ECE 2105: Electrostatics Master Cheatsheet & Exam Diagnostic Guide

---

1. Core Equations Directory

Physical Quantity / Law	Point (Differential) Form	Integral Form	Units	Conditions / Remarks
Coulomb’s Law (Force)	—	${\vec{F}}_{12}={\hat{a}}_{R_{12}}\frac{q_1{q}_2}{4\pi {ϵ}_0{R}_{12}^2}$	$\text{N}$ (Newtons)	Valid for stationary point charges in free space.
Gauss’s Postulate	$\nabla \cdot \vec{D}={\rho }_v$	${\oint }_S\vec{D}\cdot d\vec{s}=Q_{free}$	${\text{C/m}}^2$ (Flux Density)	Fundamental Postulate. Relates flux density to free charge sources.
Conservative Postulate	$\nabla \times \vec{E}=0$	${\oint }_C\vec{E}\cdot d\vec{l}=0$	$\text{V/m}$ (E-Field)	Fundamental Postulate. Shows field is irrotational. Yields KVL.
Field-Potential Relation	$\vec{E}=-\nabla V$	$V=-\int \vec{E}\cdot d\vec{l}$	$\text{V}$ (Volts)	Negative sign shows field points toward decreasing potential.
Electric Flux Density	$\vec{D}={ϵ}_0\vec{E}+\vec{P}$	—	${\text{C/m}}^2$	Accounts for both free charge field and material polarization.
Linear Medium Relation	$\vec{D}=ϵ\vec{E}$	—	${\text{C/m}}^2$	Valid for linear, isotropic, homogeneous media ($ϵ={ϵ}_0{ϵ}_r$).
Polarization Bound Surface	—	${\rho }_{ps}=\vec{P}\cdot {\hat{a}}_n$	${\text{C/m}}^2$	Bound surface charge density appearing on dielectric boundaries.
Polarization Bound Volume	${\rho }_p=-\nabla \cdot \vec{P}$	—	${\text{C/m}}^3$	Bound volume charge density inside polarized dielectrics.
Poisson’s Equation	${\nabla }^{2}V=-\frac{{\rho }_v}{ϵ}$	—	${\text{V/m}}^2$	Used to find potential $V$ given a volume charge distribution.
Laplace’s Equation	${\nabla }^{2}V=0$	—	—	Poisson’s equation for charge-free regions (${\rho }_v=0$).
Parallel Plate Capacitance	—	$C=\frac{ϵA}{d}$	$\text{F}$ (Farads)	Ignores fringing fields (plates are assumed very large).
Cylindrical Capacitance	—	$C=\frac{2\pi ϵL}{\ln (b/a)}$	$\text{F}$ (Farads)	Valid for coaxial cylinders of inner radius $a$, outer radius $b$.
Electrostatic Energy	—	$W_e=\frac{1}{2}{\sum }_{i=1}^k{q}_i{V}_i$	$\text{J}$ (Joules)	Assembly energy of point charges brought from infinity.

---

2. Visual Exam Traps & Sketches

In ECE 2105 exams, drawing clean, distinguishable 2D sketches of field lines ($\vec{E}$) and equipotential contours ($V$) is a high-yield mark generator. Always remember: $\vec{E}$ lines and $V$ lines must intersect at exactly $90^{\circ }$ (orthogonality).

Sketch 1: The $-{\rho }_v$ Volume Charge Sphere Cloud (The Mashuk Sir CT Trap)

• The Trap: Drawing arrows pointing outward. For negative charge clouds, fields point inward!
• Physical Behavior:
  • Inside ($r<a$): Field grows linearly inward: ${\vec{E}}_{in}=-{\hat{a}}_r\frac{{\rho }_{v}r}{3{ϵ}_0}$.
  • Outside ($r>a$): Field decays quadratically inward: ${\vec{E}}_{out}=-{\hat{a}}_r\frac{{\rho }_v{a}^3}{3{ϵ}_0{r}^2}$.

                 NEGATIVE SPHERICAL CLOUD SKETCH (-ρ_v)
                 
                           E-Field Lines: INWARD (solid)
                           Equipotentials: concentric circles (dashed)
                                     
                                     \   |   /
                                      v  |  v
                                   .-----|-----.
                                  /   \  |  /   \
                                 /     v | v     \
                                |------> (O) <----|
                                 \     ^ | ^     /
                                  \   /  |  \   /
                                   '-----|-----'
                                      ^  |  ^
                                     /   |   \

Sketch 2: The Electric Dipole ($+q$ and $-q$)

• Field Lines: Solid lines with arrows starting at $+q$ and terminating at $-q$.
• Equipotentials: Dashed oval-like loops enclosing each charge, with a straight vertical zero-potential boundary plane exactly halfway between them.

                           ELECTRIC DISTANCE DIPOLE SKETCH
                           
                             Equipotential (dashed)
                              /   E-Field (solid)   \
                             /   /    |    \   \     \
                            v   v     v     v   v     v
                         .---. .---.  |  .---. .---.
                        / +q  \     \ | /     / -q  \
                        | (O) |------>|------>| (O) |
                        \     /     / | \     \     /
                         '---' '---'  |  '---' '---'
                            ^   ^     ^     ^   ^     ^
                             \   \    |    /   /     /
                                      |
                            Zero-Potential Plane (V=0)

---

3. Core Concept Comparison Tables

Conductor vs. Dielectric in Static Fields

Feature	Perfect Conductor	Perfect Dielectric (Insulator)
Free Charge Carrier State	Abundant, completely free to move.	Zero free charges; carriers are bound to atoms.
Internal E-Field (${\vec{E}}_{in}$)	Exactly zero (${\vec{E}}_{in}=0$).	Non-zero; external field is only weakened by polarization.
Charge Distribution	Exists strictly as surface charge (${\rho }_s$).	Bound charges redistribute as volume (${\rho }_p$) and surface (${\rho }_{ps}$).
Internal Potential ($V$)	Constant (Equipotential volume).	Varies spatially throughout the material.
Constitutive Relation	Conductivity $\sigma \to \infty$.	Permittivity $ϵ={ϵ}_0{ϵ}_r$ ($\sigma \to 0$).

---

4. ECE 2105 Exam Diagnostic Quiz

Test your exam readiness with these 5 high-yield exam problems. Cover the answers, solve them on paper, and check your work step-by-step.

Question 1: Postulate to Law Derivation

Problem: Derive Kirchhoff’s Voltage Law ($\sum V=0$) from the fundamental electrostatic Curl postulate.

• Marks: 5

Click to view step-by-step solution

1. State the differential Curl postulate: $\nabla \times \vec{E}=0$
2. Integrate over an open surface $S$ bounded by a closed loop $C$: ${\int }_S(\nabla \times \vec{E})\cdot d\vec{s}=0$
3. Apply Stokes’s Theorem to convert the surface integral of the curl into a closed line integral around loop $C$: ${\oint }_C\vec{E}\cdot d\vec{l}={\int }_S(\nabla \times \vec{E})\cdot d\vec{s}$
4. Equate the expressions: ${\oint }_C\vec{E}\cdot d\vec{l}=0$
5. Relate to KVL: The line integral of electric field intensity $\vec{E}$ around a closed path represents the sum of voltage drops (potential differences) around the loop: ${\oint }_C\vec{E}\cdot d\vec{l}=-\sum V=0⟹\sum V=0\text{(KVL)}■$

---

Question 2: The $-{\rho }_v$ Cloud Field Calculation

Problem: An isolated spherical cloud of radius $a=5\text{ m}$ carries a uniform volume charge density of ${\rho }_v=-3{\text{ nC/m}}^3$. Calculate the magnitude and direction of the electric field intensity $\vec{E}$ at a radial distance $r=2\text{ m}$ from the center.

• Marks: 10

Click to view step-by-step solution

1. Identify the region: Since $r=2\text{ m}<a=5\text{ m}$, the point lies inside the charge distribution.
2. Apply Gauss’s Law to a concentric Gaussian sphere of radius $r=2\text{ m}$: ${\oint }_S\vec{D}\cdot d\vec{s}=Q_{enclosed}$ $D_r(4\pi r^2)={\rho }_v\left(\frac{4}{3}\pi r^3\right)$
3. Solve for $D_r$: $D_r=\frac{{\rho }_{v}r}{3}$
4. Convert to electric field intensity $\vec{E}$: $E_r=\frac{D_r}{{ϵ}_0}=\frac{{\rho }_{v}r}{3{ϵ}_0}$
5. Substitute the numerical values:
   • ${\rho }_v=-3\times 10^{-9}{\text{ C/m}}^3$
   • $r=2\text{ m}$
   • ${ϵ}_0\approx 8.854\times 10^{-12}\text{ F/m}$ $E_r=\frac{(-3\times 10^{-9})\times 2}{3\times (8.854\times 10^{-12})}=\frac{-6\times 10^{-9}}{2.6562\times 10^{-11}}=-225.88\text{ V/m}$
6. Write the vector answer: $\vec{\mathbf{E}}=-{\hat{\mathbf{a}}}_{\mathbf{r}}225.9\text{ V/m}\text{or}\vec{\mathbf{E}}={\hat{\mathbf{a}}}_{\mathbf{r}}(-225.9)\text{ V/m}$
   • Direction check: The negative sign indicates that the electric field points radially inward toward the center of the cloud, which is physically correct since the charge is negative.

---

Question 3: Boundary Condition Refraction

Problem: A charge-free boundary separates two dielectric media. Medium 1 has ${ϵ}_{r1}=2$ and Medium 2 has ${ϵ}_{r2}=8$. The electric field in Medium 1 makes an angle of ${\alpha }_1=30^{\circ }$ with the normal to the boundary. Calculate the angle ${\alpha }_2$ that the field in Medium 2 makes with the normal.

• Marks: 8

Click to view step-by-step solution

1. State the boundary conditions at a charge-free boundary (${\rho }_s=0$):
   • Tangential components are continuous: $E_{1t}=E_{2t}⟹E_1\sin {\alpha }_1=E_2\sin {\alpha }_2$
   • Normal components of flux density are continuous: $D_{1n}=D_{2n}⟹{ϵ}_1{E}_1\cos {\alpha }_1={ϵ}_2{E}_2\cos {\alpha }_2$
2. Divide the tangential equation by the normal equation to find the law of refraction: $\frac{E_1\sin {\alpha }_1}{{ϵ}_1{E}_1\cos {\alpha }_1}=\frac{E_2\sin {\alpha }_2}{{ϵ}_2{E}_2\cos {\alpha }_2}⟹\frac{\tan {\alpha }_1}{{ϵ}_1}=\frac{\tan {\alpha }_2}{{ϵ}_2}$ $\frac{\tan {\alpha }_1}{\tan {\alpha }_2}=\frac{{ϵ}_1}{{ϵ}_2}=\frac{{ϵ}_{r1}}{{ϵ}_{r2}}$
3. Solve for $\tan {\alpha }_2$: $\tan {\alpha }_2=\tan {\alpha }_1\left(\frac{{ϵ}_{r2}}{{ϵ}_{r1}}\right)$
4. Substitute the numerical values:
   • ${\alpha }_1=30^{\circ }⟹\tan (30^{\circ })=\frac{1}{\sqrt{3}}\approx 0.5774$
   • ${ϵ}_{r1}=2,{ϵ}_{r2}=8$ $\tan {\alpha }_2=0.5774\times \left(\frac{8}{2}\right)=0.5774\times 4=2.3094$
5. Calculate the angle ${\alpha }_2$: ${\alpha }_2={\tan }^{-1}(2.3094)\approx 66.58^{\circ }$ Answer: The electric field line bends away from the normal, making an angle of $66.6^{\circ }$ in Medium 2.

---

Question 4: Work Done Sign Pitfall

Problem: An electrostatic field is given by $\vec{E}={\hat{a}}_{x}2x-{\hat{a}}_{y}4y\text{ V/m}$. Determine the work done by an external force to carry a charge of $q=-5\mu \text{C}$ along a straight path from $P_1(0,0,0)$ to $P_2(2,3,0)$.

• Marks: 10

Click to view step-by-step solution

1. State the formula for work done by an external force: $W=-q{\int }_{P_1}^{P_2}\vec{E}\cdot d\vec{l}$
   • The Trap: Forgetting the negative sign in the external work definition. The negative sign is crucial because the external agent must exert a force equal and opposite to the field: ${\vec{F}}_{ext}=-q\vec{E}$.
2. Formulate the line integral: $\vec{E}\cdot d\vec{l}=({\hat{a}}_{x}2x-{\hat{a}}_{y}4y)\cdot ({\hat{a}}_{x}dx+{\hat{a}}_{y}dy+{\hat{a}}_{z}dz)=2xdx-4ydy$
3. Evaluate the integral (note that $2xdx-4ydy$ is a perfect differential $d(x^2-2y^2)$): ${\int }_{(0,0)}^{(2,3)}(2xdx-4ydy)={\left[x^2-2y^2\right]}_{(0,0)}^{(2,3)}$ Evaluate at endpoints:
   • At $P_1(0,0)$: $0^2-2(0{)}^2=0$
   • At $P_2(2,3)$: $2^2-2(3{)}^2=4-18=-14\text{ V}$ ${\int }_{P_1}^{P_2}\vec{E}\cdot d\vec{l}=-14\text{ V}$
4. Calculate the work done: $W=-q{\int }_{P_1}^{P_2}\vec{E}\cdot d\vec{l}=-(-5\times 10^{-6}\text{ C})\times (-14\text{ V})=-7\times 10^{-5}\text{ Joules}=-70\mu \text{J}$ Answer: The work done by the external force is $-70\mu \text{J}$.
   • Physical interpretation: Since the work is negative, the electrostatic field itself does work to move the charge along this path. The external agent must hold the charge back (do negative work) to keep it from accelerating.

---

Question 5: Polarization Charge Densities

Problem: A dielectric slab in free space has a polarization vector given by $\vec{P}={\hat{a}}_{x}3x^2{\text{ nC/m}}^2$. Calculate the volume polarization bound charge density (${\rho }_p$) inside the slab at $x=2\text{ m}$.

• Marks: 5

Click to view step-by-step solution

1. State the relationship between polarization and volume bound charge density: ${\rho }_p=-\nabla \cdot \vec{P}$
2. Evaluate the divergence of $\vec{P}$ in Cartesian coordinates: $\nabla \cdot \vec{P}=\frac{\partial P_x}{\partial x}+\frac{\partial P_y}{\partial y}+\frac{\partial P_z}{\partial z}$ Since $P_y=0$ and $P_z=0$: $\nabla \cdot \vec{P}=\frac{\partial }{\partial x}\left(3x^2\right)=6x{\text{ nC/m}}^3$
3. Apply the negative sign: ${\rho }_p=-6x{\text{ nC/m}}^3$
4. Evaluate at $x=2\text{ m}$: ${\rho }_p=-6(2)=-12{\text{ nC/m}}^3=-1.2\times 10^{-8}{\text{ C/m}}^3$ Answer: The volume bound charge density at $x=2\text{ m}$ is $-12{\text{ nC/m}}^3$.