}{2}{\sum }_{i=1}^n{\sum }_{\begin{array}{c}j=1\\ j\ne i\end{array}}^n\frac{q_i{q}_j}{4\pi {ϵ}_0{R}_{ij}}$

Step 6 — Recognise the inner sum as the potential $V_i$ at the location of $q_i$ due to all other charges: $V_i=\frac{1}{4\pi {ϵ}_0}{\sum }_{\begin{array}{c}j=1\\ j\ne i\end{array}}^n\frac{q_j}{R_{ij}}$

Step 7 — Final result: $\boxed{W_e=\frac{1}{2}\sum _{i=1}^n{q}_i{V}_i\text{(J)}}■$

Explain the factor of $\frac{1}{2}$ — it is a marked point

“The factor $\frac{1}{2}$ arises because the double summation counts every interacting pair twice — once as $(i,j)$ and once as $(j,i)$ — whereas each pair contributes only one physical interaction energy.”

$V_i$ excludes the charge itself

$V_i$ is the potential at $q_i$‘s location produced by every other charge, not including $q_i$‘s own field (which would be infinite for a point charge). Writing $j\ne i$ under the sum is essential.

1.2 Continuous Distributions and Energy Density

For continuous charge, the sum becomes an integral: $W_e=\frac{1}{2}{\int }_{V^{\prime }}{\rho }_{v}Vd{v}^{\prime }$

Substituting ${\rho }_v=\nabla \cdot \vec{D}$ and applying the Divergence Theorem (with the surface term vanishing at infinity) converts this into a field-only form: $W_e=\frac{1}{2}{\int }_{V^{\prime }}\vec{D}\cdot \vec{E}d{v}^{\prime }$

Electrostatic Energy Density

$w_e=\frac{1}{2}\vec{D}\cdot \vec{E}=\frac{1}{2}ϵE^2=\frac{D^2}{2ϵ}{\text{(J/m}}^3\text{)}$

Energy stored in a capacitor follows immediately: $W_e=\frac{1}{2}C{V}^2=\frac{1}{2}QV=\frac{Q^2}{2C}$

The conceptual shift worth mentioning

$W_e=\frac{1}{2}\sum q_i{V}_i$ locates the energy in the charges; $W_e=\frac{1}{2}\int \vec{D}\cdot \vec{E}dv$ locates it in the field, spread through all space. Both give the same number, but the field picture is the one that survives into time-varying electromagnetics, where energy demonstrably travels through space as a wave. This links directly to the Poynting vector in Chapter 4.

---

2. Solved Numerical 1 — Energy Stored in Four Point Charges [PYQ: 2019, 2020, 2023]

PYQ — 2019, 2023 (08/10 marks)

What is electrostatic energy? What energy is stored in the field with point charges $1,-2,-3$ and $4\mu$C located on the x-axis at $x=1,2,3,4$ metres respectively?

(The 2020 variant uses charges $-2,1,3,-4\mu$C at the same positions — same method, different arithmetic.)

Solution — 2019 / 2023 version

Charges and positions:

Charge	Value ($\mu$C)	Position $x$ (m)
$q_1$	$+1$	1
$q_2$	$-2$	2
$q_3$	$-3$	3
$q_4$	$+4$	4

Step 1 — Use the pair-sum form (easier than computing four separate $V_i$): $W_e=\frac{1}{4\pi {ϵ}_0}{\sum }_{i<j}\frac{q_i{q}_j}{R_{ij}}=9\times 10^9{\sum }_{i<j}\frac{q_i{q}_j}{R_{ij}}$

Step 2 — Tabulate all six pairs. With $q$ in $\mu$C, each product $q_i{q}_j$ carries $10^{-12}$:

Pair	$q_i{q}_j$ ($10^{-12}$)	$R_{ij}$ (m)	$q_i{q}_j/R_{ij}$ ($10^{-12}$)
$(1,2)$	$(1)(-2)=-2$	1	$-2.000$
$(1,3)$	$(1)(-3)=-3$	2	$-1.500$
$(1,4)$	$(1)(4)=4$	3	$+1.333$
$(2,3)$	$(-2)(-3)=6$	1	$+6.000$
$(2,4)$	$(-2)(4)=-8$	2	$-4.000$
$(3,4)$	$(-3)(4)=-12$	1	$-12.000$
		Sum	$-12.167$

Step 3 — Multiply: $W_e=9\times 10^9\times (-12.167\times 10^{-12})=-0.1095\text{J}$

$\boxed{W_e\approx -0.110\text{J}=-109.5\text{mJ}}$

Interpreting a negative energy — worth a mark

A negative total energy means the configuration is bound: net attractive interactions dominate, and an external agent would have to supply $+0.110$ J to disassemble the system back to infinity. Say this; do not just leave a bare negative number.

Use pairs, not the $\frac{1}{2}\sum q_i{V}_i$ form, under exam pressure

The pair form has six terms; the $V_i$ form has twelve (each counted twice) plus a factor of $\frac{1}{2}$. Both give the same answer, but the pair table is faster and much harder to get wrong. Just state the equivalence: $W_e=\frac{1}{2}{\sum }_i{q}_i{V}_i={\sum }_{i<j}\frac{q_i{q}_j}{4\pi {ϵ}_0{R}_{ij}}$.

---

3. Solved Numerical 2 — Work Done in a Non-Uniform Field [PYQ: 2022, 2025]

PYQ — 2022, 2025 (12 marks)

Determine the work done in carrying a $-2\mu$C charge from $P_1(2,1,-1)$ to $P_2(8,2,-1)$ in the field $\vec{E}={\hat{a}}_{x}y+{\hat{a}}_{y}x$ along the straight line joining $P_1$ and $P_2$.

Solution

Step 1 — Formula for external work: $W=-q{\int }_{P_1}^{P_2}\vec{E}\cdot d\vec{l}$

Step 2 — Form the integrand. With $d\vec{l}={\hat{a}}_{x}dx+{\hat{a}}_{y}dy+{\hat{a}}_{z}dz$ and $z$ constant ($dz=0$): $\vec{E}\cdot d\vec{l}=({\hat{a}}_{x}y+{\hat{a}}_{y}x)\cdot ({\hat{a}}_{x}dx+{\hat{a}}_{y}dy)=ydx+xdy$

Step 3 — Recognise the perfect differential. Note that $ydx+xdy=d(xy)$ so the integral is path-independent — confirming the field is conservative. (Check: $\nabla \times \vec{E}={\hat{a}}_z\left[\frac{\partial x}{\partial x}-\frac{\partial y}{\partial y}\right]={\hat{a}}_z[1-1]=0$ ✓)

Step 4 — Evaluate at the endpoints: ${\int }_{P_1}^{P_2}\vec{E}\cdot d\vec{l}=[xy{]}_{P_1}^{P_2}=(8)(2)-(2)(1)=16-2=14\text{V}$

Step 5 — Compute the work with $q=-2\mu$C $=-2\times 10^{-6}$ C: $W=-(-2\times 10^{-6})(14)=2.8\times 10^{-5}\text{J}$

$\boxed{W=28\mu \text{J}}$

The elegant shortcut — but show you earned it

Because $\vec{E}$ is conservative you never need the equation of the straight line. However, the question says “along the straight line joining $P_1$ and $P_2$”, so demonstrate that you noticed: state $\nabla \times \vec{E}=0$, conclude path independence, then use the endpoint values.

If you prefer the explicit path: the line through $(2,1)$ and $(8,2)$ is $x=6y-4$, so $dx=6dy$, and ${\int }_1^2[y(6dy)+(6y-4)dy]={\int }_1^2(12y-4)dy=[6y^2-4y{]}_1^2=(24-8)-(6-4)=14✓$

The sign trap

$W=-q\int \vec{E}\cdot d\vec{l}$ is the work done by an external agent. Here both $q$ and the sign are negative, so $W$ comes out positive — meaning you must push the negative charge against its natural tendency. Students who drop the leading minus get $-28\mu$J.

---

4. Solved Numerical 3 — Spherical Cloud with Variable Density [PYQ: 2021]

PYQ — 2021 (10 marks)

A spherical uniform charge distribution in free space has ${\rho }_v=2r$ nC/m³ for $0\le r\le 10$ m and zero otherwise. Calculate $\vec{E}$ at $r=2$ m and $r=12$ m.

Read the question carefully

Despite the word “uniform”, ${\rho }_v=2r$ varies with radius. It must stay inside the integral. Writing $Q={\rho }_v\cdot \frac{4}{3}\pi r^3$ loses the entire question.

Solution

Part (a) — Inside the cloud, $r=2$ m $<10$ m

Apply Gauss’s law over a concentric sphere of radius $r$: $D_r(4\pi r^2)={\int }_0^r{\rho }_{v}d{v}^{\prime }={\int }_0^r(2r^{\prime })(4\pi r^{\prime 2}d{r}^{\prime })=8\pi {\int }_0^r{r}^{\prime 3}d{r}^{\prime }=8\pi {\left[\frac{r^{\prime 4}}{4}\right]}_0^r=2\pi r^4$

Divide through: $D_r=\frac{2\pi r^4}{4\pi r^2}=\frac{r^2}{2}{\text{nC/m}}^2$

At $r=2$ m: $D_r=\frac{2^2}{2}=2{\text{nC/m}}^2=2\times 10^{-9}{\text{C/m}}^2$ $E_r=\frac{D_r}{{ϵ}_0}=\frac{2\times 10^{-9}}{8.854\times 10^{-12}}=225.9\text{V/m}$

$\boxed{\vec{E}(2\text{m})={\hat{a}}_{r}225.9\text{V/m (radially outward)}}$

Part (b) — Outside the cloud, $r=12$ m $>10$ m

All charge is now enclosed. Integrate to the cloud radius $a=10$ m: $Q_{\text{total}}={\int }_0^{10}(2r^{\prime })(4\pi r^{\prime 2}d{r}^{\prime })=2\pi [r^{\prime 4}{]}_0^{10}=2\pi (10^4)=20000\pi \text{nC}$

$D_r=\frac{Q_{\text{total}}}{4\pi r^2}=\frac{20000\pi \times 10^{-9}}{4\pi (12{)}^2}=\frac{20000\times 10^{-9}}{576}=3.472\times 10^{-8}{\text{C/m}}^2$

$E_r=\frac{3.472\times 10^{-8}}{8.854\times 10^{-12}}=3921.4\text{V/m}$

$\boxed{\vec{E}(12\text{m})={\hat{a}}_{r}3921.4\text{V/m (radially outward)}}$

---

5. Solved Numerical 4 — Field and Potential from Three Point Charges [PYQ: 2016, 2018]

PYQ — 2016, 2018 (09/10 marks)

A negative point charge of magnitude $2\mu$C is situated in air at the origin, and two positive point charges of $1\mu$C each are at $y=\pm 2$ m. Calculate the electric field strength and electric potential at a point 4 m from the origin on the x-axis.

Solution

Setup: $q_1=-2\mu$C at $(0,0,0)$; $q_2=+1\mu$C at $(0,2,0)$; $q_3=+1\mu$C at $(0,-2,0)$. Observation point $P(4,0,0)$.

Distances: $R_1=4\text{m},R_2=R_3=\sqrt{4^2+2^2}=\sqrt{20}=2\sqrt{5}\approx 4.472\text{m}$

Part (a) — Potential (scalar, so just add)

$V=\frac{1}{4\pi {ϵ}_0}\left(\frac{q_1}{R_1}+\frac{q_2}{R_2}+\frac{q_3}{R_3}\right)=9\times 10^9\left(\frac{-2\times 10^{-6}}{4}+\frac{2\times 10^{-6}}{2\sqrt{5}}\right)$ $=9\times 10^3\left(-0.5+\frac{1}{\sqrt{5}}\right)=9\times 10^3(-0.5+0.4472)=9\times 10^3(-0.0528)$ $\boxed{V=-475.2\text{V}}$

Part (b) — Field (vector — use symmetry)

By symmetry, the $y$-components from $q_2$ and $q_3$ cancel exactly; only $x$-components survive.

From $q_1$ (negative, so the field points toward it, i.e. $-{\hat{a}}_x$): $E_1=\frac{9\times 10^9\times 2\times 10^{-6}}{4^2}=1125\text{V/m}⟹{\vec{E}}_1=-{\hat{a}}_{x}1125$

From $q_2$ and $q_3$ (each at distance $\sqrt{20}$, with $\cos \theta =4/\sqrt{20}$): $E_2=E_3=\frac{9\times 10^9\times 1\times 10^{-6}}{20}=450\text{V/m}$ $E_{2x}=E_{3x}=450\times \frac{4}{\sqrt{20}}=450\times 0.8944=402.5\text{V/m}$

Wait — apply the cosine to the correct component. The $x$-component of each is $E\cos \theta$ where $\cos \theta =4/4.472=0.8944$: $E_{2x}=E_{3x}=450\times 0.8944=402.5\text{V/m}(+{\hat{a}}_x\text{ direction, since charges are positive})$

Sum: $\vec{E}={\hat{a}}_x(-1125+402.5+402.5)=-{\hat{a}}_{x}320\text{V/m}$

$\boxed{\vec{E}\approx -{\hat{a}}_{x}320\text{V/m}\text{(directed toward the origin)}}$

Note on a discrepancy in older notes

Some circulating versions of this solution report $-1044.5$ V/m, obtained by mistakenly dividing by $\sqrt{20}$ twice (once in $E=kq/R^2$ with $R^2=20$, and again when resolving the component). The correct $x$-component is $E\cos \theta =450\times (4/\sqrt{20})=402.5$ V/m, giving a net $-320$ V/m. Verify this yourself in the exam by checking that $|{\vec{E}}_2|=450$ V/m and that its $x$-component must be less than 450.

---

6. Solved Numerical 5 — Potential at the Centre of a Rectangle [PYQ: 2020]

PYQ — 2020 (10 marks)

What is the potential at the centre of a rectangle whose sides are $a=2.0$ m and $b=1.0$ m, with charges $q_1=+1.0\times 10^{-8}$ C, $q_2=-2.0\times 10^{-8}$ C, $q_3=+3\times 10^{-8}$ C and $q_4=+2.0\times 10^{-8}$ C at the corners?

Solution

Step 1 — All four corners are equidistant from the centre. The half-diagonal is: $R=\frac{1}{2}\sqrt{a^2+b^2}=\frac{1}{2}\sqrt{4+1}=\frac{\sqrt{5}}{2}=1.118\text{m}$

Step 2 — Potential is a scalar, so with a common $R$ it factors out: $V=\frac{1}{4\pi {ϵ}_{0}R}{\sum }_{i=1}^4{q}_i$

Step 3 — Sum the charges: $\sum q_i=(1-2+3+2)\times 10^{-8}=4\times 10^{-8}\text{C}$

Step 4 — Evaluate: $V=\frac{9\times 10^9\times 4\times 10^{-8}}{1.118}=\frac{360}{1.118}=322.0\text{V}$

$\boxed{V\approx 322\text{V}}$

Why this one is nearly free marks

Because all four charges sit at the same distance from the centre, you never need geometry beyond the half-diagonal, and you never need vectors — potential is scalar. Spot this and the question takes ninety seconds. (Note: the electric field at the centre would require full vector addition and is a much harder question — read carefully which one is asked.)

---

7. Solved Numerical 6 — $E$ and $V$ vs. Radius for a Spherical Shell [PYQ: 2025]

PYQ — 2025 (12 marks)

A positive point charge $Q$ is at the centre of a spherical conducting shell of inner radius $R_i$ and outer radius $R_o$. Illustrate the variation of electric field intensity $\vec{E}$ and electric potential $V$ as a function of radial distance $R$.

Solution — four regions

The shell is a conductor, so induced charge $-Q$ appears on its inner surface and $+Q$ on its outer surface (total charge on the shell is zero).

Region	$Q_{\text{enc}}$	$E(R)$	$V(R)$
$0<R<R_i$	$Q$	$\frac{Q}{4\pi {ϵ}_0{R}^2}$	$\frac{Q}{4\pi {ϵ}_0}\left(\frac{1}{R}-\frac{1}{R_i}+\frac{1}{R_o}\right)$
$R_i<R<R_o$ (inside conductor)	$Q+(-Q)=0$	$0$	$\frac{Q}{4\pi {ϵ}_0{R}_o}$ (constant)
$R>R_o$	$Q$	$\frac{Q}{4\pi {ϵ}_0{R}^2}$	$\frac{Q}{4\pi {ϵ}_{0}R}$

Key features to mark on the sketch:

1. $E\propto 1/R^2$ for $R<R_i$, rising steeply as $R\to 0$.
2. $E$ drops discontinuously to zero at $R=R_i$ and stays zero throughout the conductor.
3. $E$ jumps back to $Q/4\pi {ϵ}_0{R}_o^2$ at $R=R_o$ and resumes its $1/R^2$ decay.
4. $V$ is continuous everywhere — it never jumps, even where $E$ does.
5. $V$ is flat (constant) across the conductor, since $E=0$ there — the shell is an equipotential body.
6. $V\propto 1/R$ outside and rises as $1/R$ inside the cavity.

[GRAPH: Two stacked plots against $R$. Top — $E$ vs $R$: a $1/R^2$ curve from near the origin dropping to $Q/4\pi {ϵ}_0{R}_i^2$ at $R_i$, then a vertical drop to zero, a flat zero segment from $R_i$ to $R_o$, a vertical jump up to $Q/4\pi {ϵ}_0{R}_o^2$ at $R_o$, then a $1/R^2$ decay. Bottom — $V$ vs $R$: a smooth $1/R$-type curve falling from the centre, flattening into a horizontal plateau at value $Q/4\pi {ϵ}_0{R}_o$ between $R_i$ and $R_o$, then resuming a $1/R$ decay beyond $R_o$. Governing equations as tabulated above.]

The two facts the examiner is checking

$E$ can be discontinuous (it jumps at a surface charge), but $V$ is always continuous (a jump in $V$ would imply infinite $E$). And $V$ being constant inside the conductor is the direct consequence of $E=0$ there.

---

8. PYQ Coverage for This Note

Question (verbatim, condensed)	Marks	Year(s)
What is electrostatic energy? Derive an equation for electrostatic energy to assemble $k$ charges one by one.	12	2018, 2022
Define electrostatic energy. Determine the electrostatic energy for assembling $k$ charges one by one.	12	2023
Define electrostatic energy. Derive the expression of electrostatic energy for a system of direct charges.	12	2024
Define electrostatic potential energy. Derive the expression for the work done in assembling $n$ point charges one after another from infinity.	13	2025
What energy is stored in the field with point charges $1,-2,-3,4\mu$C on the x-axis at $x=1,2,3,4$ m?	08/10	2019, 2023
Calculate the energy stored with charges $-2,1,3,-4\mu$C at $x=1,2,3,4$ m.	10	2020
Determine the work done in carrying a $-2\mu$C charge from $P_1(2,1,-1)$ to $P_2(8,2,-1)$ in $\vec{E}={\hat{a}}_{x}y+{\hat{a}}_{y}x$.	12	2022, 2025
Spherical distribution ${\rho }_v=2r$ nC/m³ — calculate $\vec{E}$ at $r=2$ m and $r=12$ m.	10	2021
Negative $2\mu$C at origin, two $+1\mu$C at $y=\pm 2$ — calculate $\vec{E}$ and $V$ at $x=4$ m.	09/10	2016, 2018
Positive $4\mu$C at the origin — calculate $\vec{E}$ at $z=8$ m on the z-axis.	07	2017
Potential at the centre of a rectangle $a=2.0$, $b=1.0$ with four corner charges.	10	2020
Positive $Q$ at the centre of a spherical conducting shell — illustrate $\vec{E}$ and $V$ versus $R$.	12	2025

The 2017 one-liner

Positive point charge $4\mu$C at the origin in air; find $\vec{E}$ at $z=8$ m. $E=\frac{9\times 10^9\times 4\times 10^{-6}}{8^2}=\frac{36\times 10^3}{64}=562.5\text{V/m}⟹\vec{E}={\hat{a}}_{z}562.5\text{V/m}$

---

9. Exam Hacks & Traps

Key Exam Checkpoints

1. Always explain the $\frac{1}{2}$ in $W_e=\frac{1}{2}\sum q_i{V}_i$ — double-counting of pairs. It is an explicit mark.
2. $V_i$ excludes $q_i$‘s own contribution. Write $j\ne i$.
3. For numericals, use the pair form ${\sum }_{i<j}$ — six terms instead of twelve, far fewer sign errors.
4. Interpret a negative $W_e$: the configuration is bound; external work is needed to disassemble it.
5. $W=-q\int \vec{E}\cdot d\vec{l}$ for external work. Keep the leading minus. A negative charge in a positive potential difference gives positive external work.
6. Check for conservative fields. If $\nabla \times \vec{E}=0$, use endpoint values and say so — it saves several minutes.
7. Non-uniform ${\rho }_v$ stays inside the integral. The 2021 question says “uniform” but gives ${\rho }_v=2r$; trust the equation, not the adjective.
8. Potential is scalar, field is vector. Never vector-add potentials; never scalar-add fields. Exploit symmetry to cancel components before computing.
9. $E$ can jump, $V$ cannot. Essential for the 2025 shell sketch.
10. Use $1/4\pi {ϵ}_0=9\times 10^9$ — memorised, it saves time and avoids calculator slips.

---

10. Self-Check

1. Derive $W_e=\frac{1}{2}\sum q_i{V}_i$ and justify the factor $\frac{1}{2}$.
2. Write the two forms of energy density and the three capacitor-energy formulas.
3. Compute the energy of $1,-2,-3,4\mu$C at $x=1,2,3,4$ m. (≈ $-0.110$ J)
4. Why is the work integral for $\vec{E}={\hat{a}}_{x}y+{\hat{a}}_{y}x$ path-independent?
5. In the 2021 cloud problem, why can’t you write $Q={\rho }_v\cdot \frac{4}{3}\pi r^3$?
6. Sketch $E$ and $V$ versus $R$ for a point charge inside a conducting shell, and state which one is continuous.

---

Back to: 02 Chapter Map - Electrostatics & Boundary Conditions | Next chapter: 3.01 Fundamental Postulates of Magnetostatics & Lorentz Force Equation

---

📄 Section: 00 Chapter 2 Active-Recall Diagnostic Quiz

00 Chapter 2 Active-Recall Diagnostic Quiz (Electrostatics)

Overview: Test your conceptual understanding and mathematical recall of electrostatic postulates, Gauss’s law applications, electric dipole derivations, dielectric polarization, boundary conditions, and Poisson/Laplace equations.

Question 1: Electrostatic Fundamental Postulates

State the differential and integral forms of the two fundamental postulates of electrostatics in free space.

Solution:

1. Gauss’s Postulate: ∇ · D = ρv ⟺ ∮S D · ds = Qenclosed
2. Conservative Postulate: ∇ × E = 0 ⟺ ∮C E · dl = 0

Question 2: Field Inside a Spherical Charge Cloud

What is the electric field intensity E at radius r inside a uniformly charged spherical cloud of radius a and volume charge density ρv (r < a)?

Solution:

Ein = âr (ρv r) / (3ϵ0)

The field grows linearly with radius r from 0 at the center to maximum (ρv a) / (3ϵ0) at r=a.

Question 3: Electric Dipole Potential Formula

Write the expression for the electric potential V at a distant point P(r, θ) due to an electric dipole with dipole moment p = qd.

Solution:

V = (p cosθ) / (4πϵ0 r²) = (p · âr) / (4πϵ0 r²)

Potential decays quadratically (1/r²) with distance.

Question 4: Polarization Charge Densities

Define surface bound charge density ρps and volume bound charge density ρp in terms of the polarization vector P.

Solution:

• Surface Bound Charge Density: ρps = P · ân
• Volume Bound Charge Density: ρp = -∇ · P

Question 5: Total Flux Density in Dielectrics

Write the constitutive relation relating total electric flux density D, electric field E, and polarization P.

Solution:

D = ϵ0 E + P

Question 6: Tangential Boundary Condition

State the boundary condition for the tangential component of electric field intensity E across a dielectric interface.

Solution:

E1t = E2t (or D1t / ϵ1 = D2t / ϵ2)

The tangential component of E is continuous across the boundary.

Question 7: Normal Boundary Condition

State the boundary condition for the normal component of electric flux density D across a dielectric interface with surface charge density ρs.

Solution:

D1n - D2n = ρs (or ϵ1 E1n - ϵ2 E2n = ρs)

If no free surface charge exists (ρs = 0), D1n = D2n (normal D is continuous).

Question 8: Dielectric Refraction Law

State the law of refraction for electric field lines passing across a charge-free boundary between medium 1 (ϵ1) and medium 2 (ϵ2).

Solution:

tanα1 / tanα2 = ϵ1 / ϵ2

Where α1, α2 are the angles the field lines make with the normal to the interface.

Question 9: Poisson’s and Laplace’s Equations

Write Poisson’s equation and Laplace’s equation for electric potential V.

Solution:

• Poisson’s Equation: ∇² V = -ρv / ϵ
• Laplace’s Equation: ∇² V = 0 (for charge-free region ρv = 0)

Question 10: Electrostatic Energy Density

Write the formula for the electrostatic energy density we stored in an electric field.

Solution:

we = 1/2 ϵ E² = 1/2 D · E [J/m³]