Chapter 1
Coulomb's Law & the Electric Field
Coulomb's Law
Force between two point charges
\( F = k\dfrac{q_1 q_2}{r^2} \)
Unit anatomy of the result: N (Newton)
N·m²/C²[k] force·area per charge²
×
C·C[q₁q₂] charge times charge
÷
m²[r²] area of separation
→
Nforce
m² cancels m² · C² cancels C² → only N survives. The r² in the denominator geometrically represents the surface area of a sphere (∝ 4πr²) — the force "dilutes" over that sphere, exactly like light intensity.
A Newton is about the weight of a small apple (100 g in Earth gravity). The r² means: go twice as far → feel ¼ the force. The same 1/r² geometry appears in gravity, sound intensity, and light brightness — any quantity that spreads uniformly in 3D space over a sphere.
Real world: Two electrons 1 Å apart (10⁻¹⁰ m, about an atomic bond length): F = 9×10⁹ × (1.6×10⁻¹⁹)² / (10⁻¹⁰)² ≈ 23 nN. That's the actual force holding an ionic bond together. Nanoscale but structurally enormous — it's why salt crystals are rigid.
Electric Field Definition
E = F/q — force per unit charge
\( E = \dfrac{F}{q_0} \)
Unit anatomy of the result: N/C = V/m
N[F] force on test charge
÷
C[q₀] the test charge
→
N/C= V/m
N/C literally means "Newtons per Coulomb" — force that one Coulomb of charge would feel here. V/m means "Volts per meter" — how fast voltage changes with distance. They are exactly equal: 1 V/m = 1 N/C. Proof: V = J/C = N·m/C, so V/m = N·m/(C·m) = N/C ✓
N/C intuition: "If I put 1 coulomb here, it would feel this many Newtons of force." V/m intuition: "Voltage drops by this many Volts per meter of distance." Both describe the same field — N/C is the force perspective, V/m is the energy-landscape perspective.
Real world: Earth's atmospheric electric field ≈ 100 V/m pointing downward on a fair day. That means every meter of air holds a 100V drop. In a thunderstorm it can reach 10,000 V/m — enough to ionize air and create lightning. The field at a smartphone antenna ≈ 1–10 V/m.
Point Charge Field
E = kq/r² — field in space
\( E = k\dfrac{q}{r^2} \)
Unit anatomy: N/C (same result as above)
N·m²/C²[k]
×
C[q] source charge
÷
m²[r²] sphere area ∝ r²
→
N/Cfield
m² / m² = 1 · C / C² = 1/C → N/C remains. The m² in numerator (from k) and m² in denominator (from r²) annihilate each other. One C in numerator (q) cancels one of the two C in denominator, leaving N/C.
The r² denominator is geometric: field lines from q spread over a sphere of area 4πr². Same total "flux" — bigger sphere, weaker field per area. This is why all inverse-square laws exist: 3D geometry.
Real world: The electric field at the surface of a proton (r ≈ 10⁻¹⁵ m): E = 9×10⁹ × 1.6×10⁻¹⁹ / (10⁻¹⁵)² ≈ 1.4×10²¹ V/m. This is the field that holds the nucleus together — incomprehensibly large at nuclear scales.
Chapter 2
Gauss's Law
Gauss's Law
Φ_E = Q_enc / ε₀ — total flux through closed surface
\( \Phi_E = \oint \vec{E}\cdot d\vec{A} = \dfrac{Q_{enc}}{\varepsilon_0} \)
Unit anatomy of flux Φ_E: N·m²/C = V·m
N/C[E] field strength
×
m²[dA] area of surface
→
N·m²/C= V·m
Right side: Q [C] ÷ ε₀ [C²/(N·m²)] = C × (N·m²/C²) = N·m²/C ✓. Both sides match. V·m form: since V = J/C = N·m/C, we get (N·m/C)·m = N·m²/C = V·m. Flux has units of "Volts times meters" — voltage multiplied by a length scale.
What is N·m²/C geometrically? Think of E as a force-per-charge density. Multiply by area → you get total "force-area" per charge passing through that surface. It counts how many field lines pierce the surface, each line carrying N·m²/C of "weight." V·m means: the field integrated over a surface area — like multiplying the steepness of a hill (V/m) by the total hillside area (m²) to get total "hill strength."
Real world: A 1 µC charge at center of a 1 m radius sphere: Φ = Q/ε₀ = 10⁻⁶/(8.85×10⁻¹²) ≈ 113,000 V·m. That's the total "voltage-area" passing through any closed surface around it — regardless of the surface's shape. This is Gauss's power: shape is irrelevant, only enclosed charge matters.
Gauss → Infinite Plane
E = σ / 2ε₀ — field from a charged sheet
\( E = \dfrac{\sigma}{2\varepsilon_0} \)
Unit anatomy: C/m² ÷ C²/(N·m²) → N/C ✓
C/m²[σ] charge per area
÷
C²/(N·m²)[ε₀]
→
N/Cfield
(C/m²) ÷ (C²/N·m²) = (C/m²) × (N·m²/C²) = C·N·m² / (m²·C²) = N/C ✓. Notice no r anywhere — E is constant regardless of distance from the plane. The m² from ε₀ exactly cancels σ's m², leaving pure N/C.
Why no distance dependence? For a point charge, the field lines spread over a sphere (area ∝ r²) → field weakens. For an infinite plane, the "Gaussian pillbox" has fixed area caps — as you move away, you see more of the plane but at the same angle, exactly compensating. The m² cancels completely out of the equation — distance literally doesn't appear.
Real world: A capacitor plate with σ = 1 µC/m² gives E = 10⁻⁶/(8.85×10⁻¹²) ≈ 113,000 V/m between the plates regardless of the gap size. This is why the field between capacitor plates is uniform — it's the same as two infinite planes, each contributing σ/2ε₀, adding to σ/ε₀ between them.
Chapter 3
Electric Potential & Energy
Electric Potential
V = U/q — energy per charge
\( V = \dfrac{U}{q} \qquad [V] = \dfrac{J}{C} \)
Unit anatomy: J/C = Volt
J[U] energy (Joules)
÷
C[q] charge (Coulombs)
→
V= J/C
1 Volt = 1 Joule / 1 Coulomb. This is the definition of the Volt. Expanding J: 1 J = 1 N·m = 1 kg·m²/s². So 1 V = 1 N·m/C = 1 kg·m²/(A·s³). Each time you see "Volts" on a battery, you're seeing Joules-per-Coulomb — energy delivered per unit of charge pushed through.
J/C is the "height" of the electric landscape. Just as gravitational PE = mgh tells you energy per mass at height h (with g as the conversion), electrical PE = qV tells you energy per charge at potential V. The Volt is the "electric altitude" — moving from 0V to 9V gives every coulomb 9 joules of energy.
Real world: A 9V battery: every coulomb of charge that flows from − to + terminal gains exactly 9 J of energy. Your phone battery is ~3.7V × 10 Ah = 3.7 × 36,000 C ≈ 133,200 J = 133 kJ of stored energy. The 3.7 is joules per coulomb, 36,000 is total coulombs.
Point Charge Potential
V = kq/r — voltage falls as 1/r
\( V = k\dfrac{q}{r} \)
Unit anatomy: (N·m²/C²)·C/m → V = J/C
N·m²/C²[k]
×
C[q]
÷
m[r] — only ONE meter!
→
N·m/C= J/C = V
(N·m²/C²) × C / m = N·m²·C / (C²·m) = N·m/C. Then N·m = J (force × distance = energy). So N·m/C = J/C = Volts ✓. Key difference from E: E uses r² → N/C. V uses r¹ → N·m/C = J/C. One extra meter in numerator converts force-per-charge into energy-per-charge.
The extra "m" in V vs E: E is a force field (N/C). V is an energy field (J/C = N·m/C). The difference is exactly one meter — because energy = force × distance. Integrating E (N/C) over a path dr (m) gives V (J/C). That's why V has one fewer power of r than E: integrating 1/r² gives 1/r.
Real world: A 1 µC charge at 1 m distance: V = 9×10⁹ × 10⁻⁶ / 1 = 9,000 V. At 2 m: 4,500 V. At 10 m: 900 V. The voltage drops as 1/r — slower than the field E (which dropped to 1/4 when you doubled the distance, V only drops to 1/2). This is why V is easier to add from multiple charges — scalar, no direction.
E from Voltage
E = −dV/dr — field is steepness of voltage
\( E = -\dfrac{dV}{dr} \qquad \left[\dfrac{V}{m}\right] = \left[\dfrac{N}{C}\right] \)
Unit anatomy: V/m — Volts per meter
V[V] potential (J/C)
÷
m[dr] infinitesimal distance
→
V/m= N/C
V/m = (J/C)/m = (N·m/C)/m = N/C ✓. The meter in the denominator (from dr, the spatial step) cancels one of the meters in Joules (N·m), leaving just N/C. V·m ÷ m = V — V/m is literally "how many volts change per meter of distance traveled."
V/m is the slope of the voltage landscape. Topographic map analogy: contour lines = equipotentials. The closer the contour lines, the steeper the slope, the stronger E. If voltage drops 100 V over 1 cm (0.01 m), E = 100/0.01 = 10,000 V/m. The minus sign: E points downhill (from high V to low V), like water flows downhill.
Real world: Inside a car battery (12V across ~1 cm electrode gap): E = 12 V / 0.01 m = 1,200 V/m inside. In a wall outlet circuit: 120V across a 1 mm gap in a light switch = 120,000 V/m — enough to arc if the contacts are dirty. V/m tells you when air will ionize and spark (dielectric breakdown of air ≈ 3×10⁶ V/m).
Electric Potential Energy
U = qV — energy stored in position
\( U = qV \)
Unit anatomy: C × J/C → J (Joules)
C[q] charge
×
J/C[V] = Volts
→
Jenergy
C × (J/C) = J. The Coulombs cancel exactly — C × J/C = J. This is elegant: V is "joules per coulomb," so multiplying by the actual coulombs gives you actual joules. Electron-volt (eV): 1 eV = 1e × 1V = 1.6×10⁻¹⁹ C × 1 J/C = 1.6×10⁻¹⁹ J. A convenient atomic-scale energy unit.
C × J/C = J is the most important unit cancellation in E&M. Voltage is the "price per coulomb." Charge is "how many coulombs." Energy = price × quantity — the Coulombs cancel because you're literally converting price-per-item × number-of-items into total cost in Joules.
Real world: A lightning bolt moves about 5 C of charge through 300 MV: U = 5 × 3×10⁸ = 1.5×10⁹ J = 1.5 GJ. That's enough to power a house for a month — but it lasts microseconds, so the average power is enormous but the total energy is relatively modest.
Chapter 4
Capacitors
Capacitance
C = Q/V — charge stored per volt
\( C = \dfrac{Q}{V} \qquad [F] = \dfrac{C}{V} = \dfrac{C^2}{J} \)
Unit anatomy: C/V = F (Farad)
C[Q] Coulombs of charge
÷
V[V] Volts = J/C
→
F= C/V = C²/J
C ÷ (J/C) = C²/J. So 1 Farad = 1 C²/J. Expanding: F = C/(J/C) = C²/J = C²/(N·m). This form tells you: a 1 F capacitor stores 1 C² of charge-squared per Joule of energy. Alternative: F = s/Ω = s·A/V (useful for RC circuits where τ = RC has units Ω·F = Ω·s/Ω = s ✓).
C/V means: Coulombs per Volt — how much charge accumulates per volt of applied pressure. A large capacitor is "easy to fill" — one volt pushes in many coulombs. C²/J means: charge² per unit energy — reflecting that energy = Q²/2C, so C = Q²/(2U), which rearranges to show Farads are inherently about the ratio of charge² to energy.
Real world: A camera flash capacitor: 100 µF at 300V stores Q = CV = 100×10⁻⁶ × 300 = 0.03 C (30 mC). Energy = ½CV² = ½ × 100×10⁻⁶ × 90,000 = 4.5 J. That 4.5 J dumps in ~1 ms → 4,500 W of instantaneous power for the flash. The Farad rating tells you exactly how much charge per volt is waiting.
Parallel Plate Capacitor
C = ε₀A/d — capacitance from geometry
\( C = \dfrac{\varepsilon_0 A}{d} \)
Unit anatomy: (F/m)·m²/m → F (Farad)
F/m[ε₀] = C²/(N·m²)
×
m²[A] plate area
÷
m[d] plate gap
→
FFarads
(F/m) × m² / m = F × m × m² / (m × m²) = F ✓. Using ε₀ = C²/(N·m²): [C²/(N·m²)] × m² / m = C²/(N·m) = C²/J = F ✓. The area m² and gap m combine to give m²/m = m of "effective length," which ε₀ (F/m) converts to Farads.
ε₀ has units F/m — Farads per meter. This means: one square meter of parallel plate, separated by one meter, in vacuum, gives you ε₀ = 8.85 pF of capacitance. It's the geometric "base rate" of capacitance for vacuum. Scale up A or shrink d to get more. The m²/m = m in A/d is the "effective depth" of the electric field — a thinner gap concentrates the field over less distance, boosting charge storage.
Real world: A DRAM memory cell: plate area ≈ 0.01 µm² = 10⁻¹⁴ m², gap ≈ 5 nm = 5×10⁻⁹ m, κ ≈ 25 (high-k dielectric): C = 25 × 8.85×10⁻¹² × 10⁻¹⁴ / 5×10⁻⁹ ≈ 0.44 fF. One bit of RAM is stored in a ~0.44 femtofarad capacitor. Your 16 GB RAM chip has ~128 billion of these.
Energy in Capacitor
U = ½CV² — energy stored in the field
\( U = \tfrac{1}{2}CV^2 \)
Unit anatomy: F × V² → J (Joules)
F[C] = C/V = C²/J
×
V²[V²] = (J/C)²
→
Jenergy
(C/V) × V² = C·V = C × (J/C) = J ✓. Or: (C²/J) × (J/C)² = (C²/J) × (J²/C²) = J ✓. The Coulombs and the "per-Coulombs" annihilate, leaving energy. The ½: comes from integrating V·dQ = (Q/C)·dQ from 0 to Q → Q²/2C = ½CV². Voltage ramps up as charge builds, so average voltage = V_final/2.
F·V² = (C/V)·V² = C·V = J. Read it as: "capacitance (charge per volt) times voltage squared (volts times volts)" — one volt cancels, leaving charge times volts = coulombs × joules-per-coulomb = joules. The ½ reflects that you're filling the cap like a triangle: energy = ½ × base × height on a Q-V graph.
Real world: A defibrillator: C = 32 µF charged to 5,000 V: U = ½ × 32×10⁻⁶ × 25×10⁶ = 400 J. That 400 J is discharged in about 10 ms = 40,000 W peak power — enough to restart a heart. The F·V² unit calculation is literally telling you: 32 millionths of a coulomb-per-volt × 25 million volts² = 400 joules.
Field Energy Density
u = ½ε₀E² — energy per cubic meter of field
\( u = \tfrac{1}{2}\varepsilon_0 E^2 \)
Unit anatomy: (F/m)·(V/m)² → J/m³
F/m[ε₀]
×
(V/m)²[E²]
→
J/m³energy density
(F/m) × V²/m² = (C/V)/m × V²/m² = C·V/m³ = (C × J/C)/m³ = J/m³ ✓. The V in F (Farads = C/V) and the V² from E² combine: one V cancels the V in the denominator of F, leaving C·V/m³ = J/m³. J/m³ means: Joules per cubic meter — energy density, how much energy is packed into each cubic meter of space where the field exists.
J/m³ is energy density. Compare: a compressed spring has elastic energy density ½·(stress)·(strain) in Pa = N/m² = J/m³. Same unit — fields store energy just like compressed materials. The electric field IS a form of stored energy distributed through space, not localized in the charges. This is the crucial insight for understanding EM waves.
Real world: Sunlight at Earth's surface has E ≈ 600 V/m (peak): u = ½ × 8.85×10⁻¹² × (600)² ≈ 1.6 µJ/m³. That's 1.6 microjoules in every cubic meter of sunlit space. Tiny — but light travels at 3×10⁸ m/s, so power = u × c ≈ 470 W/m² ≈ solar constant. The field energy sweeps through space at light speed, delivering 470 watts per square meter to Earth.
Chapter 5
Resistance & Ohm's Law
Ohm's Law
V = IR — voltage, current, resistance
\( V = IR \qquad [\Omega] = \dfrac{V}{A} = \dfrac{J \cdot s}{C^2} \)
Unit anatomy: A × Ω → V
A[I] Amperes = C/s
×
Ω[R] Ohms = V/A
→
VVolts
A × (V/A) = V ✓ — trivially, since Ohm is defined as V/A. Deeper: (C/s) × (J·s/C²) = J/C = V ✓. The seconds cancel, one Coulomb cancels, leaving J/C = Volts. Ω = J·s/C²: literally "energy × time per charge squared" — resistance is how much time-energy it takes to push charge² through the material.
Ω = V/A = (J/C)/(C/s) = J·s/C². This weird unit J·s/C² tells you something deep: resistance links energy dissipation rate (J/s = W) to current squared (A² = C²/s²). Rearranging: R = W/A² = power-per-amp-squared — how many watts of heat per amp² of current. A 1Ω resistor with 1A through it dissipates exactly 1W as heat.
Real world: A 60W light bulb at 120V: R = V²/P = 14,400/60 = 240 Ω when hot. At room temperature (cold) it's ~20 Ω — 12× less. When you flip the switch, it briefly draws 120/20 = 6A instead of its rated 120/240 = 0.5A. This cold-start surge (6× overcurrent) is why bulbs blow when switched on.
Resistance from Geometry
R = ρL/A — shape determines resistance
\( R = \dfrac{\rho L}{A} \)
Unit anatomy: (Ω·m)·m/m² → Ω
Ω·m[ρ] resistivity
×
m[L] wire length
÷
m²[A] cross section
→
ΩOhms
Ω·m × m / m² = Ω·m²/m² = Ω ✓. The two meters from L and one meter from ρ (Ω·m) combine to give m², which cancels the area m². Ω·m geometrically: resistivity is "how many Ohms does one cubic meter of this material present?" Slice that cube into L/A pieces and you get R.
Ω·m is resistance × length = "resistance per unit length per unit cross section." Imagine a 1m × 1m × 1m cube of copper (ρ ≈ 1.7×10⁻⁸ Ω·m). Face-to-face resistance = ρ × 1m / 1m² = 1.7×10⁻⁸ Ω. That tiny resistance is why thick bus bars run current in power plants. Stretch the cube into a thin wire (L↑, A↓) and R = ρL/A skyrockets.
Real world: A standard 12 AWG copper wire (A = 3.31 mm² = 3.31×10⁻⁶ m²): R per meter = ρ/A = 1.68×10⁻⁸ / 3.31×10⁻⁶ ≈ 5 mΩ/m. A 50m extension cord (100m round trip) has R = 0.5 Ω. At 15A: voltage drop = IR = 15 × 0.5 = 7.5V. You "lose" 7.5V before it even reaches your appliance — that's why long extension cords reduce power to devices.
Current Density
J = I/A — microscopic current per area
\( J = \dfrac{I}{A} = \sigma E \qquad [A/m^2] \)
Unit anatomy: A/m² — Amperes per square meter
A[I] total current
÷
m²[A] cross-section
→
A/m²current density
Simple: A/m² stays as is. Check J = σE: [S/m]×[V/m] = [(1/Ω)/m]×[V/m] = [V/(Ω·m²)] = [(V/Ω)/m²] = [A/m²] ✓. A/m² means: charge flow rate per unit area — like water flow (m³/s) divided by pipe cross-section (m²) gives flow speed (m/s). J is the "speed" of charge flow, scaled by charge density.
A/m² is how densely current flows. A/m² = (C/s)/m² — coulombs crossing each square meter per second. Geometrically: imagine slicing the wire perpendicular to current flow. J tells you how many coulombs per second pass through each square meter of that cross-section. High J in thin wires → overheating → fuses blow.
Real world: Safe continuous current density for copper wire ≈ 3×10⁶ A/m² (3 A/mm²). An MRI superconducting coil runs at J ≈ 10⁸ A/m² — 30× denser — only possible because superconductors have zero resistance and don't heat up. Beyond the critical J, superconductivity collapses and the magnet quenches catastrophically.
Chapter 6
Energy & Power — Mechanical ↔ Electrical
Mechanical Power
P = Fv = τω — rate of doing work
\( P = Fv = \tau\omega \qquad [W] = [N \cdot m/s] \)
Unit anatomy: N × m/s → W (Watt)
N[F] force
×
m/s[v] velocity
→
W= N·m/s = J/s
N × m/s = N·m/s. Since N·m = J (force × displacement = energy), N·m/s = J/s = Watts ✓. For rotation: τω = (N·m)(rad/s) — radians are dimensionless, so N·m/s = J/s = W. Watts = Joules per second — power is always energy divided by time.
N·m/s = (force)·(distance per second) = energy per second. This is saying: power is how fast you're doing work. A N of force applied to something moving at m/s — the faster it moves, the more work per second. At 0 velocity (stalled motor), F is huge but v = 0, so P = 0. Maximum power delivery requires matching force to speed.
Real world: A car engine at highway speed: thrust force ≈ 1,000 N, speed = 30 m/s → P = 1000 × 30 = 30,000 W = 30 kW ≈ 40 hp. Electric motors deliver τω: an EV motor at τ = 400 N·m and ω = 300 rad/s (≈ 2,900 rpm): P = 400 × 300 = 120 kW = 160 hp. Same watts, different physical manifestation.
Electrical Power
P = VI = I²R = V²/R — three forms
\( P = VI = I^2R = \dfrac{V^2}{R} \)
Unit anatomy: V × A → W (Watt)
V[V] = J/C "pressure"
×
A[I] = C/s "flow"
→
W= J/s
(J/C) × (C/s) = J·C/(C·s) = J/s = W ✓. The Coulombs cancel perfectly — "joules per coulomb" times "coulombs per second" = joules per second. For I²R: (C/s)² × (J·s/C²) = C²/s² × J·s/C² = J/s = W ✓. For V²/R: (J/C)² / (J·s/C²) = J²/C² × C²/(J·s) = J/s = W ✓.
V × A = (J/C) × (C/s): "joules per coulomb" × "coulombs per second" — the Coulombs are the common currency. You're paying J per C, and C per s flow through — result is J/s. Mechanically: P = F·v where F↔V and v↔I, both multiply "push" by "flow" to get power. I²R: dissipation grows as current squared — double the current, quadruple the heat.
Real world: A Tesla Model 3 motor: 250 kW peak. At 400V battery: I = P/V = 250,000/400 = 625 A. At 625A through ~0.1Ω motor resistance: P_loss = I²R = 625² × 0.1 = 39,000 W = 39 kW lost as heat. That's 15.6% copper loss — why EVs have thermal management systems. The unit story: 625² [A²] × 0.1 [Ω = J·s/C²] = 390,625 × 0.1 J·s·C²/(s²·C²) = 39,063 J/s = 39 kW.
Energy = Power × Time
U = Pt = VIt — cumulative energy
\( U = Pt = VIt = I^2Rt \)
Unit anatomy: W × s → J (Joules)
W[P] = J/s power
×
s[t] seconds of time
→
J= W·s
(J/s) × s = J·s/s = J ✓. The most intuitive cancellation: power (J/s) times time (s) = energy (J). kWh: 1 kWh = 1000 W × 3600 s = 3.6×10⁶ J = 3.6 MJ. The "hour" in kWh is just a convenient time unit — 1 kWh is still just Joules: 3,600,000 of them.
J/s × s = J: rate × time = total. Exactly like: (miles/hour) × hours = miles. The "per second" and the "seconds" cancel, leaving the accumulated quantity. This is why your electricity bill is in kWh not kW — they're charging you for total energy consumed, not instantaneous power.
Real world: US average home uses 877 kWh/month = 877 × 3.6 MJ = 3,157 MJ = 3.157 GJ per month. At $0.12/kWh that's $105/month. Broken down: your fridge (150W × 720h = 108 kWh), water heater (4kW × 2h/day × 30 = 240 kWh), AC (3.5 kW × 8h/day × 30 = 840 kWh). The unit W × h = Wh tracks each one exactly.
Work-Energy Theorem
W = qΔV — mechanical meets electrical
\( W = q\Delta V = \Delta KE = \tfrac{1}{2}mv_f^2 - \tfrac{1}{2}mv_i^2 \)
Unit anatomy: C × V → J, and kg·(m/s)² → J
C[q] charge
×
V[ΔV] = J/C
=
kg[m] mass
×
(m/s)²[v²] = m²/s²
→
Jboth = Joules
Left: C × (J/C) = J ✓. Right: kg × m²/s² = kg·m²/s² = J ✓ (since 1 J = 1 kg·m²/s² = 1 N·m). Both sides are Joules — this is the conservation of energy equation. The electrical work done on the charge (C × V = J) exactly equals the mechanical kinetic energy gained (kg·m²/s²).
C × J/C = kg·m²/s² is the bridge between E&M and mechanics. The left side is "electrical language" for energy. The right side is "mechanical language" for the same energy. They must be equal — energy is energy, regardless of what form it's in. This equivalence is why electron-volts work: 1 eV = 1.6×10⁻¹⁹ C × 1 V = 1.6×10⁻¹⁹ J.
Real world: An electron gun (CRT monitor, electron microscope): electron accelerated through 10,000 V. W = qV = 1.6×10⁻¹⁹ × 10,000 = 1.6×10⁻¹⁵ J = 10,000 eV. Setting ½mv² = 1.6×10⁻¹⁵: v = √(2 × 1.6×10⁻¹⁵/9.11×10⁻³¹) = 5.9×10⁷ m/s ≈ 0.2c. One equation, two unit systems — same Joules both ways.
Chapter 7
Resistivity & Temperature Dependence
Resistivity
ρ — the Ω·m unit dissected
\( \rho = \dfrac{RA}{L} \qquad [\Omega \cdot m] \)
Unit anatomy of Ω·m: Ohm-meter
Ω[R] resistance
×
m²[A] cross-section
÷
m[L] length
→
Ω·mresistivity
Ω × m²/m = Ω·m ✓. One meter from cross-section cancels one meter from length, leaving one meter times Ohm. What does Ω·m mean physically? The resistance of a 1m × 1m × 1m cube of the material, measured face to face: R = ρ × 1m / 1m² = ρ. So Ω·m is literally "the resistance of a 1-meter cube of this material."
Ω·m = resistance of a unit cube. Imagine a 1m cube of material. Apply voltage across two opposite faces (area 1 m², length 1 m): resistance = ρ Ω. This is the "atomic fingerprint" of the material independent of its shape. Slice it thinner (small d) and wider (large A) to make a capacitor; stretch it long and thin (large L, small A) to make a resistor.
Real world: Copper: ρ = 1.68×10⁻⁸ Ω·m. A 1m copper cube has R = 1.68×10⁻⁸ Ω face-to-face. Silicon: ρ ≈ 640 Ω·m. A 1m silicon cube: R = 640 Ω. The ratio is 640/1.68×10⁻⁸ ≈ 4×10¹⁰ — silicon is 40 billion times more resistive than copper. This 10-decade span of ρ is why we can make both conductors and insulators from the same periodic table.
Temperature Coefficient α
R(T) = R₀[1 + α(T−T₀)] — R vs temperature
\( R(T) = R_0[1 + \alpha(T-T_0)] \qquad [\alpha] = \dfrac{1}{°C} \)
Unit anatomy of α: 1/°C (per degree Celsius)
Ω[R] at temp T
=
Ω[R₀] reference
×
[1 + (1/°C)(°C)]dimensionless factor
→
Ωsame unit as R₀
R₀ [Ω] × [1 + α (1/°C) × ΔT (°C)] — the °C in α's denominator cancels the °C of the temperature difference. The bracket becomes dimensionless. R₀ × (dimensionless) = Ω ✓. α units: literally "fractional change in resistance per degree" — if α = 3.9×10⁻³/°C, that's 0.39% per degree.
1/°C means "per degree Celsius." α = 0.004/°C means: for every degree temperature rises, resistance increases by 0.4% of its reference value R₀. It's a fractional rate — like %/°C but in decimal. The 1 in [1 + α·ΔT] is the "starting point" (100% of R₀) and α·ΔT is the fractional deviation from it. When α is negative (semiconductors), the bracket is less than 1 — resistance decreases with temperature.
Real world: Platinum RTD thermometer (α = 3.85×10⁻³/°C, R₀ = 100Ω at 0°C). At 100°C: R = 100 × [1 + 0.00385 × 100] = 100 × 1.385 = 138.5 Ω. Measure R → compute T with precision of 0.01°C. Used in industrial furnaces, medical equipment, and weather stations. The 1/°C unit is doing exactly what you'd expect: scaling the Ohms by degrees.
Conductivity
σ = 1/ρ — the S/m unit dissected
\( \sigma = \dfrac{1}{\rho} \qquad [S/m] = \dfrac{1}{\Omega \cdot m} \)
Unit anatomy: 1/(Ω·m) = S/m (Siemens per meter)
1reciprocal
÷
Ω·m[ρ] resistivity
→
S/mconductivity
1/(Ω·m) = S/m since 1 S = 1/Ω (Siemens is the reciprocal Ohm, sometimes called "mho" — Ohm spelled backwards). Check via J = σE: [S/m × V/m] = [(1/Ω)/m × V/m] = [V/(Ω·m²)] = [A/m²] ✓ since V/Ω = A.
S/m = (1/Ω)/m = "conductance per meter." While ρ asks "how much resistance per unit cube?", σ asks "how much conductance per unit cube?" A high σ means a 1m cube conducts easily. S/m is the material answer — J = σE says: given the local field E (V/m), the resulting current density J (A/m²) is σ times bigger if the material is more conductive.
Real world: Seawater σ ≈ 4 S/m (salty ions carry current). Pure water σ ≈ 5.5×10⁻⁶ S/m (almost no ions). Human body ≈ 0.2–2 S/m (depends on tissue). This is why being struck by lightning in the ocean is more dangerous — seawater spreads the current over a huge volume (high σ), and your body becomes a low-resistance path. The S/m unit directly tells you how efficiently each cubic meter of material passes current.
Chapter 8
Key Combos — How Units Tell the Story
Each combo shows the step-by-step unit cancellation — the units aren't just bookkeeping, they reveal the physical logic of how quantities combine.
Combo 1
Gauss's Law → Coulomb's Law
\( \oint E\,dA = \frac{Q}{\varepsilon_0} \;\xrightarrow{\text{sphere, symmetry}}\; E \cdot 4\pi r^2 = \frac{Q}{\varepsilon_0} \;\Rightarrow\; E = \frac{kQ}{r^2} \)
1
[E × 4πr²] = [N/C × m²] = N·m²/C
Flux: field strength × area. The 4π is dimensionless (solid angle of sphere).
2
[Q/ε₀] = C ÷ [C²/(N·m²)] = N·m²/C
Both sides = N·m²/C ✓ — Gauss is dimensionally consistent by construction.
3
E = Q/(4πε₀r²) = kQ/r² → [N·m²/C] / m² = N/C ✓
Dividing flux by area recovers field. The m² from r² cancels the m² in the flux unit, leaving N/C — the field unit. Coulomb's Law is just Gauss solved for E on a sphere.
The units tell you Gauss and Coulomb must agree. N·m²/C (flux) ÷ m² (sphere area) = N/C (field). The r² in the denominator of Coulomb's Law is simply the sphere's surface area appearing as a unit-cancelling denominator.
Combo 2
E = −dV/dr ↔ V = −∫E dr
\( E\;[\text{N/C}] = -\frac{dV}{dr}\;[\text{V/m}] \qquad V\;[\text{V}] = -\int E\,dr\;[\text{N/C} \times \text{m}] \)
1
V/m = (J/C)/m = (N·m/C)/m = N/C ✓
V/m and N/C are the same unit. The extra meter in V (= N·m/C) cancels the /m from the derivative. Differentiation removes one meter from the numerator.
2
∫E dr: [N/C] × m = N·m/C = J/C = V ✓
Integration adds one meter to the numerator. N/C × m = N·m/C = J/C = Volts. The meter added by dr is exactly what promotes force-per-charge (N/C) into energy-per-charge (V).
3
E ↔ V: one differentiation (÷m) or integration (×m) apart
The entire E↔V relationship is encoded in the units: they differ by exactly one power of meters. d/dr removes a meter; ∫dr adds one.
The unit story: E [N/C] and V [J/C = N·m/C] differ by one meter in the numerator. Differentiation (÷dr) removes that meter: V/m → N/C. Integration (×dr) adds it: N/C × m → N·m/C = V. The calculus operation is literally unit-tracking.
Combo 3
Gauss → E → V → C = ε₀A/d
\( E = \frac{\sigma}{\varepsilon_0} \xrightarrow{\times d} V = Ed \xrightarrow{C=Q/V} C = \frac{Q}{Ed} = \frac{\varepsilon_0 A}{d} \)
1
E = σ/ε₀: [C/m²] / [C²/(N·m²)] = N/C ✓
Gauss applied to a plate. C/m² ÷ C²/(N·m²) = N/C — the field between the plates.
2
V = E·d: [N/C] × m = N·m/C = J/C = V ✓
Uniform field × distance = voltage drop. The meter "promotes" N/C to J/C = Volts. This is ∫E dr with constant E.
3
C = Q/V = (σA)/(σd/ε₀) = ε₀A/d: [F/m × m²/m] = F ✓
σ cancels top and bottom. ε₀ [F/m] × A [m²] / d [m] = F·m²/m² = F. Every meter cancels and we're left with pure Farads — the geometry completely determines C.
Unit chain: C/m² (charge density) → N/C (field, via ε₀) → V (voltage, via ×d) → F (capacitance, via Q/V). Each step is a unit transformation: ÷ε₀ converts charge density to field; ×m converts field to voltage; ÷voltage converts charge to capacitance.
Combo 4
∫V dQ → ½CV² — why the ½ appears
\( U = \int_0^Q V\,dQ = \int_0^Q \frac{q}{C}dq = \frac{Q^2}{2C} = \frac{1}{2}CV^2 \)
1
∫V dQ: [V] × [C] = [J/C] × C = J ✓
Integrand V·dQ has units J/C × C = J. Integrating (summing infinitesimal energies) preserves Joules. This is work done pushing each dQ against rising voltage V = q/C.
2
Q²/2C: [C²] / [F] = [C²] / [C/V] = [C·V] = [C·J/C] = J ✓
C² ÷ (C/V) = C·V = J. The two Coulombs (one from Q², one from the denominator of C = C/V) partially cancel — one remains to multiply V = J/C, giving J.
3
½CV²: [F] × [V²] = [C/V] × [V²] = [C·V] = J ✓
F × V² = (C/V) × V² = C·V = J. One V from V² cancels the V in the denominator of F = C/V, leaving C·V = J. Three forms, one unit story.
The ½ is a calculus result, not a unit result — but all three forms (∫V dQ, Q²/2C, ½CV²) must give Joules. The Coulombs always partially cancel with the Volts (since V = J/C contains C⁻¹), leaving pure Joules. This cancellation pattern — C × J/C = J — is the universal unit story of electric energy.
Combo 5
Ohm's Law + P = VI → P = I²R
\( P = VI \xrightarrow{V=IR} P = (IR) \cdot I = I^2R \)
1
P = VI: [V × A] = [(J/C) × (C/s)] = J/s = W ✓
Coulombs cancel: J/C × C/s = J/s. Power is always Joules per second regardless of the form.
2
V = IR: [A × Ω] = [(C/s) × (V/A)] = V — Amps cancel
(C/s) × (V·s/C) = V. Both s and C cancel, cleanly giving Volts from Ohm's Law.
3
I²R: [A² × Ω] = [(C/s)² × J·s/C²] = J/s = W ✓
A² = C²/s². Ω = J·s/C². Product: C²/s² × J·s/C² = J·C²·s/(s²·C²) = J/s. Both C² and one s cancel, leaving J/s = Watts.
The unit story of I²R: [A²][Ω] = [C²/s²][J·s/C²]. The C² in Amperes squared and the C² in Ohms cancel. One of the two seconds cancels, leaving J/s = W. Ohm's unit (J·s/C²) is precisely engineered so that squaring current (C²/s²) and multiplying gives Watts.
Combo 6
τ = RC — why Ohms × Farads = seconds
\( \tau = RC \qquad [\Omega][F] = \left[\frac{V}{A}\right]\left[\frac{C}{V}\right] = \frac{C}{A} = \frac{C}{C/s} = s \)
1
R [Ω] = V/A (by Ohm's definition)
Resistance is defined as voltage per ampere — pressure per flow rate.
2
C [F] = C/V (by capacitance definition)
Capacitance is charge per volt — storage per pressure.
3
RC: (V/A)(C/V) = C/A = C/(C/s) = s ✓
Volts cancel (V top and V bottom). Left with C/A = C/(C/s) = s. The "pressure" unit (V) completely drops out, leaving charge ÷ current = time. Makes deep sense: τ is how long it takes current to drain the stored charge.
τ = RC = seconds is profound. R slows the drain (high R → low I); C holds more charge (high C → more Q to drain). τ = Q / I = C·V / (V/R) = RC — literally the time to empty the capacitor at the initial current rate. Volts cancel because they're the shared "pressure" between R and C.
Combo 7
J = E/ρ → R = ρL/A (micro to macro Ohm)
\( J = \frac{E}{\rho} \;\xrightarrow{J=I/A,\;E=V/L}\; \frac{I}{A} = \frac{V/L}{\rho} \;\Rightarrow\; R = \frac{\rho L}{A} \)
1
J = E/ρ: [N/C] / [Ω·m] = [V/m] / [V·m/A] = A/m² ✓
Ω·m = V·m/A (since Ω = V/A). So (V/m) / (V·m/A) = A/m². The Volts cancel, m/m² = 1/m, leaving A/m².
2
Substitute J=I/A, E=V/L: I/A = V/(ρL) → V/I = ρL/A
Geometry substitution — I/A is microscopic (density), I is macroscopic (total). V/L assumes uniform field along a conductor of length L.
3
R = V/I = ρL/A: [Ω·m × m / m²] = Ω ✓
Ω·m × m / m² = Ω·m² / m² = Ω. The two geometry meters (L and A) reduce to a dimensionless factor, and Ω·m → Ω exactly.
The unit story of micro→macro: J [A/m²] × A [m²] = I [A] — area "promotes" density to total. E [V/m] × L [m] = V [V] — length "promotes" field to voltage. These two geometry multiplications transform the material law (J = E/ρ) into the circuit law (V = IR). Meters appear twice and cancel twice.
Combo 8
u = ½ε₀E² → U = ½CV² (field ↔ circuit)
\( U = u \times \text{Vol} = \tfrac{1}{2}\varepsilon_0 E^2 \cdot (Ad) \xrightarrow{E=V/d,\;C=\varepsilon_0 A/d} \tfrac{1}{2}CV^2 \)
1
u × Vol: [J/m³] × m³ = J ✓
Energy density times volume = total energy. The m³ cancel perfectly. Volume of the capacitor gap = A × d.
2
½ε₀(V/d)²·Ad = ½ε₀V²A/d = ½(ε₀A/d)V² = ½CV²
Substitute E = V/d: d² in denominator, d in volume → net d in denominator. A/d with ε₀ is exactly C. The geometry of the volume and the geometry of E = V/d conspire to produce C.
3
Units: [F/m][V²/m²][m³] = [F/m × V²·m] = F·V² → J ✓
F/m × V²/m² × m³: the three meters multiply to m³, cancel the m in F/m and m² in V²/m², leaving F·V² = (C/V)·V² = C·V = J.
Field picture and circuit picture of the same energy: J/m³ × m³ = J, and F × V² = J. The geometry (Ad = volume, A/d = C) is the bridge. The d in the denominator of E = V/d partially cancels the d in the volume Ad — leaving exactly the A/d ratio that defines C. Units track every geometric substitution automatically.