Exhaustive Textbook Study Guide: Cheng Examples 3-5, 3-6, and 3-7

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Study Guide Overview

In David K. Cheng’s Field and Wave Electromagnetics (Section 3-4 and 3-5), the transition from vector point postulates to practical field configurations is built through key textbook examples.

To ensure absolute alignment with both Cheng’s official textbook numbering and course syllabus checklists, this guide provides a deep, mathematically complete, and highly scannable breakdown of three prime exam-tested configurations:

1. Cheng Example 3-5 (Planar Sheet): Field of an Infinite Surface Charge Sheet via Gauss’s Law. [PYQ: 2015, 2023]
2. Cheng Example 3-6 (Spherical Cloud): Field of a Symmetrical Spherical Electron Cloud. [PYQ: 2018, 2019, 2020, 2022]
3. Cheng Example 3-7 (Dipole streamlines): Equipotential and Electric Field Lines of an Electric Dipole. [PYQ: 2016, 2023]

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1. Cheng Example 3-5: Infinite Planar Charge Sheet [PYQ: 2015, 2023]

Verbatim Question

Determine the electric field intensity $\vec{E}$ caused by an infinite planar sheet lying in the $xy$-plane with a uniform surface charge density ${\rho }_s$ in free space.

                     +z  ^
                         |      Curved/Side Walls (dS_side is perpendicular to E)
                         |      ==> E . dS_side = 0 (No Flux Leakage)
                    .----+----.
                   /   | | |   \  <-- Top Face (Area A)
                  /    | | |    \     dS_top = +a_z ds  ==> E . dS = E_z A
                 |==== v v v ====|
                 |               |
   --------------|-------o-------|---------------> +y
  /              |               |              /   <-- Infinite Sheet (z = 0)
 /               |==== ^ ^ ^ ====|             /        Surface Charge Density ρ_s
                 |    | | |    |
                  \    | | |    / <-- Bottom Face (Area A)
                   \   | | |   /      dS_bottom = -a_z ds ==> E . dS = E_z A
                    '----+----'
                         |
                         | -z

🧠 The Intuitive Concept

An infinite planar sheet of charge has infinite planar symmetry. Because the sheet is infinite in the $x$ and $y$ directions, the electric field lines cannot have any components parallel to the sheet (any tangential components from individual charge elements cancel out perfectly due to symmetry).

Thus, the electric field $\vec{E}$ must point strictly normal to the sheet (along $+z$ above the sheet, and along $-z$ below it).

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📐 Step-by-Step Gauss’s Law Derivation

Step 1: Construct the Gaussian Surface

Choose a rectangular “pillbox” or a closed cylinder of cross-sectional area $A$ oriented perpendicular to the sheet, extending symmetrically from $-z$ to $+z$.

Step 2: Set Up the Closed Surface Integral

The closed surface integral consists of three distinct faces: the top face ($S_{\text{top}}$), the bottom face ($S_{\text{bottom}}$), and the side walls ($S_{\text{side}}$):

$${\oint }_S\vec{E}\cdot d\vec{s}={\int }_{S_{\text{top}}}\vec{E}\cdot d\vec{s}+{\int }_{S_{\text{bottom}}}\vec{E}\cdot d\vec{s}+{\int }_{S_{\text{side}}}\vec{E}\cdot d\vec{s}$$

• For the Side Walls ($S_{\text{side}}$): The normal vector $d\vec{s}$ points outward in the radial direction (${\hat{a}}_r$), which is perpendicular to the electric field $\vec{E}=\pm {\hat{a}}_z{E}_z$. $$\vec{E}\cdot d{\vec{s}}_{\text{side}}=0⟹{\int }_{S_{\text{side}}}\vec{E}\cdot d\vec{s}=0$$
• For the Top Face ($S_{\text{top}}$ at $+z$): Here, $\vec{E}={\hat{a}}_z{E}_z$ and $d\vec{s}={\hat{a}}_{z}ds$. $${\int }_{S_{\text{top}}}\vec{E}\cdot d\vec{s}={\int }_{S_{\text{top}}}(E_z{\hat{a}}_z)\cdot ({\hat{a}}_{z}ds)=E_{z}A$$
• For the Bottom Face ($S_{\text{bottom}}$ at $-z$): Here, $\vec{E}=-{\hat{a}}_z{E}_z$ and $d\vec{s}=-{\hat{a}}_{z}ds$. $${\int }_{S_{\text{bottom}}}\vec{E}\cdot d\vec{s}={\int }_{S_{\text{bottom}}}(-E_z{\hat{a}}_z)\cdot (-{\hat{a}}_{z}ds)=E_{z}A$$

Adding the active fluxes together:

$${\oint }_S\vec{E}\cdot d\vec{s}=E_{z}A+E_{z}A=2E_{z}A$$

Step 3: Apply Gauss’s Law

The net charge enclosed by the Gaussian surface is simply the surface area of the sheet captured inside the box multiplied by the surface charge density:

$$Q_{\text{encl}}={\rho }_{s}A$$

By Gauss’s Law:

$${\oint }_S\vec{E}\cdot d\vec{s}=\frac{Q_{\text{encl}}}{{ϵ}_0}⟹2E_{z}A=\frac{{\rho }_{s}A}{{ϵ}_0}$$

Dividing out the arbitrary area $A$:

$$E_z=\frac{{\rho }_s}{2{ϵ}_0}$$

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🏆 Final Vector Formulation

$$\vec{E}={\hat{a}}_z\frac{{\rho }_s}{2{ϵ}_0}(\text{for }z>0);\vec{E}=-{\hat{a}}_z\frac{{\rho }_s}{2{ϵ}_0}(\text{for }z<0)\text{[V/m]}$$

Exam Pitfall: The Independent Distance Behavior

Note that the final field equation does not contain distance ($z$). The electric field of an infinite planar sheet of charge is completely uniform and does not decay with distance!

Always state this physical insight explicitly to secure full credit from your examiner.

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2. Cheng Example 3-6: Spherical Electron Cloud Field

Verbatim Question

Determine the $\vec{E}$ field caused by a spherical cloud of electrons with a uniform volume charge density $\rho =-{\rho }_0$ for $0\le R\le b$ (both ${\rho }_0$ and $b$ are positive) and $\rho =0$ for $R>b$.

                                 |  E-Field [ER]
                                 |
                                 |          R = b (Surface Interface)
                      ___________|___________
                     /           |           \
                    /            |            \
                   /   Region 1: |  Region 2:  \
                  |   Inside     |   Outside    |
  ----------------|--- 0 <= R <= b -- R > b ----|------------------> Radial R
                  |              |              |
                   \  E ~ R      |   E ~ 1/R^2 /
                    \ (Linear)   |   (Point)  /
                     \___________|___________/
                                 |
                                 |

📐 Step-by-Step Derivation

Due to perfect spherical symmetry, we construct concentric spherical Gaussian surfaces of radius $R$ centered at the origin. On these surfaces, the electric field is strictly radial and constant in magnitude:

$$\vec{E}={\hat{a}}_R{E}_R\text{and}d\vec{s}={\hat{a}}_{R}ds⟹{\oint }_S\vec{E}\cdot d\vec{s}=E_R(4\pi R^2)$$

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🔴 Case A: Inside the Cloud ($0\le R\le b$)

Step 1: Construct the Internal Gaussian Surface

Construct a hypothetical spherical surface $S_i$ with radius $R<b$.

Step 2: Compute Enclosed Volume Charge

Since the charge is uniformly distributed throughout the volume of the cloud, the charge enclosed within our Gaussian radius $R$ is:

$$Q_{\text{encl}}={\int }_V\rho dv={\int }_0^R(-{\rho }_0)(4\pi R^{\prime 2})d{R}^{\prime }=-{\rho }_0\left(\frac{4\pi }{3}{R}^3\right)$$

Step 3: Apply Gauss’s Law

$${\oint }_{S_i}\vec{E}\cdot d\vec{s}=\frac{Q_{\text{encl}}}{{ϵ}_0}⟹E_R(4\pi R^2)=-\frac{{\rho }_0}{{ϵ}_0}\left(\frac{4\pi }{3}{R}^3\right)$$

Dividing both sides by $4\pi R^2$:

$$E_R=-\frac{{\rho }_{0}R}{3{ϵ}_0}$$

Vector Form:

$${\vec{E}}_{\text{in}}=-{\hat{a}}_R\frac{{\rho }_{0}R}{3{ϵ}_0}\text{[V/m]}$$

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🔵 Case B: Outside the Cloud ($R>b$)

Step 1: Construct the External Gaussian Surface

Construct a spherical surface $S_o$ with radius $R>b$.

Step 2: Compute Total Enclosed Charge

Since the Gaussian surface completely encloses the entire electron cloud, the total enclosed charge is constant and equal to the total charge of the cloud:

$$Q_{\text{total}}={\int }_0^b(-{\rho }_0)(4\pi R^{\prime 2})d{R}^{\prime }=-{\rho }_0\left(\frac{4\pi }{3}{b}^3\right)$$

Step 3: Apply Gauss’s Law

$$E_R(4\pi R^2)=\frac{Q_{\text{total}}}{{ϵ}_0}⟹E_R(4\pi R^2)=-\frac{{\rho }_0{b}^3}{3{ϵ}_0}$$

Dividing both sides by $4\pi R^2$:

$$E_R=-\frac{{\rho }_0{b}^3}{3{ϵ}_0{R}^2}$$

Vector Form:

$${\vec{E}}_{\text{out}}=-{\hat{a}}_R\frac{{\rho }_0{b}^3}{3{ϵ}_0{R}^2}\text{[V/m]}$$

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📈 Visualizing the Field Profile

   Electric Field E_R
         ^
         |  
         |  
   ------o-----------------------------------------> Radius R
         | \                   . ' ` .
         |  \              . '         ` . <-- Quadratic Decay (~ 1/R^2)
         |   \         . '
         |    \    . '
  -E_max |-----o  <-- Peak Field at R = b
         |     |

Mathematical Proof: Maximum Field Strength is at the Surface ( $R=b$)

• For the region inside ($R\le b$), the field magnitude is $E_{\text{in}}(R)=\frac{{\rho }_{0}R}{3{ϵ}_0}$. Taking the derivative with respect to $R$: $$\frac{d{E}_{\text{in}}}{dR}=\frac{{\rho }_0}{3{ϵ}_0}>0⟹\text{Field grows monotonically with }R.$$
• For the region outside ($R>b$), the field magnitude is $E_{\text{out}}(R)=\frac{{\rho }_0{b}^3}{3{ϵ}_0{R}^2}$. Taking the derivative with respect to $R$: $$\frac{d{E}_{\text{out}}}{dR}=-\frac{2{\rho }_0{b}^3}{3{ϵ}_0{R}^3}<0⟹\text{Field decays monotonically with }R.$$

Because the field increases continuously up to $R=b$ and decreases continuously thereafter, the absolute maximum field intensity must lie strictly on the boundary interface ($R=b$):

$$\boxed{|\vec{E}{|}_{\text{max}}=\frac{{\rho }_{0}b}{3{ϵ}_0}\text{[V/m]}}$$

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3. Cheng Example 3-7: Equipotential & Streamlines of a Dipole [PYQ: 2016, 2023]

Verbatim Question

Derive the equations for the equipotential surfaces and the electric field lines (streamlines) of an electric dipole centered at the origin along the $z$-axis. Sketch the 2D field profile.

                            ^ Z-axis (Dipole Axis)
                            |
                         +q o  (z = d/2)
                         .  |  .
                       .    |    .
                      .     |     .  <-- Streamlines (Solid loops)
                     .      |      .     R = C_c sin^2(θ)
                     |      |      |
  -------------------x------o------x-------------------> Orthogonal Equipotentials (Dashed)
                     |      |      |                     R = C_v sqrt(cos(θ))
                     .      |      .
                      .     |     .
                       .    |    .
                         -q o  (z = -d/2)
                            |

📐 Step-by-Step Derivation of Field Profiles

Part A: Equipotential Surfaces

The electrostatic potential $V$ at a far-field point $P(R,\theta )$ due to an electric dipole centered at the origin is given by:

$$V=\frac{p\cos \theta }{4\pi {ϵ}_0{R}^2}$$

To find the equation of an equipotential surface, we set $V$ equal to a constant $c_V$:

$$\frac{p\cos \theta }{4\pi {ϵ}_0{R}^2}=c_V⟹R^2=\frac{p}{4\pi {ϵ}_0{c}_V}\cos \theta$$

Taking the square root of both sides:

$$\boxed{R=c_V^{\prime }\sqrt{\cos \theta }\text{where }{c}_V^{\prime }=\sqrt{\frac{p}{4\pi {ϵ}_0{c}_V}}}$$

• Physical Significance: For any positive constant $c_V$, $R$ reaches its maximum value directly along the dipole axis ($\theta =0^{\circ }⟹\cos \theta =1$) and shrinks to zero at the equatorial plane ($\theta =90^{\circ }⟹\cos \theta =0$). This forms symmetrical lobe structures above and below the horizontal plane.

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Part B: Electric Field Lines (Streamlines)

By definition, a streamline is a path whose tangent at any point matches the direction of the electric field vector $\vec{E}$ at that point. In spherical coordinates, the differential displacement vector $d\vec{l}={\hat{a}}_{R}dR+{\hat{a}}_{\theta }Rd\theta +{\hat{a}}_{\varphi }R\sin \theta d\varphi$ must be parallel to $\vec{E}={\hat{a}}_R{E}_R+{\hat{a}}_{\theta }{E}_{\theta }$.

Thus, the components must satisfy the proportion:

$$\frac{dR}{E_R}=\frac{Rd\theta }{E_{\theta }}=\frac{R\sin \theta d\varphi }{E_{\varphi }}$$

Since $E_{\varphi }=0$ for a symmetrical dipole, $d\varphi =0$ (all streamlines lie in flat planes of constant azimuth $\varphi$). This reduces our streamline differential equation to:

$$\frac{dR}{E_R}=\frac{Rd\theta }{E_{\theta }}$$

Step 1: Substitute the Dipole Field Components

The electric field components of a dipole are:

$$E_R=\frac{p\cos \theta }{2\pi {ϵ}_0{R}^3}\text{and}{E}_{\theta }=\frac{p\sin \theta }{4\pi {ϵ}_0{R}^3}$$

Substituting these into the proportional relationship:

$$\frac{dR}{\frac{p\cos \theta }{2\pi {ϵ}_0{R}^3}}=\frac{Rd\theta }{\frac{p\sin \theta }{4\pi {ϵ}_0{R}^3}}⟹\frac{dR}{2\cos \theta }=\frac{Rd\theta }{\sin \theta }$$

Step 2: Separate Variables

$$\frac{1}{R}dR=2\frac{\cos \theta }{\sin \theta }d\theta$$

Step 3: Integrate Both Sides

$$\int \frac{1}{R}dR=2\int \frac{\cos \theta }{\sin \theta }d\theta$$ $$\ln R=2\ln (\sin \theta )+\ln c_C$$ $$\ln R=\ln ({\sin }^2\theta )+\ln c_C⟹\ln R=\ln (c_C{\sin }^2\theta )$$

Taking the exponential of both sides yields the elegant streamline equation:

$$\boxed{R=c_C{\sin }^2\theta }$$

• Physical Significance: The integration constant $c_C$ represents the maximum radial distance of a streamline loop, which occurs exactly at the equatorial plane ($\theta =90^{\circ }⟹{\sin }^2\theta =1$).

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📊 Summary Comparison of Static Geometries

Configuration	Charge Source	Gaussian Surface Shape	E-Field Mathematical Decay
Infinite Planar Sheet	Surface Charge ${\rho }_s$	Rectangular Pillbox / Box	Independent of distance ($E\propto \text{const}$)
Spherical Cloud (Inside)	Volume Charge $-{\rho }_0$	Spherical Shell ($R<b$)	Linear growth with radius ($E\propto R$)
Spherical Cloud (Outside)	Total Volume Charge $Q$	Spherical Shell ($R>b$)	Inverse-Square Decay ($E\propto 1/R^2$)
Electric Dipole (Far-field)	Symmetrical Charge Pair	N/A (Derived via gradient $-\nabla V$)	Inverse-Cube Decay ($E\propto 1/R^3$)

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💡 Active-Recall Study Prompts

1. Why does the Gaussian surface of an infinite planar sheet produce zero flux through its side walls?
2. At what specific radial point inside a spherical electron cloud is the electric field zero, and why?
3. What is the physical meaning of the constant $c_C$ in the dipole streamline equation $R=c_C{\sin }^2\theta$?