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.
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.
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.
| Concept | Mechanical | Electrical |
|---|---|---|
| "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 check | kg · m/s² · m = J | C · 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.
The Core Constants
Charge & Coulomb's Law
Coulomb's Law — all forms
| Symbol | SI Unit | Dimensions | What it represents |
|---|---|---|---|
| $F$ | N | kg·m/s² | Force on each charge — Newton's 3rd law: equal and opposite |
| $k$ | N·m²/C² | kg·m³/(A²·s⁴) | Proportionality constant — how strong electricity is in this universe |
| $q_1, q_2$ | C | A·s | Charges — positive or negative; sign determines attract/repel |
| $r$ | m | m | Center-to-center separation between the charges |
| $\varepsilon_0$ | C²/(N·m²) | A²·s⁴/(kg·m³) | Permittivity of free space — $k = 1/4\pi\varepsilon_0$ |
$\hat{r}_{12}$ 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.
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.
Electric Field
Definition and forms
| Symbol | SI Unit | Dimensions | What it represents |
|---|---|---|---|
| $\vec{E}$ | N/C = V/m | kg·m/(A·s³) | Force per unit positive charge — the field that exists at a point in space |
| $q_0$ | C | A·s | Test charge — hypothetically small so it doesn't disturb the source field |
| $q$ | C | A·s | Any real charge placed in the field — the force it feels is $q\vec{E}$ |
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.
| Symbol | Unit | Meaning |
|---|---|---|
| $\sigma$ | C/m² | Surface charge density — coulombs per square meter on the sheet |
| $E$ | N/C | Field 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.
| Symbol | Unit | Meaning |
|---|---|---|
| $\oiint \vec{E}\cdot d\vec{A}$ | N·m²/C | Electric flux — counts how many field lines pierce outward through the closed surface |
| $Q_\text{enc}$ | C | Total charge enclosed inside the Gaussian surface |
| $\varepsilon_0$ | C²/(N·m²) | Permittivity — converts charge to flux units |
Electric Potential & Energy
Potential energy of a system of charges
| Symbol | Unit | Meaning |
|---|---|---|
| $U$ | J | Potential 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) |
| $r$ | m | Separation — note this is $1/r$, not $1/r^2$. Integrating $F \propto 1/r^2$ over distance gives $U \propto 1/r$ |
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.
| Symbol | Unit | Dimensions | Meaning |
|---|---|---|---|
| $V$ | V (volt) | J/C = kg·m²/(A·s³) | Potential energy per unit charge — the "height" of the energy landscape |
| $V_{ab}$ | V | J/C | Potential of $a$ relative to $b$ — work per unit charge to move from $b$ to $a$ |
| $\nabla V$ | V/m | J/(C·m) = N/C | Gradient of potential — rate of change of voltage in space. The field IS this gradient, negated. |
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
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.
Capacitors
| Symbol | Unit | Dimensions | Meaning |
|---|---|---|---|
| $C$ | F (farad) | C/V = A²·s⁴/(kg·m²) | Charge stored per volt applied — pure geometric property of the conductor arrangement |
| $A$ | m² | m² | Plate area — larger plates hold more charge for same voltage |
| $d$ | m | m | Plate separation — closer plates mean stronger field and more charge per volt |
| $\kappa$ | dimensionless | — | Dielectric constant — how much the insulating material boosts capacitance by polarizing |
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.
| Symbol | Unit | Meaning |
|---|---|---|
| $u$ | J/m³ | Energy per unit volume stored in the field — the field itself carries energy through space |
| $E$ | V/m | Electric 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.