Young & Freedman · Complete Reference

Every Constant, Every Equation,
Every Unit — Explained

Not just what the symbols mean, but what the units actually are in physical reality — with every algebraic variation shown.

00

Start Here: What Is a Joule?

Before any E&M equation makes sense, you need a visceral feel for the joule. It is the unit of energy — and the same joule appears in mechanics, electricity, thermodynamics, and chemistry. It is the universal currency of physics.

Dimensional Analysis — The Joule
1 J = 1 N · m = 1 kg · m/s² · m = kg · m² / s²
A newton is kg·m/s² (mass times acceleration = force). Multiplying by meters gives you force times distance = work. So a joule is literally "one kilogram moved at one meter per second squared, through one meter." That's the definition of mechanical work.

The joule in mechanical work

Mechanical work is $W = F \cdot d$ — force times displacement along the force direction. Lift a 1 kg apple by 10 cm against gravity ($g = 9.8\ \text{m/s}^2$): $$W = mgh = 1 \times 9.8 \times 0.1 = 0.98\ \text{J}$$ You just did about one joule of work. That energy is now stored as gravitational potential energy.

Physical scale of a joule

Lifting a medium apple 10 cm ≈ 1 J. A 60-watt bulb uses 60 J every second. The kinetic energy of a 1 kg ball at 1.4 m/s ≈ 1 J (from $\frac{1}{2}mv^2$). One food calorie (kcal) = 4184 J. A AA battery stores about 3000 J.

The joule in electrical work

In electricity, work is done when a charge moves through a potential difference. The work done moving charge $q$ through voltage $V$ is: $$W = qV$$ Check the units: $[\text{C}] \times [\text{V}] = [\text{C}] \times [\text{J/C}] = [\text{J}]$. ✓

This is the exact same joule as mechanics. Moving a charge through a voltage is the electrical analog of lifting a mass through a height. The charge is the "mass," the voltage is the "height," and the work is the energy transferred.

Mechanical vs Electrical — Side by Side
ConceptMechanicalElectrical
"What moves"mass $m$charge $q$
"What drives it"gravitational field $g$electric field $\vec{E}$
"Height" (potential)height $h$voltage $V$
Potential energy$U = mgh$$U = qV$
Work done$W = F \cdot d = mgh$$W = qV$
Units checkkg · m/s² · m = JC · J/C = J

The electron-volt — a more natural unit at small scales

One electron-volt (eV) is the energy one elementary charge gains crossing 1 volt: $$1\ \text{eV} = e \times 1\ \text{V} = 1.602\times10^{-19}\ \text{C} \times 1\ \text{J/C} = 1.602\times10^{-19}\ \text{J}$$ This is tiny — but so is an electron. In atomic physics and semiconductor engineering, joules are absurdly large units, so eV is the natural language.

01

The Core Constants

$e$
Elementary Charge
$1.602 \times 10^{-19}\ \text{C}$
The indivisible quantum of charge. Every proton carries $+e$, every electron $-e$. A coulomb is $1/e \approx 6.24 \times 10^{18}$ elementary charges. It takes 6.24 quintillion electrons to make one coulomb. Charge is quantized — you can't have half an $e$ in ordinary matter.
$k$
Coulomb's Constant
$8.988 \times 10^{9}\ \text{N·m}^2/\text{C}^2$
Sets the "loudness" of the electric force. Compare to Newton's gravitational constant $G = 6.67\times10^{-11}\ \text{N·m}^2/\text{kg}^2$. The ratio $k/G \approx 10^{20}$ tells you electricity is roughly $10^{20}$ times stronger than gravity between elementary particles. That's why electromagnetic forces dominate chemistry and materials. Note: $k = 1/(4\pi\varepsilon_0)$ — they are the same constant, just written differently.
$\varepsilon_0$
Permittivity of Free Space
$8.854 \times 10^{-12}\ \text{C}^2/(\text{N·m}^2)\ =\ \text{F/m}$
Measures how easily the vacuum permits an electric field to exist. A higher $\varepsilon_0$ would mean the vacuum is more "electrically flexible" — fields would be weaker for the same charge. Think of it as the spring constant of empty space resisting electric field lines. It appears naturally in Gauss's law and the energy density of electric fields. Units F/m = farads per meter appear when computing capacitance of geometric configurations.
$\mu_0$
Permeability of Free Space
$4\pi \times 10^{-7}\ \text{T·m/A}\ \approx 1.257\times10^{-6}\ \text{H/m}$
The magnetic analog of $\varepsilon_0$ — measures how easily a magnetic field forms in vacuum. Crucially, $c = 1/\sqrt{\varepsilon_0\mu_0}$ gives the speed of light. Maxwell discovered this in 1865 and realized light is an electromagnetic wave. The product $\varepsilon_0\mu_0$ encodes the speed limit of the universe.
$c$
Speed of Light
$2.998 \times 10^{8}\ \text{m/s}\ \approx 3\times10^8\ \text{m/s}$
Fundamental to E&M because it emerges from $\varepsilon_0$ and $\mu_0$: $c = 1/\sqrt{\varepsilon_0\mu_0}$. Also appears in the energy of EM radiation ($E = hf$) and relativistic corrections to fields of moving charges. At circuit speeds, we don't need it — but it sets the scale of everything electromagnetic.
$m_e$
Electron Mass
$9.109 \times 10^{-31}\ \text{kg}$
Appears when computing the dynamics of electrons in fields — how fast they accelerate, their kinetic energy, drift velocity in conductors. The charge-to-mass ratio $e/m_e$ determines how easily electrons respond to fields.
$\kappa$
Dielectric Constant (relative permittivity)
dimensionless (water: 80, glass: 5–10, vacuum: 1)
When you insert a material between capacitor plates, its molecules polarize in the field, partially canceling it. $\kappa = \varepsilon/\varepsilon_0$ tells you how much more permittive the material is than vacuum. Water's $\kappa = 80$ means its molecular dipoles reduce the field to $1/80$ of its vacuum value — which is why water is such a good solvent (it shields ionic charges from each other).
02

Charge & Coulomb's Law

Coulomb's Law — all forms

Core Coulomb's Law — magnitude
$$F = k\frac{|q_1||q_2|}{r^2} = \frac{|q_1||q_2|}{4\pi\varepsilon_0 r^2}$$
SymbolSI UnitDimensionsWhat it represents
$F$Nkg·m/s²Force on each charge — Newton's 3rd law: equal and opposite
$k$N·m²/C²kg·m³/(A²·s⁴)Proportionality constant — how strong electricity is in this universe
$q_1, q_2$CA·sCharges — positive or negative; sign determines attract/repel
$r$mmCenter-to-center separation between the charges
$\varepsilon_0$C²/(N·m²)A²·s⁴/(kg·m³)Permittivity of free space — $k = 1/4\pi\varepsilon_0$
Dimensional Check — Coulomb's Law
F = k · C² / m² = (N·m²/C²) · C² / m² = N ✓
The units of $k$ are specifically chosen so that when you multiply charge squared and divide by distance squared, you get newtons. This is not a coincidence — $k$ is defined this way to make the formula work in SI.
Vector Coulomb's Law — vector form
$$\vec{F}_{1\to2} = k\frac{q_1 q_2}{r^2}\,\hat{r}_{12}$$

$\hat{r}_{12}$ is the unit vector pointing from charge 1 toward charge 2. If $q_1 q_2 > 0$ (like charges), $\vec{F}$ points away from $q_1$ — repulsion. If $q_1 q_2 < 0$ (unlike charges), $\vec{F}$ points toward $q_1$ — attraction. The sign takes care of direction automatically.

Why $1/r^2$? — The Geometry Answer

Imagine charge $q_1$ radiating its "influence" equally in all directions, like a light bulb. At distance $r$, that influence is spread over a sphere of surface area $4\pi r^2$. Double $r$ → 4× more surface area → force at each point is $\frac{1}{4}$ as strong. The $1/r^2$ falloff is a geometric inevitability for anything that spreads in 3D. Gravity follows the same law for the same reason.

03

Electric Field

Definition and forms

Definition Electric Field from Force
$$\vec{E} = \frac{\vec{F}}{q_0} \qquad \Longleftrightarrow \qquad \vec{F} = q\vec{E}$$
SymbolSI UnitDimensionsWhat it represents
$\vec{E}$N/C = V/mkg·m/(A·s³)Force per unit positive charge — the field that exists at a point in space
$q_0$CA·sTest charge — hypothetically small so it doesn't disturb the source field
$q$CA·sAny real charge placed in the field — the force it feels is $q\vec{E}$
What does N/C = V/m actually mean?
1 N/C = 1 newton per coulomb = 1 (J/m) / C = 1 J/(C·m) = 1 V/m ✓
N/C interpretation: "Place 1 coulomb here and it feels 1 newton of force." That's the direct mechanical reading — field as force-per-charge.

V/m interpretation: "The voltage changes by 1 volt for every meter you travel in this direction." That's the energy-landscape reading — field as the slope (gradient) of potential.

Both say the same thing. The field IS the rate of change of potential. A steep voltage hill = strong field. A flat voltage plain = zero field. This is why charges sitting on a conductor feel no net force — the surface of a conductor is an equipotential, so the internal field is zero.
Point Charge Field of a Point Charge
$$E = k\frac{Q}{r^2} = \frac{Q}{4\pi\varepsilon_0 r^2}$$
solve for Q
$$Q = \frac{Er^2}{k}$$
solve for r
$$r = \sqrt{\frac{kQ}{E}}$$
force on charge q
$$F = qE = k\frac{qQ}{r^2}$$
Infinite Plane Field of an Infinite Sheet of Charge
$$E = \frac{\sigma}{2\varepsilon_0}$$
SymbolUnitMeaning
$\sigma$C/m²Surface charge density — coulombs per square meter on the sheet
$E$N/CField is uniform — doesn't depend on distance! The field lines are parallel and never spread out.

Between two parallel plates with opposite charges: fields from each plate add, giving $E = \sigma/\varepsilon_0$. Outside the plates: fields cancel, giving $E = 0$. This is exactly the parallel-plate capacitor geometry.

Gauss's Law Gauss's Law
$$\oiint \vec{E} \cdot d\vec{A} = \frac{Q_\text{enc}}{\varepsilon_0}$$
SymbolUnitMeaning
$\oiint \vec{E}\cdot d\vec{A}$N·m²/CElectric flux — counts how many field lines pierce outward through the closed surface
$Q_\text{enc}$CTotal charge enclosed inside the Gaussian surface
$\varepsilon_0$C²/(N·m²)Permittivity — converts charge to flux units
Electric Flux — What is N·m²/C?
Φ_E = E · A = (N/C) · m² = N·m²/C
Flux is the product of field strength and area. Think of field lines as a river current — flux counts how many liters per second pass through a net. A stronger current (E) or bigger net (A) means more flux. The dot product $\vec{E}\cdot d\vec{A}$ means only the component perpendicular to the surface counts — a net held parallel to the current catches nothing.
04

Electric Potential & Energy

Potential energy of a system of charges

Energy Potential Energy — Two Point Charges
$$U = k\frac{q_1 q_2}{r} = \frac{q_1 q_2}{4\pi\varepsilon_0 r}$$
work to assemble
$$W_\text{ext} = \Delta U = U_f - U_i$$
as $r\to\infty$
$$U \to 0 \text{ (reference)}$$
many charges
$$U = \sum_{i
SymbolUnitMeaning
$U$JPotential energy of the two-charge system — positive means repulsive (stored energy you'd release pushing them apart), negative means attractive (energy you must add to separate them)
$r$mSeparation — note this is $1/r$, not $1/r^2$. Integrating $F \propto 1/r^2$ over distance gives $U \propto 1/r$
Why 1/r for energy but 1/r² for force?

Force is $F = -dU/dr$ — force is the derivative (slope) of potential energy. If $U = k q_1 q_2 / r$, then: $$F = -\frac{dU}{dr} = -\frac{d}{dr}\left(\frac{kq_1q_2}{r}\right) = -kq_1q_2\cdot\left(-\frac{1}{r^2}\right) = \frac{kq_1q_2}{r^2}$$ So force being $1/r^2$ and energy being $1/r$ are not independent facts — one follows from the other by calculus. Force = negative gradient of energy. This is the same relationship as in mechanics: $F = -dU/dx$ for a spring, gravity, etc.

Potential Electric Potential $V$ — all forms
$$V = \frac{U}{q} \qquad \Longleftrightarrow \qquad U = qV$$ $$V = k\frac{Q}{r} \quad \text{(point charge)} \qquad V = \sum_i k\frac{q_i}{r_i} \quad \text{(superposition)}$$ $$V_{ab} = V_a - V_b = \int_a^b \vec{E}\cdot d\vec{\ell} \qquad \vec{E} = -\nabla V = -\frac{dV}{dr}\hat{r}$$
SymbolUnitDimensionsMeaning
$V$V (volt)J/C = kg·m²/(A·s³)Potential energy per unit charge — the "height" of the energy landscape
$V_{ab}$VJ/CPotential of $a$ relative to $b$ — work per unit charge to move from $b$ to $a$
$\nabla V$V/mJ/(C·m) = N/CGradient of potential — rate of change of voltage in space. The field IS this gradient, negated.
What does a Volt actually mean?
1 V = 1 J/C = 1 N·m / C = 1 W/A = 1 kg·m²/(A·s³)
J/C reading: "Each coulomb of charge carries 1 joule of energy through this potential." A 9-V battery gives each coulomb 9 joules of energy. That's the battery's job.

W/A reading: Volts = watts per ampere. A 100W device on 10V pulls 10A. This connects to Ohm's law directly — voltage is what converts power to current.

Intuitive picture: Voltage is to electricity what height is to water. Water flows downhill (from high to low gravitational PE). Current flows from high to low voltage. The voltage difference is the "pressure" pushing current through the circuit.

Work, potential and the path independence

Work Work done by Electric Force vs External Agent
$$W_\text{elec} = q(V_a - V_b) = -\Delta U \qquad \text{(work done BY the field)}$$ $$W_\text{ext} = q(V_b - V_a) = +\Delta U \qquad \text{(work done AGAINST the field)}$$

When the field does positive work on a charge, the charge loses potential energy (like a ball falling — gravity does work, PE decreases). When you push a charge against the field, you do positive work and the PE increases. These are always equal and opposite — energy is conserved.

05

Capacitors

Definition Capacitance — all forms
$$C = \frac{Q}{V} \qquad \Longleftrightarrow \qquad Q = CV \qquad \Longleftrightarrow \qquad V = \frac{Q}{C}$$
parallel plate
$$C = \kappa\frac{\varepsilon_0 A}{d}$$
series
$$\frac{1}{C_\text{eq}} = \sum_i \frac{1}{C_i}$$
parallel
$$C_\text{eq} = \sum_i C_i$$
SymbolUnitDimensionsMeaning
$C$F (farad)C/V = A²·s⁴/(kg·m²)Charge stored per volt applied — pure geometric property of the conductor arrangement
$A$Plate area — larger plates hold more charge for same voltage
$d$mmPlate separation — closer plates mean stronger field and more charge per volt
$\kappa$dimensionlessDielectric constant — how much the insulating material boosts capacitance by polarizing
What is a Farad?
1 F = 1 C/V = 1 C² / J = A²·s⁴ / (kg·m²)
A farad stores one coulomb for every volt applied. Since 1 coulomb is already a huge amount of charge, a farad is a gigantic unit. Practical capacitors are µF ($10^{-6}$ F), nF ($10^{-9}$ F), or pF ($10^{-12}$ F). The iPhone's touch screen uses femtofarad-scale ($10^{-15}$ F) capacitors in each pixel sensor. Supercapacitors in electric vehicles can reach thousands of farads by using nanometer-thick gaps over enormous surface areas.
Energy Energy stored in a Capacitor — all three forms
$$U_C = \frac{1}{2}CV^2 = \frac{Q^2}{2C} = \frac{1}{2}QV$$
use when V known
$$U = \frac{1}{2}CV^2$$
use when Q known
$$U = \frac{Q^2}{2C}$$
general
$$U = \frac{1}{2}QV$$

Derivation: As you add charge $dq$ at voltage $q/C$, work $dW = (q/C)dq$. Integrate from 0 to $Q$: $U = \int_0^Q \frac{q}{C}dq = \frac{Q^2}{2C}$. The factor of $\frac{1}{2}$ arises because voltage starts at 0 and builds up — you're doing less work at the start.

Field Energy Energy Density of an Electric Field
$$u = \frac{1}{2}\varepsilon_0 E^2$$
SymbolUnitMeaning
$u$J/m³Energy per unit volume stored in the field — the field itself carries energy through space
$E$V/mElectric field strength at that point

Total energy in capacitor: $U = u \cdot \text{Vol} = \frac{1}{2}\varepsilon_0 E^2 \cdot (Ad)$. Since $E = V/d$ and $C = \varepsilon_0 A/d$, this gives $U = \frac{1}{2}CV^2$ exactly. The energy isn't on the plates — it's in the field between them. This idea scales up: EM waves carry energy through empty space via oscillating $E$ and $B$ fields.

06

Current & Drift Velocity

Definition Electric Current — all forms
$$I = \frac{dq}{dt} \qquad \Longleftrightarrow \qquad q = \int I\,dt$$ $$I = nqv_d A$$
SymbolUnitDimensionsMeaning
$I$A (ampere)C/sCharge flowing past a cross-section per second
$n$m⁻³1/m³Number density of charge carriers (electrons per cubic meter)
$q$CA·sCharge per carrier (for electrons: $-e$, but we use magnitude)
$v_d$m/sm/sDrift velocity — the slow net motion of electrons. In copper at 1A through 1mm² wire, $v_d \approx 0.1$ mm/s! Electrons move slowly; the field signal travels at ~$c$.
$A$Cross-sectional area of the conductor
What is an Ampere?
1 A = 1 C/s = 6.24 × 10¹⁸ electrons/s
One ampere is one coulomb per second. The household circuit breaker trips at 15-20A. An LED draws ~20 mA. Lightning is ~20,000 A for a few microseconds. A phone charger is 1–3 A. The key insight: a 1A current in a wire doesn't mean electrons are sprinting. They drift at mm/s. But there are so many electrons ($n \sim 10^{28}/\text{m}^3$ in copper) that even a tiny drift velocity carries 1 coulomb per second past any cross-section.
Density Current Density $\vec{J}$
$$\vec{J} = \frac{I}{A}\hat{\ell} = nq\vec{v}_d \qquad \text{and} \qquad \vec{J} = \sigma\vec{E}$$
SymbolUnitMeaning
$J$A/m²Current per unit area — tells you how "dense" the current flow is, not just total amps
$\sigma$S/m = 1/(Ω·m)Electrical conductivity — inverse of resistivity. High $\sigma$ = easy current flow = metal. Low $\sigma$ = insulator.
07

Resistance & Ohm's Law

Ohm's Law Ohm's Law — all forms & variations
$$V = IR \qquad \Longleftrightarrow \qquad I = \frac{V}{R} \qquad \Longleftrightarrow \qquad R = \frac{V}{I}$$
solve for $V$
$$V = IR$$
solve for $I$
$$I = \frac{V}{R}$$
solve for $R$
$$R = \frac{V}{I}$$
SymbolUnitDimensionsMeaning
$V$V (volt)J/CPotential difference across the resistor — the "pressure" driving current
$I$A (ampere)C/sCurrent through the resistor — the "flow rate"
$R$Ω (ohm)V/A = kg·m²/(A²·s³)Resistance — the "friction" or "narrowness" opposing current flow
What is an Ohm?
1 Ω = 1 V/A = 1 J·s/C² = kg·m²/(A²·s³)
V/A reading: "It takes 1 volt of push to drive 1 ampere through 1 ohm." A higher ohm means more voltage needed per amp of current.

Typical values: Household wire: ~0.01 Ω/m. Light bulb filament: 240 Ω (hot). Human body: 1,000–100,000 Ω (varies with moisture). A dry resistor in your lab kit: 100–10,000 Ω. A short circuit: ideally 0 Ω. An open circuit: ideally ∞ Ω.
Resistivity Resistance from Geometry & Material
$$R = \rho\frac{L}{A}$$
solve for $\rho$
$$\rho = \frac{RA}{L}$$
solve for $L$
$$L = \frac{RA}{\rho}$$
temperature dependence
$$\rho(T) = \rho_0[1+\alpha(T-T_0)]$$
SymbolUnitMeaning
$\rho$Ω·mResistivity — intrinsic material property. Copper: $1.7\times10^{-8}\ \Omega\cdot\text{m}$. Silicon: $\sim10^{3}$. Glass: $\sim10^{12}$. A 15-order-of-magnitude range!
$L$mLength of conductor — longer wire = more collisions = higher resistance
$A$Cross-sectional area — fatter wire = more parallel paths = lower resistance
$\alpha$1/°C or 1/KTemperature coefficient of resistivity — metals get more resistive when hot (more atomic vibrations to scatter electrons)
Combinations Resistors in Series and Parallel
Series
$$R_\text{eq} = R_1 + R_2 + \cdots + R_n$$

Same $I$, voltages add.
$R_\text{eq} > $ any individual $R$

Parallel
$$\frac{1}{R_\text{eq}} = \frac{1}{R_1} + \frac{1}{R_2} + \cdots$$

Same $V$, currents add.
$R_\text{eq} < $ any individual $R$

Two parallel resistors
$$R_\text{eq} = \frac{R_1 R_2}{R_1 + R_2}$$

Product-over-sum formula — shortcut for exactly two

08

Power & Energy in Circuits

Power Electric Power — all three forms
$$P = IV = I^2R = \frac{V^2}{R}$$
use when $I$ and $V$ known
$$P = IV$$
use when $I$ and $R$ known
$$P = I^2R$$
use when $V$ and $R$ known
$$P = V^2/R$$
SymbolUnitDimensionsMeaning
$P$W (watt)J/s = kg·m²/s³Rate of energy delivery or dissipation — joules per second
$IV$A·V = W(C/s)·(J/C) = J/sCurrent times voltage — amps of flow times joules per coulomb = joules per second
$I^2R$A²·Ω = W(C/s)²·(V/A) = J/sJoule heating — power dissipated as heat. Bigger current or bigger resistance = more heat. This is why high-voltage transmission lines use high $V$ and low $I$ — to minimize $I^2R$ losses.
What is a Watt?
1 W = 1 J/s = 1 V·A = kg·m²/s³
A watt is a joule per second — the rate of energy use, not the total energy. Your phone charger is 5–20 W. A microwave is 1000 W. A human in moderate exercise: ~100 W. A car engine: ~100,000 W (100 kW). The sun's output: $3.8\times10^{26}$ W.

Your electricity bill charges you in kilowatt-hours (kWh): $1\ \text{kWh} = 1000\ \text{W} \times 3600\ \text{s} = 3.6 \times 10^6\ \text{J} = 3.6\ \text{MJ}$. Running a 1000 W microwave for 1 hour uses 1 kWh.
EMF EMF and Terminal Voltage of a Battery
$$V_\text{terminal} = \mathcal{E} - Ir \qquad \Longleftrightarrow \qquad \mathcal{E} = V_\text{terminal} + Ir$$ $$P_\text{delivered} = \mathcal{E}I - I^2r = IV_\text{terminal}$$
SymbolUnitMeaning
$\mathcal{E}$VEMF (electromotive force) — the open-circuit voltage the battery can supply; energy per unit charge the battery's chemistry provides
$r$ΩInternal resistance of the battery — causes terminal voltage to sag under load
$Ir$VVoltage drop inside the battery itself — wasted energy heating the battery
$V_\text{terminal}$VWhat you actually measure at the battery terminals while current flows
09

Kirchhoff's Rules

KCL Kirchhoff's Current Law — Junction Rule
$$\sum_\text{junction} I = 0 \qquad \Longleftrightarrow \qquad \sum I_\text{in} = \sum I_\text{out}$$

Deep meaning: Conservation of charge. In steady state (DC), charge cannot accumulate at a node. Every coulomb that arrives must leave. Units: every term is in amperes [A = C/s]. The equation says charge flow rates balance.

How to apply KCL

1. Label every branch current (direction can be assumed — a negative result just means your assumed direction was backward).
2. At each junction: sum currents entering = sum currents leaving.
3. $N$ junctions give $N-1$ independent KCL equations (the last is redundant).

KVL Kirchhoff's Voltage Law — Loop Rule
$$\sum_\text{closed loop} \Delta V = 0$$

Deep meaning: Conservation of energy. The electric potential is a well-defined single-valued function. If you walk around any closed path and return to start, the total voltage change must be zero — you can't gain energy for free. Units: every term is in volts [V = J/C]. The equation says energy per unit charge is conserved per loop.

ElementTraversed with currentTraversed against current
Resistor $R$$-IR$ (voltage drop)$+IR$ (voltage rise)
Battery $\mathcal{E}$$+\mathcal{E}$ (from − to + terminal)$-\mathcal{E}$ (from + to − terminal)
Internal resistance $r$$-Ir$$+Ir$
Capacitor $C$$-Q/C$ (drop toward + plate)$+Q/C$
How to apply KVL

1. Choose a loop direction (clockwise or counterclockwise — doesn't matter, just be consistent).
2. Walk around the loop, writing $\Delta V$ for each element using the sign table above.
3. Set the sum to zero. Each independent loop gives one KVL equation.
4. $B$ branches − $N$ nodes + 1 = number of independent KVL loops (mesh analysis).

Solving a circuit — the systematic approach

Circuit-Solving Recipe

For a circuit with $B$ branches, $N$ nodes, and $L$ loops:

  1. Assign a current variable and direction to each branch.
  2. Write $(N-1)$ independent KCL equations at junctions.
  3. Write $L = B - N + 1$ independent KVL equations around meshes.
  4. You now have $B$ equations in $B$ unknowns — solve the linear system.
  5. Negative current means your assumed direction was reversed — just flip the arrow.

This is the same matrix equation $A\vec{x} = \vec{b}$ you solve in linear algebra. KCL and KVL together form a complete, solvable system — that's the power of Kirchhoff.

10

All Units — Intuition Guide

Every SI unit in E&M can be expressed in base SI units: kg, m, s, A. Here's the complete translation:

UnitSymbolEqualsBase SIPhysical intuition
CoulombCA·sA·s Amount of charge. 1 C = 6.24×10¹⁸ elementary charges. Enormous — practical currents involve fractions of a coulomb per second.
VoltVJ/Ckg·m²/(A·s³) Energy per unit charge. "Height" in the electrical landscape. The pressure that drives current. 9V battery gives each coulomb 9 joules.
AmpereAC/sA (base unit) Charge flow rate. 1A = 6.24×10¹⁸ electrons/second past a cross-section. Your phone charger: ~2A.
OhmΩV/Akg·m²/(A²·s³) Opposition to current flow. Ratio of voltage applied to current produced. Higher Ω = more voltage needed per amp.
WattWJ/s = V·Akg·m²/s³ Rate of energy transfer. Your body at rest: ~80W. A 100W bulb: 100 joules per second converted to heat and light.
FaradFC/VA²·s⁴/(kg·m²) Charge stored per volt. 1F is huge — stores 1 coulomb per volt. Most circuits use µF or pF. Supercapacitors: 1–3000 F.
N/C = V/mE fieldN/C = V/mkg·m/(A·s³) Force per coulomb (N/C) = voltage gradient (V/m). Same quantity, two ways to think about it. Near a Van de Graaff at 100kV in 10cm: E = 10⁶ V/m = 1 MV/m.
N·m²/CΦ_EElectric fluxkg·m³/(A·s³) Field strength × area. Counts how many field lines pierce through a surface. Used in Gauss's law.
J/m³uEnergy densitykg/(m·s²) Energy stored per cubic meter of field. The electric field itself carries energy — this is the energy density formula $u = \frac{1}{2}\varepsilon_0 E^2$.
Ω·mρResistivitykg·m³/(A²·s³) Intrinsic resistance of a material regardless of geometry. Copper: 1.7×10⁻⁸ Ω·m. Glass: ~10¹² Ω·m. 20 orders of magnitude between best conductor and insulator.
A/m²JCurrent densityA/m² Current per unit cross-sectional area. Safe limit for copper wire: ~10⁷ A/m². Fuses blow when current density exceeds the wire's thermal limit.
eVeV1.602×10⁻¹⁹ Jkg·m²/s² Energy gained by one elementary charge across 1V. Atomic bond energies: 1–10 eV. X-ray photons: 10³–10⁶ eV. This unit is natural at atomic scales where joules are absurdly large.

The Ultimate Dimensional Sanity Check

When you're unsure if an equation is right, check units. Every valid physics equation must have the same dimensions on both sides. Here's the most useful chain to memorize:

J= N·m= C·V= A·V·s= W·s= kg·m²/s²
V= J/C= W/A= Ω·A= N·m/C= kg·m²/(A·s³)

If both sides of your equation have the same unit expression in base SI, the equation is dimensionally consistent. It might still be wrong (missing a factor of 2, wrong sign, etc.), but it can't be fundamentally broken.