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.

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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"]

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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.

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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.

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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.

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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 ${\rho }_s$.

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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.

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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.

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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.

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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

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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.

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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.

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Next: 2.02 Electric Potential, Equipotential Contours & Dipole Derivations