02 electrostatics_master_notes

Topic 2: Electrostatics — Complete A+ Study Notes

This module is the cornerstone of the first half of ECE 2105: Electromagnetic Fields and Waves. It bridges the gap between basic Coulomb interaction and advanced boundary value problems using vector calculus. These notes merge the official syllabus, Mashuk Sir’s Class Test topics, David K. Cheng’s textbook, and topper notes (01 Static Electric Field.pdf, 02 Solution of Electrostatic.pdf).

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1. Fundamental Postulates of Electrostatics

The behavior of electrostatic fields in free space is completely defined by two fundamental postulates, written in both differential (point) and integral forms.

A. The Postulates in Free Space

Point / Differential Form	Integral Form	Physical Law / Principle
$\nabla \cdot \vec{D}={\rho }_v$	${\oint }_S\vec{D}\cdot d\vec{s}=Q_{enclosed}$	Gauss’s Law (Electric flux has sources in charges)
$\nabla \times \vec{E}=0$	${\oint }_C\vec{E}\cdot d\vec{l}=0$	Conservative Nature of E-Field (Kirchhoff’s Voltage Law)

Where:

• $\vec{E}$ = Electric field intensity ($\text{V/m}$)
• $\vec{D}={ϵ}_0\vec{E}$ = Electric flux density (${\text{C/m}}^2$)
• ${\rho }_v=\frac{dQ}{dv}$ = Volume charge density (${\text{C/m}}^3$)
• $Q_{enclosed}$ = Total charge enclosed by surface $S$ ($\text{C}$)

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B. Mathematical Derivations (Differential $\leftrightarrow$ Integral)

Derivation 1: Divergence Postulate $\to$ Gauss’s Law

1. Start with the differential postulate: $\nabla \cdot \vec{D}={\rho }_v$
2. Integrate both sides over an arbitrary volume $V$: ${\int }_V(\nabla \cdot \vec{D})dv={\int }_V{\rho }_{v}dv$
3. Apply the Divergence Theorem to the left-hand side to convert the volume integral of a divergence into a closed surface flux integral: ${\int }_V(\nabla \cdot \vec{D})dv={\oint }_S\vec{D}\cdot d\vec{s}$
4. Recognize that the volume integral of volume charge density represents the total enclosed charge $Q$: ${\int }_V{\rho }_{v}dv=Q_{enclosed}$
5. Equating the results yields the integral form: ${\oint }_S\vec{D}\cdot d\vec{s}=Q_{enclosed}$

Derivation 2: Curl Postulate $\to$ Conservative Path Postulate (KVL)

1. Start with the differential postulate: $\nabla \times \vec{E}=0$
2. Integrate both sides over an arbitrary open surface $S$ bounded by a closed contour $C$: ${\int }_S(\nabla \times \vec{E})\cdot d\vec{s}=0$
3. Apply Stokes’s Theorem to the left-hand side to convert the surface integral of a curl into a closed line integral: ${\int }_S(\nabla \times \vec{E})\cdot d\vec{s}={\oint }_C\vec{E}\cdot d\vec{l}$
4. This directly yields the integral form: ${\oint }_C\vec{E}\cdot d\vec{l}=0$

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C. Physical Significance of the Postulates

1. Gauss’s Law ($\nabla \cdot \vec{D}={\rho }_v$): This tells us that static electric fields are divergent and have their sources and sinks strictly at electric charges. Positive charges act as sources (flux lines diverge outward), while negative charges act as sinks (flux lines converge inward).
2. Conservative Postulate ($\nabla \times \vec{E}=0$): This means the static electric field is completely irrotational (curl-free). It forms no closed loops on its own. Physically, this means that the work done in moving a unit charge along any closed path in an electrostatic field is exactly zero. This is the field-theory equivalent of Kirchhoff’s Voltage Law (KVL).

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2. Coulomb’s Law & Continuous Charge Distributions

A. Coulomb’s Law (Point Charges)

The electrostatic force ${\vec{F}}_{12}$ exerted on a point charge $q_2$ by another point charge $q_1$ in free space is: ${\vec{F}}_{12}={\hat{a}}_{R_{12}}\frac{q_1{q}_2}{4\pi {ϵ}_0{R}_{12}^2}$

Where:

• ${\vec{R}}_{12}={\vec{R}}_2-{\vec{R}}_1$ is the vector from $q_1$ to $q_2$.
• $R_{12}=|{\vec{R}}_{12}|$ is the scalar distance.
• ${\hat{a}}_{R_{12}}=\frac{{\vec{R}}_{12}}{R_{12}}$ is the unit vector pointing from $q_1$ to $q_2$.

Consequently, the electric field intensity $\vec{E}$ generated by a single point charge $q$ is: $\vec{E}={\hat{a}}_R\frac{q}{4\pi {ϵ}_0{R}^2}$

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B. Continuous Charge Distributions

If charge is distributed continuously over a body, we replace the point charge $q$ with a differential charge element $dq$ and integrate over the geometry.

[Figure: Continuous Charge Distributions - Line (ρ_L dl'), Surface (ρ_S ds'), and Volume (ρ_v dv') source elements pointing to an observation point P. Found in David K. Cheng, Chapter 3, Figure 3-4]

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

1. Line Charge Distribution (${\rho }_L$): $d{q}^{\prime }={\rho }_{L}d{l}^{\prime }⟹\vec{E}=\frac{1}{4\pi {ϵ}_0}{\int }_{L^{\prime }}{\hat{a}}_R\frac{{\rho }_{L}d{l}^{\prime }}{R^2}(\text{Unit of }{\rho }_L:\text{C/m})$

   High-Yield Derivation: Electric Field of an Infinite Line Charge via Direct Integration Problem: Derive the electric field intensity $\vec{E}$ at a distance $r$ from an infinitely long, straight line charge of uniform density ${\rho }_L$ along the $z$-axis.

   1. Geometry and Coordinate Setup (Write in Exam):

   • Place the line charge along the $z$-axis. The observation point is $P(r,0,0)$.
   • A differential charge element $d{q}^{\prime }={\rho }_{L}d{z}^{\prime }$ is at $(0,0,z^{\prime })$.
   • Define the vectors:
     • Source position vector: ${\vec{R}}^{\prime }={\hat{a}}_z{z}^{\prime }$
     • Observation position vector: $\vec{R}={\hat{a}}_{r}r$
     • Vector distance: $\vec{R}-{\vec{R}}^{\prime }={\hat{a}}_{r}r-{\hat{a}}_z{z}^{\prime }$
     • Distance magnitude: $R=\sqrt{r^2+z^{\prime 2}}$
     • Unit vector: ${\hat{a}}_R=\frac{{\hat{a}}_{r}r-{\hat{a}}_z{z}^{\prime }}{\sqrt{r^2+z^{\prime 2}}}$

   2. Set up the Integral (Write in Exam): Using Coulomb’s Law, the differential field is: $d\vec{E}=\frac{d{q}^{\prime }}{4\pi {ϵ}_0{R}^2}{\hat{a}}_R=\frac{{\rho }_L({\hat{a}}_{r}r-{\hat{a}}_z{z}^{\prime })d{z}^{\prime }}{4\pi {ϵ}_0(r^2+z^{\prime 2}{)}^{3/2}}$ Integrate from $z^{\prime }=-\infty$ to $+\infty$: $\vec{E}=\frac{{\rho }_L}{4\pi {ϵ}_0}{\int }_{-\infty }^{\infty }\frac{{\hat{a}}_{r}r-{\hat{a}}_z{z}^{\prime }}{(r^2+z^{\prime 2}{)}^{3/2}}d{z}^{\prime }$

   3. Symmetry Cancellation (Write in Exam): Due to the symmetry of the infinite line charge along the $z$-axis, the longitudinal $z$-components of the field from symmetric charge elements cancel out completely. $E_z=-\frac{{\rho }_L}{4\pi {ϵ}_0}{\int }_{-\infty }^{\infty }\frac{z^{\prime }d{z}^{\prime }}{(r^2+z^{\prime 2}{)}^{3/2}}=0$ This leaves only the radial field component $E_r$: $E_r=\frac{{\rho }_{L}r}{4\pi {ϵ}_0}{\int }_{-\infty }^{\infty }\frac{d{z}^{\prime }}{(r^2+z^{\prime 2}{)}^{3/2}}$

   Study Detail: Mathematical Proof of Symmetry Cancellation $f(z^{\prime })=\frac{z^{\prime }}{(r^2+z^{\prime 2}{)}^{3/2}}$ is an odd function ($f(-z^{\prime })=-f(z^{\prime })$). When integrating any odd function over symmetric limits $(-\infty ,\infty )$, the areas under the curve on the positive and negative sides are equal and opposite, cancelling out to exactly zero: ${\int }_{-\infty }^{\infty }\frac{z^{\prime }d{z}^{\prime }}{(r^2+z^{\prime 2}{)}^{3/2}}=0■$

   The integrand

   4. Solve the Integral (Write in Exam): Evaluate the radial integral using standard formulas: $E_r=\frac{{\rho }_{L}r}{4\pi {ϵ}_0}{\left[\frac{z^{\prime }}{r^2\sqrt{r^2+z^{\prime 2}}}\right]}_{-\infty }^{\infty }=\frac{{\rho }_{L}r}{4\pi {ϵ}_0}\left(\frac{2}{r^2}\right)=\frac{{\rho }_L}{2\pi {ϵ}_{0}r}$ In vector form: $\vec{\mathbf{E}}={\hat{\mathbf{a}}}_{\mathbf{r}}\frac{{\rho }_{\mathbf{L}}}{2\pi {ϵ}_0\mathbf{r}}(\text{V/m})■$

   Study Detail: Step-by-Step Integration via Trigonometric Substitution $z^{\prime }=r\tan \theta$:

   If you want to solve the integral manually from scratch, use the substitution

   • $d{z}^{\prime }=r{\sec }^2\theta d\theta$
   • $(r^2+z^{\prime 2}{)}^{3/2}=(r^2{\sec }^2\theta {)}^{3/2}=r^3{\sec }^3\theta$
   • Limits: as $z^{\prime }\to \pm \infty$, $\theta \to \pm \pi /2$.

   Substitute these into the integral: ${\int }_{-\infty }^{\infty }\frac{d{z}^{\prime }}{(r^2+z^{\prime 2}{)}^{3/2}}={\int }_{-\pi /2}^{\pi /2}\frac{r{\sec }^2\theta d\theta }{r^3{\sec }^3\theta }=\frac{1}{r^2}{\int }_{-\pi /2}^{\pi /2}\frac{1}{\sec \theta }d\theta =\frac{1}{r^2}{\int }_{-\pi /2}^{\pi /2}\cos \theta d\theta =\frac{1}{r^2}[\sin \theta {]}_{-\pi /2}^{\pi /2}=\frac{2}{r^2}■$

   Physical Meaning: The electric field of an infinite line charge falls off inversely with the first power of distance ($1/r$), unlike a point charge which decays with the square of distance ($1/R^2$). This is because the charge extends infinitely in one dimension, concentrating the field lines cylindrically.

2. Surface Charge Distribution (${\rho }_S$): $d{q}^{\prime }={\rho }_{S}d{s}^{\prime }⟹\vec{E}=\frac{1}{4\pi {ϵ}_0}{\int }_{S^{\prime }}{\hat{a}}_R\frac{{\rho }_{S}d{s}^{\prime }}{R^2}(\text{Unit of }{\rho }_S:{\text{C/m}}^2)$

3. Volume Charge Distribution (${\rho }_v$): $d{q}^{\prime }={\rho }_{v}d{v}^{\prime }⟹\vec{E}=\frac{1}{4\pi {ϵ}_0}{\int }_{V^{\prime }}{\hat{a}}_R\frac{{\rho }_{v}d{v}^{\prime }}{R^2}(\text{Unit of }{\rho }_v:{\text{C/m}}^3)$

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3. Electric Potential ($V$) & Field-Potential Relations

A. Electric Potential Definition

The electric potential $V$ at an observation point is defined as the work done per unit positive charge by an external force to bring a test charge from infinity (reference of $0\text{ V}$) to that point: $V=-{\int }_{\infty }^P\vec{E}\cdot d\vec{l}$

The potential difference between two arbitrary points $P_1$ and $P_2$ is: $V_2-V_1=-{\int }_{P_1}^{P_2}\vec{E}\cdot d\vec{l}$

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B. Deriving the Relation $\vec{E}=-\nabla V$

We can derive the differential relation between the vector electric field and the scalar electric potential using a point charge:

1. The potential of a point charge at a distance $R$ is: $V=\frac{q}{4\pi {ϵ}_{0}R}$
2. Take the gradient of $V$ in spherical coordinates (since $V$ only varies with $R$): $\nabla V={\hat{a}}_R\frac{\partial V}{\partial R}={\hat{a}}_R\frac{\partial }{\partial R}\left(\frac{q}{4\pi {ϵ}_{0}R}\right)$
3. Perform the differentiation: $\nabla V={\hat{a}}_R\left(-\frac{q}{4\pi {ϵ}_0{R}^2}\right)=-\left({\hat{a}}_R\frac{q}{4\pi {ϵ}_0{R}^2}\right)$
4. Identify that the term in the brackets is the electric field intensity vector $\vec{E}$: $\nabla V=-\vec{E}⟹\vec{E}=-\nabla V$

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C. Electric Field Lines vs. Equipotential Lines

Distinguishing Properties:

1. Orthogonality: Electric field lines and equipotential lines are always strictly perpendicular to each other at every point.
2. Direction of Maximum Decrease: Electric field lines always point in the direction of the maximum rate of decrease of the potential (since $\vec{E}=-\nabla V$).
3. No Work: Moving a charge along an equipotential line requires zero work because $dV=0⟹dW=-qdV=0$.

2D Field Sketches:

[Figure: Left: Uniformly charged positive sphere showing radial electric field lines pointing outward (solid arrows) and concentric circles representing equipotential lines (dashed lines). Right: Electric dipole showing curved field lines flowing from positive to negative charge, and perpendicular oval-shaped equipotential contours. Found in David K. Cheng, Chapter 3, Figure 3-12]

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4. Gauss’s Law & Symmetric Charge Applications

Gauss’s law is highly effective for calculating the electric field of symmetric charge distributions.

A. Infinite Sheet of Charge

Consider an infinite flat plane with a uniform surface charge density ${\rho }_s$.

[Figure: Infinite sheet of charge in the z=0 plane. Gaussian surface is a cylinder of cross-sectional area A extending symmetrically to z = -h and z = +h. Found in David K. Cheng, Chapter 3, Figure 3-7]

1. Gaussian Surface: A circular cylinder of cross-sectional area $A$ oriented normal to the sheet, extending symmetrically to distance $z=\pm h$.
2. Left-hand side of Gauss’s Law (${\oint }_S\vec{D}\cdot d\vec{s}$): The field points only in the $\pm {\hat{a}}_z$ direction. No flux passes through the curved sides of the cylinder ($D_r=0$). ${\oint }_S\vec{D}\cdot d\vec{s}={\int }_{\text{top}}\vec{D}\cdot d\vec{s}+{\int }_{\text{bottom}}\vec{D}\cdot d\vec{s}=D_{z}A+D_{z}A=2D_{z}A$
3. Right-hand side ($Q_{enclosed}$): The charge enclosed inside the cylinder is: $Q={\rho }_{s}A$
4. Equating both sides: $2D_{z}A={\rho }_{s}A⟹D_z=\frac{{\rho }_s}{2}$
5. In terms of electric field intensity $\vec{E}$: $\vec{E}={\hat{a}}_n\frac{{\rho }_s}{2{ϵ}_0}$

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B. Infinitely Long Line Charge

Consider an infinitely long straight wire carrying a uniform line charge density ${\rho }_L$.

1. Gaussian Surface: A coaxial cylinder of radius $r$ and length $L$.
2. Flux Evaluation: The field is purely radial ($\vec{E}=E_r{\hat{a}}_r$). No flux crosses the flat end-caps. ${\oint }_S\vec{D}\cdot d\vec{s}=D_r(2\pi rL)$
3. Enclosed Charge: $Q={\rho }_{L}L$
4. Equating: $D_r(2\pi rL)={\rho }_{L}L⟹D_r=\frac{{\rho }_L}{2\pi r}⟹\vec{E}={\hat{a}}_r\frac{{\rho }_L}{2\pi {ϵ}_{0}r}$

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C. Uniformly Charged Spherical Cloud (The “Topper” Derivation)

A very high-yield topic. Consider a spherical cloud of radius $a$ containing a uniform volume charge density ${\rho }_v={\rho }_0$.

[Figure: Uniformly charged spherical cloud of radius a. Gaussian surface is a concentric sphere of radius R < a (internal) or R > a (external). Found in David K. Cheng, Chapter 3, Figure 3-8]

Case 1: Inside the Cloud ($R\le a$)

1. Choose a concentric Gaussian sphere of radius $R<a$.
2. Flux: $\oint \vec{D}\cdot d\vec{s}=D_R(4\pi R^2)$
3. Enclosed Charge: Only the charge contained within the radius $R$ is enclosed: $Q_{enclosed}={\int }_0^R{\rho }_{0}dv={\rho }_0\left(\frac{4}{3}\pi R^3\right)$
4. Equating both sides: $D_R(4\pi R^2)={\rho }_0\left(\frac{4}{3}\pi R^3\right)⟹D_R=\frac{{\rho }_{0}R}{3}$
5. Therefore: ${\vec{E}}_{in}={\hat{a}}_R\frac{{\rho }_{0}R}{3{ϵ}_0}(\text{Varies LINEARLY from center to surface})$

Case 2: Outside the Cloud ($R>a$)

1. Choose a concentric Gaussian sphere of radius $R>a$.
2. Flux: $\oint \vec{D}\cdot d\vec{s}=D_R(4\pi R^2)$
3. Enclosed Charge: The entire charge of the cloud is enclosed: $Q_{enclosed}={\rho }_0\left(\frac{4}{3}\pi a^3\right)$
4. Equating: $D_R(4\pi R^2)={\rho }_0\left(\frac{4}{3}\pi a^3\right)⟹D_R=\frac{{\rho }_0{a}^3}{3R^2}$
5. Therefore: ${\vec{E}}_{out}={\hat{a}}_R\frac{{\rho }_0{a}^3}{3{ϵ}_0{R}^2}(\text{Varies INVERSELY with }{R}^2\text{ outside})$

Core Proof: Maximum Strength at Surface ($R=a$)

Evaluating both equations at $R=a$: $E_{in}(a)=\frac{{\rho }_{0}a}{3{ϵ}_0}$ $E_{out}(a)=\frac{{\rho }_0{a}^3}{3{ϵ}_0{a}^2}=\frac{{\rho }_{0}a}{3{ϵ}_0}$ Since the internal field grows linearly and the external field decays quadratically, the field strength reaches its absolute maximum at the boundary of the cloud: $E_{max}=\frac{{\rho }_{0}a}{3{ϵ}_0}$

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5. The Electric Dipole

An electric dipole consists of two equal but opposite point charges $+q$ and $-q$ separated by a small distance $d$.

[Figure: Electric dipole with charges +q at (0, 0, d/2) and -q at (0, 0, -d/2). Point P is located at distance R and angle θ from the origin. Found in David K. Cheng, Chapter 3, Figure 3-10]

A. Derivation of Electric Potential $V$ at Distant Point $P(R,\theta )$

1. Let $+q$ be at $z=d/2$ and $-q$ be at $z=-d/2$. The potential at point $P$ is: $V=\frac{q}{4\pi {ϵ}_0}\left(\frac{1}{R_{+}}-\frac{1}{R_{-}}\right)$

2. Since the observation point $P$ is very far away ($R\gg d$), we write the approximate inverse distances using the binomial expansion (Write in Exam): $\frac{1}{R_{+}}\approx \frac{1}{R}\left(1+\frac{d}{2R}\cos \theta \right)$ $\frac{1}{R_{-}}\approx \frac{1}{R}\left(1-\frac{d}{2R}\cos \theta \right)$

   Study Detail: Geometric Derivation & Binomial Expansion Steps ${\left(\frac{d}{2R}\right)}^2\to 0$ terms: $R_{\pm }^2=R^2+(d/2{)}^2\mp Rd\cos \theta \approx R^2\left(1\mp \frac{d}{R}\cos \theta \right)$ Take the inverse root: $\frac{1}{R_{\pm }}=R^{-1}{\left(1\mp \frac{d}{R}\cos \theta \right)}^{-1/2}$ Apply the binomial expansion $(1+x{)}^n\approx 1+nx$ (where $n=-1/2$, $x=\mp \frac{d}{R}\cos \theta$): $\frac{1}{R_{\pm }}\approx \frac{1}{R}\left(1\pm \frac{d}{2R}\cos \theta \right)■$ Note that the standard discrete combinatorics notation $\left(\genfrac{}{}{0ex}{}{n}{k}\right)$ is invalid for fractional/negative powers; generalized Taylor series is used.

   Using the Law of Cosines and neglecting the higher order

3. Subtract the terms (Write in Exam): $\frac{1}{R_{+}}-\frac{1}{R_{-}}\approx \frac{1}{R}\left[\left(1+\frac{d}{2R}\cos \theta \right)-\left(1-\frac{d}{2R}\cos \theta \right)\right]=\frac{d\cos \theta }{R^2}$

4. Substitute back to get the final potential (Write in Exam): $V=\frac{qd\cos \theta }{4\pi {ϵ}_0{R}^2}$

5. Define the electric dipole moment vector as $\vec{p}=q\vec{d}={\hat{a}}_{z}qd$ (pointing from $-q$ to $+q$). Since the unit radial vector is ${\hat{a}}_R$ (where $\vec{d}\cdot {\hat{a}}_R=d\cos \theta$): $\mathbf{V}=\frac{\vec{\mathbf{p}}\cdot {\hat{\mathbf{a}}}_{\mathbf{R}}}{4\pi {ϵ}_0{\mathbf{R}}^2}(\text{V})■$

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B. Derivation of Electric Field Intensity $\vec{E}$

There are two ways to derive the electric field: taking the negative gradient of potential, or using vector algebra directly.

Method 1: Using the Gradient Relation (Spherical Coordinates)

Using $\vec{E}=-\nabla V$ in spherical coordinates: $\vec{E}=-\nabla V=-\left[{\hat{a}}_R\frac{\partial V}{\partial R}+{\hat{a}}_{\theta }\frac{1}{R}\frac{\partial V}{\partial \theta }+{\hat{a}}_{\varphi }\frac{1}{R\sin \theta }\frac{\partial V}{\partial \varphi }\right]$ Since $V$ is independent of $\varphi$ ($\frac{\partial V}{\partial \varphi }=0$):

1. Radial Component ($E_R$): $E_R=-\frac{\partial V}{\partial R}=-\frac{\partial }{\partial R}\left(\frac{p\cos \theta }{4\pi {ϵ}_0{R}^2}\right)=-\left(\frac{-2p\cos \theta }{4\pi {ϵ}_0{R}^3}\right)=\frac{2p\cos \theta }{4\pi {ϵ}_0{R}^3}$
2. Angular Component ($E_{\theta }$): $E_{\theta }=-\frac{1}{R}\frac{\partial V}{\partial \theta }=-\frac{1}{R}\frac{\partial }{\partial \theta }\left(\frac{p\cos \theta }{4\pi {ϵ}_0{R}^2}\right)=-\frac{1}{R}\left(\frac{-p\sin \theta }{4\pi {ϵ}_0{R}^2}\right)=\frac{p\sin \theta }{4\pi {ϵ}_0{R}^3}$
3. Combine into vector form: $\vec{\mathbf{E}}=\frac{\mathbf{p}}{4\pi {ϵ}_0{\mathbf{R}}^3}\left(2\cos \theta {\hat{\mathbf{a}}}_{\mathbf{R}}+\sin \theta {\hat{\mathbf{a}}}_{\theta }\right)(\text{V/m})■$

Method 2: Coordinate-Free Vector Derivation (Cartesian Preparation)

For a coordinate-free representation, we evaluate the distances vectorially. Let the dipole center be at the origin.

1. The field is: $\vec{E}=\frac{q}{4\pi {ϵ}_0}\left[\frac{\vec{R}-\vec{d}/2}{|\vec{R}-\vec{d}/2{|}^3}-\frac{\vec{R}+\vec{d}/2}{|\vec{R}+\vec{d}/2{|}^3}\right]$

2. For $R\gg d$, we approximate the denominators as (Write in Exam): $|\vec{R}\pm \vec{d}/2{|}^{-3}\approx R^{-3}\left(1\mp \frac{3}{2}\frac{\vec{R}\cdot \vec{d}}{R^2}\right)$

   Study Detail: Derivation of Denominator Vector Approximations $d^2$ terms: $|\vec{R}\pm \vec{d}/2{|}^2=R^2\pm \vec{R}\cdot \vec{d}+\frac{d^2}{4}\approx R^2\left(1\pm \frac{\vec{R}\cdot \vec{d}}{R^2}\right)$ Raise both sides to the power $-3/2$: $|\vec{R}\pm \vec{d}/2{|}^{-3}\approx R^{-3}{\left(1\pm \frac{\vec{R}\cdot \vec{d}}{R^2}\right)}^{-3/2}$ Apply Taylor series expansion $(1+x{)}^{-3/2}\approx 1-\frac{3}{2}x$: $|\vec{R}\pm \vec{d}/2{|}^{-3}\approx R^{-3}\left(1\mp \frac{3}{2}\frac{\vec{R}\cdot \vec{d}}{R^2}\right)■$

   Expand the dot products and ignore

3. Substitute back and expand (Write in Exam): $\vec{E}\approx \frac{q}{4\pi {ϵ}_0{R}^3}\left[\left(\vec{R}-\frac{\vec{d}}{2}\right)\left(1+\frac{3}{2}\frac{\vec{R}\cdot \vec{d}}{R^2}\right)-\left(\vec{R}+\frac{\vec{d}}{2}\right)\left(1-\frac{3}{2}\frac{\vec{R}\cdot \vec{d}}{R^2}\right)\right]$ Neglecting $d^2$ terms, this simplifies to: $\vec{\mathbf{E}}=\frac{1}{4\pi {ϵ}_0{\mathbf{R}}^3}\left[\frac{3(\vec{\mathbf{p}}\cdot \vec{\mathbf{R}})}{{\mathbf{R}}^2}\vec{\mathbf{R}}-\vec{\mathbf{p}}\right](\text{where }\vec{p}=q\vec{d})■$

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6. Electrostatic Boundary Conditions

When crossing the interface between two different media, the normal and tangential fields behave according to strict boundary conditions.

[Figure: Tangential path loop abcd across boundary of media 1 and 2 with height Δh → 0. Cylindrical pillbox normal surface across boundary with height Δh → 0. Found in David K. Cheng, Chapter 3, Figure 3-23]

A. Tangential Component ($E_{1t}=E_{2t}$)

1. Apply the conservative postulate ${\oint }_C\vec{E}\cdot d\vec{l}=0$ over a tiny rectangular loop $abcd$ of length $\Delta w$ and height $\Delta h$ crossing the interface.
2. Let the height $\Delta h\to 0$ so that the path integrals along the sides perpendicular to the interface vanish.
3. The remaining line integral is: $\oint \vec{E}\cdot d\vec{l}=E_{1t}\Delta w-E_{2t}\Delta w=0$
4. Dividing by $\Delta w$ yields: $E_{1t}=E_{2t}$

The tangential component of the electric field intensity is completely continuous across any interface.

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B. Normal Component ($D_{1n}-D_{2n}={\rho }_s$)

1. Apply Gauss’s Law ${\oint }_S\vec{D}\cdot d\vec{s}=Q_{enclosed}$ to a small cylindrical pillbox of cross-sectional area $\Delta s$ and height $\Delta h$ crossing the interface.
2. Let the height $\Delta h\to 0$ so that the flux passing through the curved sides of the cylinder vanishes.
3. The remaining flux is solely from the flat top and bottom faces: $\oint \vec{D}\cdot d\vec{s}=D_{1n}\Delta s-D_{2n}\Delta s=Q_{enclosed}$
4. The enclosed charge is the free surface charge on the interface: $Q_{enclosed}={\rho }_s\Delta s$
5. Equating and dividing by $\Delta s$: $D_{1n}-D_{2n}={\rho }_s$

If the boundary is charge-free (${\rho }_s=0$), then: $D_{1n}=D_{2n}$

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C. Conductor to Dielectric Boundary

Inside a perfect conductor, there can be no static electric field (${\vec{E}}_c=0$, ${\vec{D}}_c=0$). Setting medium 2 as the conductor yields:

1. Tangential: $E_t=0$
2. Normal: $D_n={\rho }_s⟹E_n=\frac{{\rho }_s}{ϵ}$

Static electric fields must always exit a conductor’s surface at a strictly perpendicular angle.

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D. Bending (Refraction) of Electric Field Lines

At a charge-free boundary (${\rho }_s=0$) between two dielectrics: $E_{1t}=E_{2t}⟹E_1\sin {\alpha }_1=E_2\sin {\alpha }_2$ $D_{1n}=D_{2n}⟹{ϵ}_1{E}_1\cos {\alpha }_1={ϵ}_2{E}_2\cos {\alpha }_2$

Dividing these two equations yields the Law of Refraction for Electric Fields: $\frac{\tan {\alpha }_1}{\tan {\alpha }_2}=\frac{{ϵ}_1}{{ϵ}_2}(\text{or }\frac{{ϵ}_{r1}}{{ϵ}_{r2}})$

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7. Conductors & Dielectrics in Static Fields

A. Physical Behavior of Conductors in Static Fields

Under electrostatic equilibrium, an ideal conductor (which contains an abundance of free electrons) exhibits the following five properties:

1. Zero Internal Electric Field (${\vec{E}}_{in}=0$): If an external electric field is applied, free charges inside the conductor move immediately until they redistribute on the surface and create an opposing internal field that completely cancels the external field.
2. Zero Internal Volume Charge Density (${\rho }_v=0$): By Gauss’s Law: $\nabla \cdot \vec{D}=\nabla \cdot (ϵ\vec{E})=0⟹{\rho }_v=0$ Hence, no net charge can exist inside the body of the conductor; all net charges must reside strictly on the outer surface.
3. Equipotential Volume: The potential difference between any two points $a$ and $b$ inside or on the surface of the conductor is: $V_b-V_a=-{\int }_a^b\vec{E}\cdot d\vec{l}=0⟹V_a=V_b$ Thus, the entire conductor body is at a single constant potential.
4. Perpendicular Surface E-Field: At the conductor boundary, the tangential electric field component is zero ($E_t=0$), and the normal electric field is: ${\vec{E}}_n={\hat{a}}_n\frac{{\rho }_s}{{ϵ}_0}$ Static electric field lines must always exit or enter a conductor surface perpendicularly.

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B. Dielectric Polarization & Bound Charges

Dielectric materials do not possess free electrons, but contain bound charges (electrons and nuclei) that displace slightly when subjected to an external electric field $\vec{E}$. This microscopic displacement of charge elements creates physical electric dipoles aligned with the field.

• Polarization Vector ($\vec{P}$): Defined as the net electric dipole moment per unit volume of the dielectric material, measured in ${\text{C/m}}^2$: $\stackrel{⃗}{P={\lim }_{\Delta v\to 0}\frac{{\sum }_{i=1}^{n\Delta v}{\vec{p}}_i}{\Delta v}}$ where ${\vec{p}}_i=q_i{\vec{d}}_i$ is the individual microscopic dipole moment vector.

C. High-Yield Derivation: Polarization Potential and Bound Charge Densities

Problem: Derive the mathematical expressions for the volume polarization charge density (${\rho }_p$) and surface polarization charge density (${\rho }_{ps}$) of a polarized dielectric.

1. Set up the Potential Integral (Write in Exam): An infinitesimal volume $d{v}^{\prime }$ of polarized dielectric behaves like a dipole of moment $d\vec{p}=\vec{P}d{v}^{\prime }$. Integrating the dipole potential over the dielectric volume $V^{\prime }$ yields: $V=\frac{1}{4\pi {ϵ}_0}{\int }_{V^{\prime }}\frac{\vec{P}\cdot {\hat{a}}_R}{R^2}d{v}^{\prime }$

2. Substitute Vector Gradient of $1/R$ (Write in Exam): Since $\frac{{\hat{a}}_R}{R^2}={\nabla }^{\prime }\left(\frac{1}{R}\right)$, substitute this into the integral: $V=\frac{1}{4\pi {ϵ}_0}{\int }_{V^{\prime }}\vec{P}\cdot {\nabla }^{\prime }\left(\frac{1}{R}\right)d{v}^{\prime }$ where $\nabla$ represents gradient operations with respect to source coordinates $(x^{\prime },y^{\prime },z^{\prime })$.

3. Vector Expansion using Product Rule (Write in Exam): Apply the vector identity ${\nabla }^{\prime }\cdot \left(\frac{\vec{P}}{R}\right)=\frac{{\nabla }^{\prime }\cdot \vec{P}}{R}+\vec{P}\cdot {\nabla }^{\prime }\left(\frac{1}{R}\right)$: $V=\frac{1}{4\pi {ϵ}_0}\left[{\int }_{V^{\prime }}{\nabla }^{\prime }\cdot \left(\frac{\vec{P}}{R}\right)d{v}^{\prime }-{\int }_{V^{\prime }}\frac{{\nabla }^{\prime }\cdot \vec{P}}{R}d{v}^{\prime }\right]$

4. Apply the Divergence Theorem (Write in Exam): Convert the first volume integral of divergence into a closed surface integral: $V=\frac{1}{4\pi {ϵ}_0}{\oint }_{S^{\prime }}\frac{\vec{P}\cdot {\hat{a}}_n}{R}d{S}^{\prime }+\frac{1}{4\pi {ϵ}_0}{\int }_{V^{\prime }}\frac{-\left({\nabla }^{\prime }\cdot \vec{P}\right)}{R}d{v}^{\prime }$ where ${\hat{a}}_n$ is the outward unit vector normal to surface $S^{\prime }$.

Study Detail: Product Rule & Divergence Theorem Steps

• Gradient Vector Proof: Note that $R$ is the distance from source $Q(x^{\prime },y^{\prime },z^{\prime })$ to observation $P(x,y,z)$. The gradient is taken with respect to the source coordinates, yielding the positive ${\hat{a}}_R$ direction: ${\nabla }^{\prime }(1/R)={\hat{a}}_R/R^2$.
• Product Rule: We use the vector identity $\nabla \cdot (u\vec{A})=u\nabla \cdot \vec{A}+\vec{A}\cdot \nabla u$ where $u=1/R$ and $\vec{A}=\vec{P}$.
• Divergence Theorem: The theorem states that ${\int }_V(\nabla \cdot \vec{F})dv={\oint }_S\vec{F}\cdot {\hat{a}}_{n}dS$. Here $\vec{F}=\vec{P}/R$, so the flux integral is ${\oint }_S\frac{\vec{P}\cdot {\hat{a}}_n}{R}dS$.

5. Match Bound Charge Densities (Write in Exam): Comparing this with the standard potential formulas for a surface charge density ${\rho }_{ps}$ and volume charge density ${\rho }_p$: $V=\frac{1}{4\pi {ϵ}_0}{\oint }_{S^{\prime }}\frac{{\rho }_{ps}}{R}d{S}^{\prime }+\frac{1}{4\pi {ϵ}_0}{\int }_{V^{\prime }}\frac{{\rho }_p}{R}d{v}^{\prime }$ We define:

1. Surface Bound Charge Density: ${\rho }_{\mathbf{ps}}=\vec{\mathbf{P}}\cdot {\hat{\mathbf{a}}}_{\mathbf{n}}({\text{C/m}}^2)■$
2. Volume Bound Charge Density: ${\rho }_{\mathbf{p}}=-\nabla \cdot \vec{\mathbf{P}}({\text{C/m}}^3)■$ (Note: We drop the coordinate-free prime symbol in final expressions). (Q.E.D.)

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D. Electric Flux Density ($\vec{D}$) & Material Parameters

In the presence of dielectrics, we separate charge density into free charges (${\rho }_v$) and bound polarization charges (${\rho }_p$). Applying Gauss’s Postulate: $\nabla \cdot \vec{E}=\frac{{\rho }_{total}}{{ϵ}_0}=\frac{{\rho }_v+{\rho }_p}{{ϵ}_0}$ Substitute ${\rho }_p=-\nabla \cdot \vec{P}$: $\nabla \cdot \vec{E}=\frac{{\rho }_v-\nabla \cdot \vec{P}}{{ϵ}_0}⟹\nabla \cdot ({ϵ}_0\vec{E}+\vec{P})={\rho }_v$ We define Electric Flux Density ($\vec{D}$) (also called displacement field) as: $\vec{\mathbf{D}}={ϵ}_0\vec{\mathbf{E}}+\vec{\mathbf{P}}({\text{C/m}}^2)■$ This yields the generalized Gauss’s Law: $\nabla \cdot \vec{D}={\rho }_v(\text{Differential form})$ ${\oint }_S\vec{D}\cdot d\vec{s}=Q_{free}(\text{Integral form})$

• Material Relations: For a linear and isotropic dielectric, the polarization vector $\vec{P}$ is directly proportional to electric field intensity $\vec{E}$: $\vec{P}={ϵ}_0{\chi }_e\vec{E}$ where ${\chi }_e$ is the dimensionless electric susceptibility of the medium. Substituting this: $\vec{D}={ϵ}_0\vec{E}+{ϵ}_0{\chi }_e\vec{E}={ϵ}_0(1+{\chi }_e)\vec{E}={ϵ}_0{ϵ}_r\vec{E}=ϵ\vec{E}$ where:
  • ${ϵ}_r=1+{\chi }_e=\frac{ϵ}{{ϵ}_0}$ is the relative permittivity (or dielectric constant).
  • $ϵ={ϵ}_0{ϵ}_r$ is the absolute permittivity of the medium.

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8. Poisson’s & Laplace’s Equations

A. Derivation

1. Start with the differential Gauss’s postulate in medium: $\nabla \cdot \vec{D}={\rho }_v$
2. Substitute the constitutive relation $\vec{D}=ϵ\vec{E}$: $\nabla \cdot (ϵ\vec{E})={\rho }_v$
3. For a homogeneous medium ($ϵ$ is constant): $ϵ(\nabla \cdot \vec{E})={\rho }_v⟹\nabla \cdot \vec{E}=\frac{{\rho }_v}{ϵ}$
4. Substitute the potential relation $\vec{E}=-\nabla V$: $\nabla \cdot (-\nabla V)=\frac{{\rho }_v}{ϵ}⟹{\nabla }^2\mathbf{V}=-\frac{{\rho }_{\mathbf{v}}}{ϵ}(\text{Poisson’s Equation})■$
5. In a charge-free region (${\rho }_v=0$): ${\nabla }^2\mathbf{V}=0(\text{Laplace’s Equation})■$

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B. Application 1: Capacitance of a Parallel Plate Capacitor

Using Laplace’s equation to find capacitance:

1. Assume two infinite plates located at $y=0$ (grounded, $V=0$) and $y=d$ (maintained at $V=V_0$).
2. Since potential only varies along the $y$-direction, Laplace’s equation reduces to: $\frac{d^{2}V}{d{y}^2}=0$
3. Integrate once: $\frac{dV}{dy}=C_1$
4. Integrate twice: $V(y)=C_{1}y+C_2$
5. Apply the boundary conditions:
   • At $y=0$, $V(0)=0⟹C_2=0$.
   • At $y=d$, $V(d)=V_0⟹C_{1}d=V_0⟹C_1=\frac{V_0}{d}$.
6. This yields the potential distribution: $V(y)=\frac{V_0}{d}y(\text{V})$
7. Calculate the electric field: $\vec{E}=-\nabla V=-{\hat{a}}_y\frac{\partial V}{\partial y}=-{\hat{a}}_y\frac{V_0}{d}(\text{V/m})$
8. Find the surface charge density ${\rho }_s$ on the conductor plate at $y=d$ (unit normal vector pointing into dielectric is $-{\hat{a}}_y$): ${\rho }_s=D_n=ϵE_n=ϵ\left(\frac{V_0}{d}\right)$
9. Find the total charge $Q$ on the plate of area $A$: $Q={\rho }_{s}A=\frac{ϵA{V}_0}{d}(\text{C})$
10. Determine the capacitance: $\mathbf{C}=\frac{\mathbf{Q}}{{\mathbf{V}}_0}=\frac{ϵ\mathbf{A}}{\mathbf{d}}(\text{F})■$

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C. Application 2: Capacitance of a Cylindrical Capacitor

Using Laplace’s equation in cylindrical coordinates:

1. Symmetry Setup: Consider an inner conductor of radius $a$ at potential $V=V_0$, and an outer conductor of inner radius $b$ at potential $V=0$, with coaxial length $L$ ($L\gg b$ to neglect fringing fields). The space is filled with dielectric $ϵ$.
2. Laplace’s Equation: Since potential only varies radially with Cylindrical coordinate $r$: ${\nabla }^{2}V=\frac{1}{r}\frac{d}{dr}\left(r\frac{dV}{dr}\right)=0$
3. Integrate once: $r\frac{dV}{dr}=C_1⟹\frac{dV}{dr}=\frac{C_1}{r}$
4. Integrate twice: $V(r)=C_1\ln (r)+C_2$
5. Apply the boundary conditions:
   • At $r=b$, $V(b)=0⟹C_1\ln (b)+C_2=0⟹C_2=-C_1\ln (b)$.
   • Substitute $C_2$ back: $V(r)=C_1\ln (r)-C_1\ln (b)=C_1\ln \left(\frac{r}{b}\right)$
   • At $r=a$, $V(a)=V_0$: $V_0=C_1\ln \left(\frac{a}{b}\right)=-C_1\ln \left(\frac{b}{a}\right)⟹C_1=-\frac{V_0}{\ln (b/a)}$
6. This yields the potential distribution: $V(r)=V_0\frac{\ln (b/r)}{\ln (b/a)}$
7. Calculate the electric field $\vec{E}$: $\vec{E}=-\nabla V=-{\hat{a}}_r\frac{dV}{dr}=-{\hat{a}}_r\left(-\frac{V_0}{r\ln (b/a)}\right)={\hat{a}}_r\frac{V_0}{r\ln (b/a)}$
8. Find the surface charge density ${\rho }_s$ on the inner conductor surface at $r=a$ (unit normal is $+{\hat{a}}_r$): ${\rho }_s=D_n=ϵE_n(a)=\frac{ϵV_0}{a\ln (b/a)}$
9. Find the total charge $Q$ on the inner cylinder of surface area $A=2\pi aL$: $Q={\rho }_{s}A=\left(\frac{ϵV_0}{a\ln (b/a)}\right)(2\pi aL)=\frac{2\pi ϵL{V}_0}{\ln (b/a)}$
10. Determine the capacitance: $\mathbf{C}=\frac{\mathbf{Q}}{{\mathbf{V}}_0}=\frac{2\pi ϵ\mathbf{L}}{\ln (b/a)}(\text{F})■$

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9. Electrostatic Energy

A. Assembling $k$ Point Charges

The work required to assemble $k$ point charges one by one from infinity is stored as electrostatic potential energy: $W_e=\frac{1}{2}{\sum }_{i=1}^k{q}_i{V}_i$

Where $V_i$ is the total electric potential at the position of charge $q_i$ due to all other charges: $V_i=\frac{1}{4\pi {ϵ}_0}{\sum }_{j\ne i}\frac{q_j}{R_{ij}}$

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10. Verbatim PYQ Question Bank (Topic 2)

• Q1 [PYQ 2015, 2016, 2017, 2020, 2022, 2025]: Write and explain the differential form of the fundamental postulates of electrostatics in free space. Also explain the integral form of fundamental postulates. State their physical significance. [10 Marks]
• Q2 [PYQ 2018, 2021]: State Coulomb’s law. Explain electric field intensity due to a continuous distribution of charge with (i) surface charge density and (ii) line charge density. [10 Marks]
• Q3 [PYQ 2016]: Define electric field intensity and electric potential. Also derive the relation between them when caused by a point charge. [10 Marks]
• Q4 [PYQ 2017]: 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. [08 Marks]
• Q5 [PYQ 2024, 2025]: Define equipotential line. 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. [07/10 Marks]
• Q6 [PYQ 2015, 2023]: State and explain Gauss’s law. Using this law, determine the electric field intensity of an infinite sheet of charge. [11/10 Marks]
• Q7 [PYQ 2018, 2019, 2020, 2022]: Show that the electric field intensity inside a uniformly charged cloud is zero at its center, varies linearly up to the surface, and is maximum at the surface of that cloud. [10/12/20 Marks]
• Q8 [PYQ 2016, 2017, 2019, 2021, 2023, 2024]: Two equal but opposite charges separated by a distance $l$ constitute an electric dipole. Derive an expression for the electric potential $V$ (or electric field intensity $\vec{E}$) at a distant point $P$ in space due to this dipole. [12/13 Marks]
• Q9 [PYQ 2016, 2019, 2021]: 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. [07/08 Marks]
• Q10 [PYQ 2017, 2019, 2022, 2025]: Two dielectric media with permittivities ${ϵ}_1$ and ${ϵ}_2$ are separated by a charge-free boundary. The electric field intensity in medium 1 makes an angle ${\alpha }_1$ with the normal. Derive the boundary conditions and determine the magnitude and direction of the electric field intensity in medium 2. [08/09 Marks]
• Q11 [PYQ 2016, 2020]: 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. [09 Marks]
• Q12 [PYQ 2015, 2017, 2022, 2023, 2025]: Derive the Poisson’s and Laplace’s equations with respect to an electric potential. [08/11 Marks]
• Q13 [PYQ 2018, 2019, 2022, 2023, 2024]: A parallel plate capacitor consists of two plates with a separation $d$ in between. The space between the conductors is filled with a dielectric of permittivity $ϵ$. Determine the capacitance of this capacitor. [12/13 Marks]
• Q14 [PYQ 2021, 2022, 2025]: A fixed voltage $V_0$ is applied across a parallel plate capacitor separated by a distance $d$. Assuming negligible fringing effect, determine the potential at any point and the surface charge density on each of the plates. [11/12 Marks]
• Q15 [PYQ 2015, 2017]: A cylindrical capacitor consists of an inner conductor of radius $a$ and outer conductor whose inner radius is $b$. The space between the conductors is filled with a dielectric of permittivity $ϵ$, and the length of the capacitor is $L$. Determine the capacitance. [10/13 Marks]
• Q16 [PYQ 2018, 2022, 2023, 2024, 2025]: Define electrostatic potential energy. Derive the expression for the work done in assembling $n$ (or $k$) point charges one after another from infinity. [12/13 Marks]

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11. Step-by-Step Solutions to Classic Exam Numericals

Numerical 1: Dipole Axis Point Charges

Question [PYQ 2016, 2018]: A negative point charge of magnitude $2\mu \text{C}$ is situated in air at the origin and two positive point charges of $1\mu \text{C}$ each are at points $y=\pm 2\text{ meters}$. Calculate the electric field strength and electric potential at a point $4\text{ meters}$ from the origin on the $x$-axis. [10 Marks]

Solution:

Let’s define our coordinates and charge positions:

• $q_1=-2\mu \text{C}=-2\times 10^{-6}\text{ C}$ at position $(0,0,0)$.
• $q_2=+1\mu \text{C}=1\times 10^{-6}\text{ C}$ at position $(0,2,0)$.
• $q_3=+1\mu \text{C}=1\times 10^{-6}\text{ C}$ at position $(0,-2,0)$.
• Observation Point: $P(4,0,0)$.

[Figure: Charge layout on Cartesian axes: q_1=-2μC at origin, q_2=+1μC at (0,2), q_3=+1μC at (0,-2). Observation point P is on the x-axis at (4,0)]

Step 1: Calculate the Electric Potential ($V$) at $P(4,0,0)$

The potential is a scalar field, so we simply sum the potentials due to each point charge: $V=\frac{1}{4\pi {ϵ}_0}\left(\frac{q_1}{R_1}+\frac{q_2}{R_2}+\frac{q_3}{R_3}\right)$

1. Distance from $q_1(0,0,0)$ to $P(4,0,0)$: $R_1=\sqrt{(4-0{)}^2+0^2+0^2}=4\text{ m}$
2. Distance from $q_2(0,2,0)$ to $P(4,0,0)$: $R_2=\sqrt{(4-0{)}^2+(0-2{)}^2+0^2}=\sqrt{16+4}=\sqrt{20}=2\sqrt{5}\text{ m}$
3. Distance from $q_3(0,-2,0)$ to $P(4,0,0)$: $R_3=\sqrt{(4-0{)}^2+(0-(-2){)}^2+0^2}=\sqrt{16+4}=\sqrt{20}=2\sqrt{5}\text{ m}$

Substitute these distances and charges into the potential formula (using $\frac{1}{4\pi {ϵ}_0}\approx 9\times 10^9$): $V=9\times 10^9\left(\frac{-2\times 10^{-6}}{4}+\frac{1\times 10^{-6}}{2\sqrt{5}}+\frac{1\times 10^{-6}}{2\sqrt{5}}\right)$ $V=9\times 10^3\left(-0.5+\frac{2}{2\sqrt{5}}\right)=9\times 10^3\left(-0.5+\frac{1}{\sqrt{5}}\right)$ Since $\frac{1}{\sqrt{5}}\approx 0.4472$: $V=9\times 10^3\left(-0.5+0.4472\right)=9\times 10^3(-0.0528)=-475.2\text{ V}$

Answer: The electric potential at $P$ is $-475.2\text{ V}$.

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Step 2: Calculate the Electric Field Strength ($\vec{E}$) at $P(4,0,0)$

The electric field is a vector field. By symmetry, the $y$-components of the fields from the symmetric charges $q_2$ and $q_3$ will completely cancel out, leaving only a net $x$-directed field. $\vec{E}={\vec{E}}_1+{\vec{E}}_2+{\vec{E}}_3$

1. Field due to $q_1$ at origin: ${\vec{E}}_1=-{\hat{a}}_x\frac{|q_1|}{4\pi {ϵ}_0{R}_1^2}=-{\hat{a}}_x\frac{9\times 10^9\times 2\times 10^{-6}}{4^2}=-{\hat{a}}_{x}1125\text{ V/m}$
2. Fields due to $q_2$ and $q_3$: The distance to both charges is $R=\sqrt{20}$. The angle $\theta$ that the distance vectors make with the $x$-axis is: $\cos \theta =\frac{\Delta x}{R}=\frac{4}{\sqrt{20}}$ The $x$-components of their fields point in the $+{\hat{a}}_x$ direction: $E_{2x}=E_{3x}=E\cos \theta =\left(\frac{1\times 10^{-6}\times 9\times 10^9}{20}\right)\times \left(\frac{4}{\sqrt{20}}\right)=450\times \frac{4}{2.236\times 20}=450\times 0.0894=40.25\text{ V/m}$
3. Summing the components: ${\vec{E}}_{total}={\hat{a}}_x(-1125+40.25+40.25)=-{\hat{a}}_{x}1044.5\text{ V/m}$

Answer: The net electric field strength at $P$ is $-{\hat{a}}_{x}1044.5\text{ V/m}$.

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Numerical 2: Variable Density Spherical Charge Cloud

Question [PYQ 2021]: A spherical uniform charge distribution in free space has ${\rho }_v=2r{\text{ nC/m}}^3$ for $0\le r\le 10\text{m}$ and zero otherwise. Calculate $\vec{E}$ at $r=2\text{m}$ and $r=12\text{m}$. [10 Marks]

Solution:

Step 1: Calculate $\vec{E}$ at $r=2\text{m}$ (Inside the distribution, $r<10\text{m}$)

We apply Gauss’s law over a concentric sphere of radius $r=2\text{m}$: ${\oint }_S\vec{D}\cdot d\vec{s}=Q_{enclosed}$ $D_r(4\pi r^2)={\int }_0^r{\rho }_{v}d{v}^{\prime }={\int }_0^r(2r^{\prime })(4\pi r^{\prime 2}d{r}^{\prime })$ $D_r(4\pi r^2)=8\pi {\int }_0^r{r}^{\prime 3}d{r}^{\prime }=8\pi {\left[\frac{r^{\prime 4}}{4}\right]}_0^r=2\pi r^4$ Divide both sides by $4\pi r^2$: $D_r=\frac{2\pi r^4}{4\pi r^2}=\frac{r^2}{2}{\text{ nC/m}}^2$ Now, substitute $r=2\text{m}$: $D_r=\frac{2^2}{2}=2{\text{ nC/m}}^2=2\times 10^{-9}{\text{ C/m}}^2$ Convert to electric field intensity $\vec{E}$: $E_r=\frac{D_r}{{ϵ}_0}=\frac{2\times 10^{-9}}{8.854\times 10^{-12}}=225.88\text{ V/m}$

Answer at $r=2\text{m}$: $\vec{E}={\hat{a}}_{r}225.9\text{ V/m}$.

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Step 2: Calculate $\vec{E}$ at $r=12\text{m}$ (Outside the distribution, $r>10\text{m}$)

We apply Gauss’s law over a concentric sphere of radius $r=12\text{m}$. The total charge enclosed is limited to the physical radius of the cloud $a=10\text{m}$: $Q_{total}={\int }_0^{10}(2r^{\prime })(4\pi r^{\prime 2}d{r}^{\prime })=2\pi [r^{\prime 4}{]}_0^{10}=2\pi (10000)=20000\pi \text{ nC}=20\pi \times 10^{-6}\text{ C}$ Using the external point charge approximation: $D_r=\frac{Q_{total}}{4\pi r^2}=\frac{20\pi \times 10^{-6}}{4\pi (12{)}^2}=\frac{5\times 10^{-6}}{144}=3.472\times 10^{-8}{\text{ C/m}}^2$ Now, convert to electric field intensity $\vec{E}$: $E_r=\frac{D_r}{{ϵ}_0}=\frac{3.472\times 10^{-8}}{8.854\times 10^{-12}}=3921.4\text{ V/m}$

Answer at $r=12\text{m}$: $\vec{E}={\hat{a}}_{r}3921.4\text{ V/m}$.

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Numerical 3: Work Done in an Inhomogeneous Field

Question [PYQ 2022, 2025]: Determine the work done in carrying a $-2\mu \text{C}$ 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 $P_2$. [12 Marks]

Solution:

The work done by an external force to carry a charge $q$ is: $W=-q{\int }_{P_1}^{P_2}\vec{E}\cdot d\vec{l}$

Step 1: Find the Equation of the Straight Line Joining $P_1$ and $P_2$

Looking at the coordinates:

• $z$ remains constant at $-1$ ($dz=0$).
• The projection in the $xy$-plane passes through $(2,1)$ and $(8,2)$.
• Using the two-point line formula: $y-y_1=\frac{y_2-y_1}{x_2-x_1}(x-x_1)⟹y-1=\frac{2-1}{8-2}(x-2)$ $y-1=\frac{1}{6}(x-2)⟹x=6y-4⟹dx=6dy$

Step 2: Evaluate the Line Integral

The differential length element is $d\vec{l}={\hat{a}}_{x}dx+{\hat{a}}_{y}dy+{\hat{a}}_{z}dz$. The integrand becomes: $\vec{E}\cdot d\vec{l}=({\hat{a}}_{x}y+{\hat{a}}_{y}x)\cdot ({\hat{a}}_{x}dx+{\hat{a}}_{y}dy)=ydx+xdy$ Notice that $ydx+xdy$ is a perfect differential: $ydx+xdy=d(xy)$ Because the field is conservative, the integral is path-independent and depends only on the endpoints! ${\int }_{P_1}^{P_2}\vec{E}\cdot d\vec{l}={\int }_{P_1}^{P_2}d(xy)=[xy{]}_{P_1}^{P_2}$ Evaluate at the boundaries $P_1(2,1,-1)$ and $P_2(8,2,-1)$:

• At $P_1$: $xy=2\times 1=2$.
• At $P_2$: $xy=8\times 2=16$. ${\int }_{P_1}^{P_2}\vec{E}\cdot d\vec{l}=16-2=14\text{ V}$

Step 3: Calculate Total Work Done

Now, substitute the charge $q=-2\mu \text{C}=-2\times 10^{-6}\text{ C}$: $W=-q{\int }_{P_1}^{P_2}\vec{E}\cdot d\vec{l}=-(-2\times 10^{-6}\text{ C})\times (14\text{ V})=2.8\times 10^{-5}\text{ Joules}=28\mu \text{J}$

Answer: The total work done is $28\mu \text{J}$ (or $2.8\times 10^{-5}\text{ J}$).

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Numerical 4: Electrohydrodynamic Pump Potential Derivation

Question [PYQ 2024]: In an electrohydrodynamic pump, the region between two electrodes is filled with a uniform charge density ${\rho }_0$. If the left electrode has a potential of $V_0$ and the right electrode has a potential of $0\text{ V}$, determine the expressions for electric potential and electric field intensity at any point between the electrodes. [13 Marks]

Solution:

Let the left electrode be at $x=0$ ($V=V_0$) and the right electrode be at $x=d$ ($V=0$). The region is filled with uniform volume charge density ${\rho }_v={\rho }_0$.

Step 1: Solve Poisson’s Equation

Since the geometry only varies along the $1\text{D}$ coordinate $x$, we write Poisson’s equation: $\frac{d^{2}V}{d{x}^2}=-\frac{{\rho }_0}{ϵ}$ Integrate once: $\frac{dV}{dx}=-\frac{{\rho }_0}{ϵ}x+C_1$ Integrate a second time to find the potential $V(x)$: $V(x)=-\frac{{\rho }_0}{2ϵ}{x}^2+C_{1}x+C_2$

Step 2: Apply Boundary Conditions

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

Step 3: Write Final Equations

Substitute $C_1$ and $C_2$ back to get the potential distribution: $V(x)=-\frac{{\rho }_0}{2ϵ}{x}^2+\left(\frac{{\rho }_{0}d}{2ϵ}-\frac{V_0}{d}\right)x+V_0$ $V(x)=V_0\left(1-\frac{x}{d}\right)+\frac{{\rho }_0}{2ϵ}x(d-x)$

Now, take the negative derivative to find the electric field intensity ($\vec{E}=E_x{\hat{a}}_x$): $E_x=-\frac{dV}{dx}=-\left[-\frac{{\rho }_0}{ϵ}x+\frac{{\rho }_{0}d}{2ϵ}-\frac{V_0}{d}\right]$ $\vec{E}(x)={\hat{a}}_x\left[\frac{V_0}{d}+\frac{{\rho }_0}{ϵ}\left(x-\frac{d}{2}\right)\right]$

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12. Exam Hacks & Common Pitfalls

1. The “Work Done” Sign Trap:
   • The Pitfall: Forgetting the negative sign in potential difference $V_2-V_1=-\int \vec{E}\cdot d\vec{l}$, or confusing work done by the field vs. work done by an external force.
   • A+ Hack: Always write the definition of external work: $W=-q\int \vec{E}\cdot d\vec{l}$. If the final path integral is negative and the charge is negative, the resulting work is positive, meaning you must push the charge against its natural physical tendency.
2. Dipole Distant Approximation:
   • The Pitfall: Trying to compute exact values for $R_{+}$ and $R_{-}$ instead of using binomial approximations when $R\gg d$.
   • A+ Hack: State clearly: “Since the observation distance is far greater than the charge separation ($R\gg d$), we apply the binomial expansion $(1\pm x{)}^{-1}\approx 1\mp x$.” This justification secures full presentation marks.
3. Units Matter:
   • A+ Hack: Never omit units.
     • Electric Field ($\vec{E}$): $\text{V/m}$ or $\text{N/C}$
     • Electric Potential ($V$): $\text{V}$ (Volts)
     • Capacitance ($C$): $\text{F}$ (Farads)
     • Electrostatic Energy ($W_e$): $\text{J}$ (Joules)
4. Gaussian Surface Justification:
   • A+ Hack: When applying Gauss’s Law to find $\vec{E}$ fields, always state: “We choose a Gaussian surface $S$ such that $\vec{D}$ is everywhere either normal or parallel to $S$.” This mathematical justification shows the examiner that you understand why the dot product simplifies to $D\cdot A$.