21
Chapter 21
Electric Charge & Coulomb's Law
21.1Coulomb's Law
$$F = k\frac{|q_1||q_2|}{r^2} = \frac{1}{4\pi\varepsilon_0}\frac{|q_1||q_2|}{r^2}$$
UNITS:
N=
N·m²/C²×
C²÷m²
N — what's physically happening when you measure it: two charged tape strips repel each other. You put one on a scale and hold the other at a fixed distance. The scale reads a deflection — that deflection is in Newtons. The same Newton your arm feels lifting a 100 g coffee mug. There's no special "electric Newton." The electric force is a push, and when it acts on a mass, that mass accelerates at $a = F/m$ exactly as if you'd hit it with your hand. The unit N is unambiguous: it's kg·m/s², the rate at which momentum is being deposited.
N·m²/C² — what $k$ is physically doing: look at $|q_1||q_2|/r^2$ before $k$ multiplies it. Its units are C²/m². That number, on its own, is not a force — it's a geometric score of "how much charge-product, per how much distance-squared." Two charges 1 m apart with 1 C each score 1 C²/m². The constant $k = 8.99\times10^9$ N·m²/C² is the universe's fixed exchange rate: for every 1 C²/m² of that score, you get $8.99\times10^9$ Newtons. The m² in $k$'s numerator cancels the m² denominator; the C² in $k$'s denominator cancels the C² from the charges. What's left is N. The unit of $k$ is literally designed to perform that cancellation.
Why $1/r^2$, geometrically: place a single charge in empty space. Its influence radiates outward in all directions equally, like light from a bulb. At radius $r$, that total influence is spread over a sphere of surface area $4\pi r^2$. The amount of influence is fixed — it can't disappear. But it's diluted over more sphere as $r$ grows. Dilution goes as $1/(4\pi r^2)$, so force per unit area — intensity — falls as $1/r^2$. Double the distance: four times the sphere area, one-quarter the force. This is not a property of electricity. It's the geometry of 3D space. Gauss's Law makes this exact: the total flux through any closed surface around the charge is constant, regardless of the surface's shape.
N·m²/C² — what $k$ is physically doing: look at $|q_1||q_2|/r^2$ before $k$ multiplies it. Its units are C²/m². That number, on its own, is not a force — it's a geometric score of "how much charge-product, per how much distance-squared." Two charges 1 m apart with 1 C each score 1 C²/m². The constant $k = 8.99\times10^9$ N·m²/C² is the universe's fixed exchange rate: for every 1 C²/m² of that score, you get $8.99\times10^9$ Newtons. The m² in $k$'s numerator cancels the m² denominator; the C² in $k$'s denominator cancels the C² from the charges. What's left is N. The unit of $k$ is literally designed to perform that cancellation.
Why $1/r^2$, geometrically: place a single charge in empty space. Its influence radiates outward in all directions equally, like light from a bulb. At radius $r$, that total influence is spread over a sphere of surface area $4\pi r^2$. The amount of influence is fixed — it can't disappear. But it's diluted over more sphere as $r$ grows. Dilution goes as $1/(4\pi r^2)$, so force per unit area — intensity — falls as $1/r^2$. Double the distance: four times the sphere area, one-quarter the force. This is not a property of electricity. It's the geometry of 3D space. Gauss's Law makes this exact: the total flux through any closed surface around the charge is constant, regardless of the surface's shape.
Concrete example — two protons at nuclear distance
Two protons, $q = 1.6\times10^{-19}$ C each, separated $r = 10^{-15}$ m (nuclear scale). $F = (8.99\times10^9)(1.6\times10^{-19})^2/(10^{-15})^2$. Unit chain: (N·m²/C²)·C²/m² = N. Result: ~230 N of electric repulsion. That's 50 pounds of force between two particles smaller than an atom. This is why atomic nuclei need the strong nuclear force to hold together — they're fighting 230 N of electric repulsion at point-blank range.
equivF — Newton · q — Coulomb · k and ε₀
Form
Derived from
Physical meaning / when to use
N = kg·m/s²
$F=ma$, base SI
Force is mass times acceleration — the mechanical definition. All Newtons are this.
N = J/m
Work $W=Fd$, so $F=W/d$
Force is energy per meter — how steeply the PE landscape slopes in space
N = C·V/m
$F=qE$, $E$ in V/m
Electric force = charge × voltage-gradient. Links force directly to voltage drop.
C = A·s
$I=dQ/dt$, so $Q=It$
1 Amp for 1 second moves 1 Coulomb past any cross-section
C²/(N·m²) = F/m
$C=\varepsilon_0 A/d$
$\varepsilon_0$ is Farads of capacitance per meter of geometric plate ratio
Why the Coulomb is such a large unit
1 C requires $6.24\times10^{18}$ proton-charges. Try to put 1 C of bare charge on a 1 m sphere in vacuum: the electric potential at the surface would be $V = kQ/r = 9\times10^9$ V — nine billion Volts. That's why free charge in nature is always nearly neutralized. "1 Coulomb of charge" shows up only in currents (where equal and opposite charges are both present), never as isolated charge.
22
Chapter 22
Electric Field
22.1Definition of the Electric Field — $\vec{E}=\vec{F}/q_0$
$$\vec{E} = \frac{\vec{F}}{q_0}\quad\Longrightarrow\quad\vec{F}=q_0\vec{E}$$
UNITS:
N/C≡V/m
— exactly equal, two physical pictures
N/C — you're measuring a force that would act on a specific amount of charge: picture a proton sitting between two charged plates. It feels a force — a physical push you could measure with a spring scale. That force in Newtons, divided by the proton's charge in Coulombs, gives the field in N/C. The field is $10^4$ N/C? That means any charge you place there feels $10^4$ Newtons per Coulomb it carries. A $2\ \mu\mathrm{C}$ charge feels $10^4 \times 2\times10^{-6} = 0.02$ N — a real, mechanical force, same as gravity on a 2-gram object. The N/C number is a rate: "Newtons of force delivered per Coulomb of charge present."
V/m — you're measuring how fast voltage changes across space: now think about the same plates, but instead of watching the proton, watch a voltmeter. Probe one position, record the voltage. Move the probe 1 cm toward the positive plate, record again. The difference in volts divided by the distance in meters is $E$ in V/m. A field of $10^4$ V/m means: for every meter you walk in the field direction, you cross $10^4$ Volts of potential difference. Walk 1 mm: cross 10 V. Walk 1 cm: cross 100 V. The V/m unit is not about forces at all — it's about how steeply the energy landscape changes per meter of distance. Yet it equals N/C exactly, because energy per charge per meter = force per charge.
Why two descriptions for one unit — and when to use each: N/C is your tool when you want to know what happens to a particle: will it accelerate? Which way? How hard? V/m is your tool when you want to know about energy: how much kinetic energy does the particle gain? What voltage difference exists across a gap? In circuits, you almost always think in V/m — the field drives current, and current flows down the voltage gradient. In particle physics, you think in N/C — what force deflects the beam?
V/m — you're measuring how fast voltage changes across space: now think about the same plates, but instead of watching the proton, watch a voltmeter. Probe one position, record the voltage. Move the probe 1 cm toward the positive plate, record again. The difference in volts divided by the distance in meters is $E$ in V/m. A field of $10^4$ V/m means: for every meter you walk in the field direction, you cross $10^4$ Volts of potential difference. Walk 1 mm: cross 10 V. Walk 1 cm: cross 100 V. The V/m unit is not about forces at all — it's about how steeply the energy landscape changes per meter of distance. Yet it equals N/C exactly, because energy per charge per meter = force per charge.
Why two descriptions for one unit — and when to use each: N/C is your tool when you want to know what happens to a particle: will it accelerate? Which way? How hard? V/m is your tool when you want to know about energy: how much kinetic energy does the particle gain? What voltage difference exists across a gap? In circuits, you almost always think in V/m — the field drives current, and current flows down the voltage gradient. In particle physics, you think in N/C — what force deflects the beam?
V/m physically — inside a AA battery
A 1.5 V AA battery, electrode gap ~5 mm. Field inside: $E = V/d = 1.5/0.005 = $ 300 V/m. In N/C terms: a charge of $1\ \mu\mathrm{C}$ at that location feels $F = qE = 10^{-6}\times300 = 3\times10^{-4}$ N of force. In V/m terms: an electron crossing the 5 mm gap gains $\Delta U = eEd = (1.6\times10^{-19})(300)(0.005) = 2.4\times10^{-19}$ J = 1.5 eV of kinetic energy. The V/m unit literally told you the energy story: 300 Joules per Coulomb per meter of travel.
V/m physically — lightning rod and air breakdown
Air breaks down (ionizes, allowing a spark) at about $3\times10^6$ V/m. A lightning rod concentrates field lines at its tip — the local $E$ near the sharp point can reach this threshold while the voltage of the cloud is still "only" ~100 MV. The V/m unit is the physically meaningful one for breakdown: it's not just about voltage (altitude), it's about how steeply that voltage changes per meter of space. Sharp geometry → steep voltage gradient → breakdown threshold reached → lightning strikes preferentially there.
equivE — N/C = V/m, proven identical
Form
Algebraic proof
Physical picture it emphasizes
N/C
$\vec{E}=\vec{F}/q$ directly
Force picture: how hard the field pushes on a unit charge. Use with $F=qE$.
V/m
V/m=(N·m/C)/m=N/C
Energy/gradient picture: how steeply voltage drops per meter. Use with $V=Ed$, circuits.
J/(C·m)
Expand V/m=(J/C)/m
Energy per charge per meter — work done per Coulomb per meter of travel
kg·m/(A·s³)
Full SI base
Dimensional proofs connecting to Maxwell's equations
The field as physically real — not just a math tool
If you suddenly move a charge, the force on a distant charge doesn't update instantly — the change in field propagates at the speed of light. The field carries energy and momentum between the charges while the signal is in transit. This is why the field is the fundamental physical object. Coulomb's Law is only valid when both charges are stationary. The field is what's always real.
22.2Charge Densities — $\lambda$, $\sigma$, $\rho$
$$dq = \lambda\,dx \quad\text{(C/m)}, \qquad dq = \sigma\,dA \quad\text{(C/m}^2\text{)}, \qquad dq = \rho\,dV \quad\text{(C/m}^3\text{)}$$
λ·dx: (C/m)·m = C ✓
σ·dA: (C/m²)·m² = C ✓
ρ·dV: (C/m³)·m³ = C ✓
$\lambda$ in C/m — spreading Coulombs along a line: a wire with $\lambda = 5\ \mu\mathrm{C/m}$ holds $5\ \mu\mathrm{C}$ per meter. Every centimeter of wire carries $0.05\ \mu\mathrm{C}$. The denominator (meters of length) is the "dilution axis." To get total charge on a 20 cm segment: $Q = \lambda L = 5\times10^{-6}\times0.2 = 1\ \mu\mathrm{C}$. The units: C/m × m = C. The meters cancel because you're "un-spreading" the charge back into a total.
$\sigma$ in C/m² — spreading Coulombs across a surface: a charged plate with $\sigma = 1\ \mu\mathrm{C/m^2}$ holds $1\ \mu\mathrm{C}$ per square meter. A hand-sized plate (0.01 m²) holds 10 nC. The $\sigma$ for a parallel-plate capacitor is directly related to the field it produces: $E = \sigma/\varepsilon_0$, so $\sigma$ (C/m²) ÷ $\varepsilon_0$ (C²/(N·m²)) = N/C ✓. The surface density is the quantity Gauss's Law directly converts into field strength.
$\rho$ in C/m³ — packing Coulombs into a volume: the ionosphere or a semiconductor depletion region has a volume charge density. $\rho = 10^{-6}$ C/m³ means one microcoulomb per cubic meter — extremely dilute. Integrating $\rho$ over a volume gives total enclosed charge for Gauss's Law: $Q_\text{enc} = \int\rho\,dV$. Units: C/m³ × m³ = C ✓.
$\sigma$ in C/m² — spreading Coulombs across a surface: a charged plate with $\sigma = 1\ \mu\mathrm{C/m^2}$ holds $1\ \mu\mathrm{C}$ per square meter. A hand-sized plate (0.01 m²) holds 10 nC. The $\sigma$ for a parallel-plate capacitor is directly related to the field it produces: $E = \sigma/\varepsilon_0$, so $\sigma$ (C/m²) ÷ $\varepsilon_0$ (C²/(N·m²)) = N/C ✓. The surface density is the quantity Gauss's Law directly converts into field strength.
$\rho$ in C/m³ — packing Coulombs into a volume: the ionosphere or a semiconductor depletion region has a volume charge density. $\rho = 10^{-6}$ C/m³ means one microcoulomb per cubic meter — extremely dilute. Integrating $\rho$ over a volume gives total enclosed charge for Gauss's Law: $Q_\text{enc} = \int\rho\,dV$. Units: C/m³ × m³ = C ✓.
23
Chapter 23
Gauss's Law & Electric Flux
23.1Electric Flux — $\Phi_E = \int\vec{E}\cdot d\vec{A}$
$$\Phi_E = \int\vec{E}\cdot d\vec{A} = \int E\cos\theta\,dA$$
UNITS:
N/C×m²
=N·m²/C
=V·m
N·m²/C — what's physically happening when flux is nonzero: you draw an imaginary closed bubble around a region of space. At every patch of that bubble, the electric field has some component punching outward through the patch. N/C is the field strength at the patch; m² is the area of the patch. Their product, summed over the whole bubble, is the total flux in N·m²/C. Physically, this number tells you how many "field lines per unit source charge" are passing through your surface. It's like counting how much wind passes through a net: intensity of wind (N/C) times effective area of net facing the wind (m²) gives total airflow. Positive flux: net flow outward — there's positive charge inside producing it. Zero flux: every field line that enters also exits — whatever's inside has zero net charge.
The $\cos\theta$ — you're projecting onto the surface's outward normal: hold a tennis racket in a wind. Face it straight into the wind: maximum air flows through. Tilt it 45°: less air flows through the strings, even though the wind hasn't changed. Turn it fully parallel to the wind: zero airflow through the strings. The $\cos\theta$ is doing exactly that projection. Only the component of $\vec{E}$ perpendicular to the surface — "trying to punch through" — contributes to flux. The field lines that run parallel to the surface glide by without contributing.
V·m — the voltage-times-length form: since N/C = V/m, we get N·m²/C = (V/m)·m² = V·m. This form has a specific physical home: boundary conditions in electrostatics. When you work with potential $V$ instead of field $E$, flux integrals naturally come out in V·m. It's the same physical quantity — total field threading a surface — expressed through the energy-per-charge language instead of the force-per-charge language.
The $\cos\theta$ — you're projecting onto the surface's outward normal: hold a tennis racket in a wind. Face it straight into the wind: maximum air flows through. Tilt it 45°: less air flows through the strings, even though the wind hasn't changed. Turn it fully parallel to the wind: zero airflow through the strings. The $\cos\theta$ is doing exactly that projection. Only the component of $\vec{E}$ perpendicular to the surface — "trying to punch through" — contributes to flux. The field lines that run parallel to the surface glide by without contributing.
V·m — the voltage-times-length form: since N/C = V/m, we get N·m²/C = (V/m)·m² = V·m. This form has a specific physical home: boundary conditions in electrostatics. When you work with potential $V$ instead of field $E$, flux integrals naturally come out in V·m. It's the same physical quantity — total field threading a surface — expressed through the energy-per-charge language instead of the force-per-charge language.
Flux through a tilted window — literal N·m²/C in action
Uniform field $E = 500$ N/C pointing in the $\hat{x}$ direction. A 0.5 m² window tilted so its normal makes $\theta = 60°$ with $\hat{x}$. Flux: $\Phi = EA\cos\theta = 500\times0.5\times\cos60° = 500\times0.5\times0.5 = $ 125 N·m²/C. Now orient the window to face the field directly ($\theta=0$): $\Phi = 500\times0.5\times1 = 250$ N·m²/C — double. Tilt it to $\theta=90°$ (parallel to field): $\Phi=0$. The unit N·m²/C is telling you: "field strength times the effective area it's threading through."
23.2Gauss's Law
$$\oint\vec{E}\cdot d\vec{A} = \frac{Q_\text{enc}}{\varepsilon_0}$$
LEFT:N·m²/C
RIGHT:
C ÷ C²/(N·m²)=N·m²/C ✓
$Q_\text{enc}/\varepsilon_0$ — what $\varepsilon_0$ is physically doing in the denominator: think of $\varepsilon_0$ as the vacuum's "generosity" with field lines. The equation says: for every Coulomb of charge enclosed, the vacuum allows $1/\varepsilon_0 \approx 1.13\times10^{11}$ N·m²/C of outward flux through any bubble surrounding it. If $\varepsilon_0$ were ten times bigger, the same charge would produce ten times less flux — the vacuum would "absorb" the field more, weakening all electric forces. The actual value $\varepsilon_0 = 8.85\times10^{-12}$ C²/(N·m²) is small, meaning the vacuum is quite "generous" — it lets charge project a lot of flux outward into space.
The closed integral $\oint$ — what you're measuring geometrically: you draw any closed bubble around a region. It doesn't have to be a sphere — it can be a potato. At every tiny patch $d\vec{A}$ of that bubble, you measure $E\cos\theta$ (the outward-pointing component of the field at that patch) times the patch area. Add all those up around the entire surface. If there's a positive charge inside, more field lines exit than enter — positive total. If there's negative charge inside, more enter than exit — negative total. Charges outside send field lines in one side and out the other, canceling to zero net contribution. This is why Gauss's Law "ignores" everything outside the surface: the math enforces it exactly.
Why this forces $E \propto 1/r^2$ for a sphere: suppose you have charge $q$ at the center of a sphere of radius $r$. By symmetry, $E$ is the same everywhere on the sphere and points radially outward. So the integral becomes simply $E \times 4\pi r^2 = q/\varepsilon_0$. Solving: $E = q/(4\pi\varepsilon_0 r^2)$. The sphere's area $4\pi r^2$ grows as $r^2$, and the total flux $q/\varepsilon_0$ is fixed — so $E$ must shrink as $1/r^2$ to compensate. The inverse-square law is not assumed — it falls out from the geometry of a sphere and the conservation of flux.
The closed integral $\oint$ — what you're measuring geometrically: you draw any closed bubble around a region. It doesn't have to be a sphere — it can be a potato. At every tiny patch $d\vec{A}$ of that bubble, you measure $E\cos\theta$ (the outward-pointing component of the field at that patch) times the patch area. Add all those up around the entire surface. If there's a positive charge inside, more field lines exit than enter — positive total. If there's negative charge inside, more enter than exit — negative total. Charges outside send field lines in one side and out the other, canceling to zero net contribution. This is why Gauss's Law "ignores" everything outside the surface: the math enforces it exactly.
Why this forces $E \propto 1/r^2$ for a sphere: suppose you have charge $q$ at the center of a sphere of radius $r$. By symmetry, $E$ is the same everywhere on the sphere and points radially outward. So the integral becomes simply $E \times 4\pi r^2 = q/\varepsilon_0$. Solving: $E = q/(4\pi\varepsilon_0 r^2)$. The sphere's area $4\pi r^2$ grows as $r^2$, and the total flux $q/\varepsilon_0$ is fixed — so $E$ must shrink as $1/r^2$ to compensate. The inverse-square law is not assumed — it falls out from the geometry of a sphere and the conservation of flux.
Recovering Coulomb's Law from units alone
Sphere of radius $r$ around point charge $q$. By symmetry, $E$ is constant and radial everywhere on the sphere. Left side: $E\times4\pi r^2$ (N/C × m²). Right side: $q/\varepsilon_0$ (C ÷ C²/(N·m²) = N·m²/C). Setting equal: $E\times4\pi r^2 = q/\varepsilon_0$, so $E = q/(4\pi\varepsilon_0 r^2)$. Unit check: C/(C²/(N·m²)·m²) = C·N·m²/(C²·m²) = N/C ✓. The $4\pi r^2$ from the sphere's area is what makes Coulomb's $1/r^2$ appear — the area grows as $r^2$, so $E$ must shrink as $1/r^2$ to keep the product (flux) constant.
equivε₀ — permittivity of free space
Form
Derived from
Physical meaning
C²/(N·m²)
$k=1/(4\pi\varepsilon_0)$
Coulomb/Gauss form — "how much flux per enclosed Coulomb does vacuum permit"
F/m
$C=\varepsilon_0 A/d$
Capacitor form — "Farads of capacitance per meter of plate geometry"
A²·s⁴/(kg·m³)
Full SI (C=A·s)
Maxwell's eqs: $c=1/\sqrt{\mu_0\varepsilon_0}$ — speed of light encoded in $\varepsilon_0$
Common Gaussian surface results
| Geometry | Surface | Result + unit check |
|---|---|---|
| Point charge $q$ | Sphere, radius $r$ | $E=q/(4\pi\varepsilon_0 r^2)$: C/(F/m·m²) = C·m/(F·m²) ... = N/C ✓ |
| Line $\lambda$ (C/m) | Cylinder, radius $r$ | $E=\lambda/(2\pi\varepsilon_0 r)$: (C/m)/((C²/(N·m²))·m) = N/C ✓ |
| Sheet $\sigma$ (C/m²) | Pillbox, area $A$ | $E=\sigma/(2\varepsilon_0)$: (C/m²)/(C²/(N·m²)) = N/C ✓ |
24
Chapter 24
Electric Potential & Potential Energy
24.1Electric Potential Energy — $U = kq_1q_2/r$
$$U = \frac{1}{4\pi\varepsilon_0}\frac{q_1 q_2}{r}$$
CHECK:
N·m²/C²×
C²/m=
N·m = J ✓
J as N·m — you're accounting for every push along the approach path: two positive charges start infinitely far apart. Bring one toward the other. At radius $r'$, Coulomb's Law says the repulsion is $F = kq_1q_2/r'^2$ Newtons. You push against that force through an infinitesimal step $dr'$: one tiny bit of work = $F\,dr'$ in N·m. Integrate those N·m contributions from $\infty$ all the way in to final separation $r$: $U = \int_\infty^r kq_1q_2/r'^2\,dr'$. The $r'^{-2}$ integrates to $-r'^{-1}$, so $U = kq_1q_2/r$. The m² in $k$'s numerator cancels one m from the $1/r$ denominator, leaving N·m — force times distance — which is Joules. This is the mechanical ledger of every push you made.
J as C·V — the landscape accounting: the same configuration, now seen differently. Charge $q_1$ creates a voltage landscape around itself: $V_1(r) = kq_1/r$ Volts at every point. That's a scalar field — you can read it off a voltmeter. Now place $q_2$ at distance $r$ in that landscape. The energy is $U = q_2 \cdot V_1(r)$: C × J/C = J. The Coulombs cancel. You didn't need to know anything about the path or the force at each step — just the voltage at the endpoint multiplied by the charge you placed there. This is why voltage is powerful: it encodes the entire force-integral into a single scalar number at each point.
Why $1/r$ not $1/r^2$ — the integration shifts the exponent by one: force goes as $r^{-2}$. Integrating force over distance: $\int r^{-2}\,dr = -r^{-1}$. Every time you integrate a power of $r$ over $r$, the exponent increases by one (becomes less negative). Force $\propto r^{-2}$ → energy $\propto r^{-1}$. Units confirm: $k$ contributes N·m²/C², the charges contribute C², and $1/r$ contributes $1/\mathrm{m}$. Product: N·m²/C² × C²/m = N·m = J. The one fewer $m$ in the denominator compared to Coulomb's Law is the unit signature of having integrated force over distance.
J as C·V — the landscape accounting: the same configuration, now seen differently. Charge $q_1$ creates a voltage landscape around itself: $V_1(r) = kq_1/r$ Volts at every point. That's a scalar field — you can read it off a voltmeter. Now place $q_2$ at distance $r$ in that landscape. The energy is $U = q_2 \cdot V_1(r)$: C × J/C = J. The Coulombs cancel. You didn't need to know anything about the path or the force at each step — just the voltage at the endpoint multiplied by the charge you placed there. This is why voltage is powerful: it encodes the entire force-integral into a single scalar number at each point.
Why $1/r$ not $1/r^2$ — the integration shifts the exponent by one: force goes as $r^{-2}$. Integrating force over distance: $\int r^{-2}\,dr = -r^{-1}$. Every time you integrate a power of $r$ over $r$, the exponent increases by one (becomes less negative). Force $\propto r^{-2}$ → energy $\propto r^{-1}$. Units confirm: $k$ contributes N·m²/C², the charges contribute C², and $1/r$ contributes $1/\mathrm{m}$. Product: N·m²/C² × C²/m = N·m = J. The one fewer $m$ in the denominator compared to Coulomb's Law is the unit signature of having integrated force over distance.
Hydrogen atom — energy in Joules and eV
Proton and electron at Bohr radius $r_0 = 5.29\times10^{-11}$ m. $U = k(-e)(e)/r_0 = -(8.99\times10^9)(1.6\times10^{-19})^2/(5.29\times10^{-11})$. Units: (N·m²/C²)(C²)/m = N·m = J. Result: $U = -4.36\times10^{-18}$ J = $-27.2$ eV. The negative sign means the electron is bound — you must supply 27.2 eV to pull it to infinity. Half that energy (13.6 eV) is the ionization energy; the other half was already converted to kinetic energy by the virial theorem. The unit eV = $1.6\times10^{-19}$ J is just $U = qV$ with $q = e$ and $V = 1$ V: one electron-charge times one Volt.
equivJ — Joule (energy)
Form
Derived from
When this form tells the story
J = N·m
$W=F\cdot d$
Work = force × distance. Integrating Coulomb force gives this.
J = C·V
$U=qV$
Charge × potential — most common E&M form. Battery, capacitor, particle.
J = kg·m²/s²
$KE=\frac{1}{2}mv^2$
Connects electric PE to kinetic energy: $qV=\frac{1}{2}mv^2$ in particle accelerators
J = W·s
$P=dU/dt$
Power × time: kWh = 3.6×10⁶ J; Joule heating $U=I^2Rt$
J = V²·F
$U=\frac{1}{2}CV^2$
Capacitor energy: F·V²=(C/V)·V²=C·V=J ✓
eV = 1.602×10⁻¹⁹ J
$U=qV$, $q=e$, $V=1$ V
Atomic energies: accelerating 1 electron through 1 V gives exactly 1 eV of KE
24.2Electric Potential (Voltage) — $V = U/q$
$$V = \frac{U}{q_0} = \frac{1}{4\pi\varepsilon_0}\frac{Q}{r}$$
UNITS:
Volt=
J/C=
N·m/C=
W/A=
Ω·A
J/C — you're measuring an energy exchange rate for charge in transit: a 9 V battery has a potential difference of 9 V across its terminals. That means: for every single Coulomb of positive charge that the battery pushes from its negative terminal to its positive terminal through the internal chemistry, 9 Joules of chemical energy are converted to electrical energy. The charge is the carrier; the Joule-per-Coulomb tells you the rate of that conversion. When the same Coulomb later flows through a light bulb, it drops 9 J of energy there as heat and light. The J/C unit is the throughput rate — energy delivered per unit of charge that carries it.
N·m/C — you're measuring the mechanical work done per Coulomb: inside the battery, chemical forces literally push ions across a gap. Those forces have a magnitude in Newtons; the ions move through a distance in meters. The work done on each ion: F × d = N·m. Per Coulomb of total charge moved: N·m/C = J/C = V. A 9 V battery does 9 N·m of work per Coulomb it pumps internally. This is why you can think of the battery's EMF as a "chemical force" — it's doing mechanical work on the charge, just at the molecular scale.
W/A — you're measuring delivered power per Ampere of current: connect that 9 V battery to a resistor. Current $I$ flows. The power delivered to the resistor is $P = IV$ Watts. Dividing: $V = P/I$ W/A. If 3 A flows and 27 W of heat is generated: $V = 27/3 = 9$ V. The W/A form is telling you: each Ampere of current passing through this device is carrying 9 Watts of power with it. Increase the current, the power scales proportionally because V is fixed.
Ω·A — you're measuring the field's resistance to current flow: Ohm's Law says the voltage across a resistor equals resistance × current. A 3 Ω resistor with 3 A through it: $V = 9$ Ω·A = 9 V. The Ω·A form is the circuit-level description: the resistor's geometry and material (Ω) multiplied by how much charge-per-second is flowing through it (A) gives the voltage the field must maintain across it to keep that current flowing. High resistance: needs lots of voltage per Ampere. Low resistance: drives current with little voltage.
N·m/C — you're measuring the mechanical work done per Coulomb: inside the battery, chemical forces literally push ions across a gap. Those forces have a magnitude in Newtons; the ions move through a distance in meters. The work done on each ion: F × d = N·m. Per Coulomb of total charge moved: N·m/C = J/C = V. A 9 V battery does 9 N·m of work per Coulomb it pumps internally. This is why you can think of the battery's EMF as a "chemical force" — it's doing mechanical work on the charge, just at the molecular scale.
W/A — you're measuring delivered power per Ampere of current: connect that 9 V battery to a resistor. Current $I$ flows. The power delivered to the resistor is $P = IV$ Watts. Dividing: $V = P/I$ W/A. If 3 A flows and 27 W of heat is generated: $V = 27/3 = 9$ V. The W/A form is telling you: each Ampere of current passing through this device is carrying 9 Watts of power with it. Increase the current, the power scales proportionally because V is fixed.
Ω·A — you're measuring the field's resistance to current flow: Ohm's Law says the voltage across a resistor equals resistance × current. A 3 Ω resistor with 3 A through it: $V = 9$ Ω·A = 9 V. The Ω·A form is the circuit-level description: the resistor's geometry and material (Ω) multiplied by how much charge-per-second is flowing through it (A) gives the voltage the field must maintain across it to keep that current flowing. High resistance: needs lots of voltage per Ampere. Low resistance: drives current with little voltage.
All four Volt-forms — one car battery example
12 V car battery powering a 4 Ω starter motor drawing 3 A.
— J/C: every Coulomb moved through the circuit does 12 J of mechanical work on the engine.
— N·m/C: if the effective path length for one Coulomb is 1 m, the average force on each Coulomb is 12 N.
— W/A: $P = IV = 3\times12 = 36$ W. Check: $V = P/I = 36/3 = 12$ V ✓
— Ω·A: $V = IR = 4\times3 = 12$ V ✓. Same 12 V, four completely different physical readings of it.
equivV — Volt (electric potential)
Form
Derived from
Physical meaning
V = J/C
$V=U/q$
Energy per charge — the "altitude" of a point in the voltage landscape
V = N·m/C
J = N·m
Work per charge — force × distance per Coulomb pushed through
V = W/A
$P=IV$
Power per Ampere — how many Joules/sec per Ampere of current
V = Ω·A
$V=IR$
Resistance × current — the circuit analysis workhorse
V = kg·m²/(A·s³)
Full SI
Base-unit expansion for dimensional proofs
24.3Field as Gradient of Potential — $\vec{E}=-\nabla V$
$$\vec{E} = -\nabla V = -\frac{\partial V}{\partial x}\hat{x} - \frac{\partial V}{\partial y}\hat{y} - \frac{\partial V}{\partial z}\hat{z}$$
UNITS:
V ÷ m=
(J/C) ÷ m=
(N·m/C) ÷ m=
N/C ✓
V/m — you're measuring how fast the voltage changes as you walk through space: take a voltmeter and probe two points 1 mm apart inside a charged capacitor. The voltage difference is, say, 100 V. That's 100 V over 0.001 m = 100,000 V/m. Now do the same measurement 1 cm apart in the weaker field near a battery terminal — maybe 1.5 V over 0.01 m = 150 V/m. The V/m number is the slope of the voltage landscape: how many Volts of potential you cross per meter you move in the field direction. A bigger slope means a stronger field. The V/m unit is not saying "there are Volts sitting in the air" — it's saying the voltage map has a certain steepness at that location, and that steepness is what pushes charges around.
The negative sign — field points where voltage falls fastest: $\vec{E} = -dV/dx$. Consider a room with a 9 V battery. The positive terminal is at high $V$; the negative terminal is at low $V$. If you measure $dV/dx$ pointing from − to +, it's positive (voltage increases in that direction). So $E_x = -dV/dx$ is negative — field points the other way, from + to −. Positive charges naturally flow in the direction of $\vec{E}$ (downhill in voltage), from + terminal through the circuit to − terminal. Electrons go the opposite way. The negative sign in $\vec{E} = -\nabla V$ is the mathematical statement that field points downhill on the voltage map.
Equipotential surfaces — where V/m is zero in that direction: on a surface where $V$ is constant, $dV = 0$ for any displacement along that surface. So $\vec{E}\cdot d\vec{l} = (-dV/dl) = 0$ — the field has zero component along the surface. The field is always strictly perpendicular to equipotential surfaces. For parallel plates, equipotentials are planes parallel to the plates, and the field is perpendicular — exactly as you'd draw it. The V/m unit enforces this geometrically: the gradient of a function is always perpendicular to the level surfaces of that function.
The negative sign — field points where voltage falls fastest: $\vec{E} = -dV/dx$. Consider a room with a 9 V battery. The positive terminal is at high $V$; the negative terminal is at low $V$. If you measure $dV/dx$ pointing from − to +, it's positive (voltage increases in that direction). So $E_x = -dV/dx$ is negative — field points the other way, from + to −. Positive charges naturally flow in the direction of $\vec{E}$ (downhill in voltage), from + terminal through the circuit to − terminal. Electrons go the opposite way. The negative sign in $\vec{E} = -\nabla V$ is the mathematical statement that field points downhill on the voltage map.
Equipotential surfaces — where V/m is zero in that direction: on a surface where $V$ is constant, $dV = 0$ for any displacement along that surface. So $\vec{E}\cdot d\vec{l} = (-dV/dl) = 0$ — the field has zero component along the surface. The field is always strictly perpendicular to equipotential surfaces. For parallel plates, equipotentials are planes parallel to the plates, and the field is perpendicular — exactly as you'd draw it. The V/m unit enforces this geometrically: the gradient of a function is always perpendicular to the level surfaces of that function.
V/m in a thundercloud — when the gradient reaches breakdown
A thundercloud bottom is at $-100$ MV relative to ground, sitting 500 m above. Average field: $E = \Delta V/d = 10^8/500 = $ $2\times10^5$ V/m. Still below the $3\times10^6$ V/m breakdown threshold — no lightning yet. But the field concentrates near sharp objects and the cloud's surface. When the local $E$ exceeds $3\times10^6$ V/m, air molecules ionize: electrons are stripped off, and the air becomes conductive. The lightning bolt follows the ionized path. The relevant quantity is V/m (the spatial derivative of voltage), not the voltage itself. Same voltage, smaller distance → higher V/m → breakdown.
24.4Potential Difference & Work — $W = q\Delta V$
$$V_a - V_b = -\int_b^a\vec{E}\cdot d\vec{l}, \qquad W_{a\to b} = q(V_a - V_b)$$
LINE INTEGRAL:
(N/C)·m = J/C = V ✓
WORK:
C·V = C·(J/C) = J ✓
The line integral units — field × path = voltage: integrating $\vec{E}$ (N/C) along a path (m) gives N·m/C = J/C = V. You're dragging a test charge along the path, measuring the field's push or resistance at each step, accumulating the total work done per unit charge. The result is the voltage difference — independent of path taken (conservative field). The V/m form of $E$ makes this especially transparent: V/m × m = V. You literally add up Volts per meter times meters.
$W = q\Delta V$ — the most important equation in all of circuits: work = charge × voltage difference. Units: C × (J/C) = J. The Coulombs cancel perfectly. This is the fundamental energy transaction: every time a Coulomb flows through a voltage drop of 1 V, 1 J of energy is transferred. In a battery: chemical energy converts to electrical energy at a rate of $V$ Joules per Coulomb pumped. In a resistor: electrical energy converts to heat at a rate of $V$ Joules per Coulomb flowing through.
Why path independence matters in units: because the electric field is conservative ($\nabla\times\vec{E}=0$), the line integral depends only on endpoints. This means voltage is a well-defined scalar at each point — not a property of how you arrived there. The unit "Volt at a point" is physically meaningful. If the field were not conservative (as in time-varying situations with inductors), the path would matter, and "voltage" would not be well-defined.
$W = q\Delta V$ — the most important equation in all of circuits: work = charge × voltage difference. Units: C × (J/C) = J. The Coulombs cancel perfectly. This is the fundamental energy transaction: every time a Coulomb flows through a voltage drop of 1 V, 1 J of energy is transferred. In a battery: chemical energy converts to electrical energy at a rate of $V$ Joules per Coulomb pumped. In a resistor: electrical energy converts to heat at a rate of $V$ Joules per Coulomb flowing through.
Why path independence matters in units: because the electric field is conservative ($\nabla\times\vec{E}=0$), the line integral depends only on endpoints. This means voltage is a well-defined scalar at each point — not a property of how you arrived there. The unit "Volt at a point" is physically meaningful. If the field were not conservative (as in time-varying situations with inductors), the path would matter, and "voltage" would not be well-defined.
W = qΔV — inside a CRT television
Old CRT TV accelerates electrons through $V = 25,000$ V. Work done: $W = qV = (1.6\times10^{-19})(25000)$. Units: C × V = C × J/C = J. $W = 4\times10^{-15}$ J = 25,000 eV of kinetic energy. This gives the electron a speed: $\frac{1}{2}mv^2 = 4\times10^{-15}$, $v\approx9\times10^7$ m/s — 30% of light speed. That fast electron hits the phosphor screen, exciting it to emit light. The C·V = J cancellation is what makes every electron gun, particle accelerator, and X-ray tube work: charge times voltage equals kinetic energy gained.
25
Chapter 25
Capacitance & Dielectrics
25.1Capacitance — $C = Q/V$ and $C = \varepsilon_0 A/d$
$$C = \frac{Q}{V} = \varepsilon_0\frac{A}{d}$$
UNITS:
F = C/V=
C²/J=
s/Ω=
A·s/V
C/V — you're measuring how much charge accumulates per unit of applied voltage pressure: connect a capacitor to a 9 V battery. Electrons flow from one plate to the other through the external circuit until the voltage across the capacitor equals 9 V and current stops. At that point, $Q = CV$ Coulombs have been separated. A $100\,\mu\mathrm{F}$ capacitor holds $Q = 100\times10^{-6}\times9 = 0.9\,\mathrm{mC}$. The Farad unit is saying: for every Volt of electrical pressure the battery applies, this device separates and holds $C$ Coulombs of charge. Double the capacitance, double the charge stored at the same voltage. The Farad is literally the "charge-storage efficiency" at a given voltage.
s/Ω — you're reading how long it takes the capacitor to charge or discharge through a resistor: this is the most important form for circuits. $\tau = RC$ must come out in seconds. Since $F = s/\Omega$: $\Omega \times F = \Omega \times s/\Omega = s$. The seconds cancel the Ohms. What that means physically: a 1 kΩ resistor limits the current into the capacitor; a 1 mF capacitor has more charge to accumulate. The product $RC = 10^3 \times 10^{-3} = 1\,\mathrm{s}$ is how long it takes the voltage to rise to 63% of its final value. This isn't a coincidence of units — it's why Ohms and Farads were defined the way they were. The time constant $\tau = RC$ is what physically governs every charging waveform you'll ever see on an oscilloscope.
A·s/V — you're measuring how much current is needed to change the voltage at a given rate: $I = C\,dV/dt$. If you want to ramp the voltage across a 100 µF capacitor at 1000 V/s, you need a current of $I = (100\times10^{-6})(1000) = 0.1\,\mathrm{A}$. Units: (A·s/V)·(V/s) = A ✓. This form makes explicit that capacitors don't block DC — they resist rapid voltage change. At DC, $dV/dt = 0$, so $I = 0$: the capacitor appears as an open circuit. At high frequency, $dV/dt$ is large, so a large current flows — the capacitor appears nearly short-circuited. The F = A·s/V unit encodes the entire frequency-response story.
s/Ω — you're reading how long it takes the capacitor to charge or discharge through a resistor: this is the most important form for circuits. $\tau = RC$ must come out in seconds. Since $F = s/\Omega$: $\Omega \times F = \Omega \times s/\Omega = s$. The seconds cancel the Ohms. What that means physically: a 1 kΩ resistor limits the current into the capacitor; a 1 mF capacitor has more charge to accumulate. The product $RC = 10^3 \times 10^{-3} = 1\,\mathrm{s}$ is how long it takes the voltage to rise to 63% of its final value. This isn't a coincidence of units — it's why Ohms and Farads were defined the way they were. The time constant $\tau = RC$ is what physically governs every charging waveform you'll ever see on an oscilloscope.
A·s/V — you're measuring how much current is needed to change the voltage at a given rate: $I = C\,dV/dt$. If you want to ramp the voltage across a 100 µF capacitor at 1000 V/s, you need a current of $I = (100\times10^{-6})(1000) = 0.1\,\mathrm{A}$. Units: (A·s/V)·(V/s) = A ✓. This form makes explicit that capacitors don't block DC — they resist rapid voltage change. At DC, $dV/dt = 0$, so $I = 0$: the capacitor appears as an open circuit. At high frequency, $dV/dt$ is large, so a large current flows — the capacitor appears nearly short-circuited. The F = A·s/V unit encodes the entire frequency-response story.
F = s/Ω — camera flash circuit, all in units
Camera flash: $C = 1000\ \mu\mathrm{F}$, charging resistor $R_1 = 10\ \mathrm{k\Omega}$, discharge resistor (flash tube) $R_2 = 0.1\ \Omega$. Charging time constant: $\tau_1 = R_1 C = 10^4\times10^{-3} = $ 10 s. That's the ~10 second wait between flashes. Discharge time constant: $\tau_2 = R_2 C = 0.1\times10^{-3} = $ 100 µs. That's how long the flash lasts. Units every step: Ω × F = (s/F) × F = s ✓. The dramatic difference in time constants (5 orders of magnitude) comes entirely from the 5-order-of-magnitude difference in resistance — capacitance is the same.
equivF — Farad (capacitance)
Form
Derived from
Physical meaning
F = C/V
$C=Q/V$
Charge per volt — Coulombs accumulated per volt of applied pressure
F = C²/J
C/V = C/(J/C) = C²/J
Energy form: $U=Q^2/(2C)$, so C²/F=J ✓
F = s/Ω
$\tau=RC$ must be seconds
RC time constant: Ω×F=s. The most circuit-critical equivalence.
F = A·s/V
C=A·s, so C/V=A·s/V
Current form: $I=C\,dV/dt$ → (A·s/V)·(V/s)=A ✓. Capacitor "resists" voltage change.
F = A²·s⁴/(kg·m²)
Full SI
Dimensional proofs; also in LC resonance $\omega=1/\sqrt{LC}$
25.2Energy Stored — $U = \frac{1}{2}CV^2 = Q^2/(2C) = \frac{1}{2}QV$
$$U = \frac{1}{2}CV^2 = \frac{Q^2}{2C} = \frac{1}{2}QV$$
ALL THREE:
(C/V)·V² = C·V = J ✓
C²/(C²/J) = J ✓
C·(J/C) = J ✓
$\frac{1}{2}CV^2$ — you're accounting for the fact that charging costs less at the beginning than at the end: picture pumping charge onto the plates. The first trickle of charge arrives with no opposing field — it costs almost nothing to place. But now the plates have a tiny voltage. The next bit of charge must be pushed against that voltage. More charge → more voltage → more work per charge to add the next bit. The total cost is $\int_0^Q (q/C)\,dq = Q^2/2C = \frac{1}{2}CV^2$. The $\frac{1}{2}$ is the "average voltage" you worked against — you went from 0 to $V$, so you paid an average of $V/2$. Units: (C/V)·V² = C·V = C·(J/C) = J. The Volts and Coulombs each cancel one level, leaving pure Joules.
$Q^2/(2C)$ — you're computing energy from what's actually on the plates: use this form when the charge is the fixed quantity — for example, if you charge a capacitor and then disconnect the battery. Now $Q$ is fixed. If you pull the plates apart, $d$ increases, $C = \varepsilon_0 A/d$ decreases, and $U = Q^2/(2C)$ increases — you had to do work against the attractive force between opposite plates. Units: C²/F = C²/(C²/J) = J. The Coulombs squared cancel against the C² buried in the Farad, leaving Joules.
$\frac{1}{2}QV$ — the most transparent form geometrically: $Q$ Coulombs were moved, each through an average voltage of $V/2$. Work = charge × voltage = $Q \times V/2$. This is exactly the area of a triangle on a $V$-vs-$Q$ graph: base $Q$, height $V$, area $\frac{1}{2}QV$. Units: C × V = C × J/C = J. The $\frac{1}{2}$ is geometric — it's the area under the straight line $V = Q/C$, which starts at the origin and rises linearly. A rectangle would be $QV$; the triangle is half that.
$Q^2/(2C)$ — you're computing energy from what's actually on the plates: use this form when the charge is the fixed quantity — for example, if you charge a capacitor and then disconnect the battery. Now $Q$ is fixed. If you pull the plates apart, $d$ increases, $C = \varepsilon_0 A/d$ decreases, and $U = Q^2/(2C)$ increases — you had to do work against the attractive force between opposite plates. Units: C²/F = C²/(C²/J) = J. The Coulombs squared cancel against the C² buried in the Farad, leaving Joules.
$\frac{1}{2}QV$ — the most transparent form geometrically: $Q$ Coulombs were moved, each through an average voltage of $V/2$. Work = charge × voltage = $Q \times V/2$. This is exactly the area of a triangle on a $V$-vs-$Q$ graph: base $Q$, height $V$, area $\frac{1}{2}QV$. Units: C × V = C × J/C = J. The $\frac{1}{2}$ is geometric — it's the area under the straight line $V = Q/C$, which starts at the origin and rises linearly. A rectangle would be $QV$; the triangle is half that.
Defibrillator — 400 J in three unit checks
$C = 32\ \mu\mathrm{F}$, charged to $V = 5000$ V. $U = \frac{1}{2}CV^2 = \frac{1}{2}(32\times10^{-6})(5000)^2 = $ 400 J. Units: (C/V)·V² = C·V = J ✓. Charge stored: $Q = CV = (32\times10^{-6})(5000) = 0.16$ C. Cross-check $Q^2/(2C) = (0.16)^2/(2\times32\times10^{-6}) = 0.0256/6.4\times10^{-5} = $ 400 J ✓. Check $\frac{1}{2}QV = \frac{1}{2}(0.16)(5000) = $ 400 J ✓. That 400 J delivered to a stopped heart in milliseconds — the unit chain C·V = J is what gets converted to mechanical cardiac energy.
25.3Energy Density — $u = \frac{1}{2}\varepsilon_0 E^2$
$$u = \frac{1}{2}\varepsilon_0 E^2 \qquad \left[\mathrm{J/m^3}\right]$$
CHECK:
C²/(N·m²)×
(N/C)²=
N/m²=
J/m³ ✓
J/m³ — you're measuring how much energy is trapped in each cubic meter of the field itself: this is the startling claim: the region of space between capacitor plates contains energy, not because there's matter there, but because there's a field there. Take a 1 cm³ cube of air between plates where $E = 10^6$ V/m. Energy in that cube: $u\times\mathrm{vol} = \frac{1}{2}(8.85\times10^{-12})(10^6)^2 \times 10^{-6} = 4.4\,\mathrm{mJ}$. That's real stored energy — discharge the capacitor and it comes out as heat and current. The J/m³ tells you the spatial density of that energy, which you then integrate over the volume of the field to get the total.
N/m² = Pa — the field pushes on surfaces with a real mechanical pressure: J/m³ = N·m/m³ = N/m². The energy density has the same dimensions as pressure. And it is pressure: the field between two oppositely charged plates pushes the plates toward each other with a force per unit area equal to $\frac{1}{2}\varepsilon_0 E^2$ Pascals. This is electrostatic pressure. If $E = 10^6$ V/m between the plates: pressure = $\frac{1}{2}(8.85\times10^{-12})(10^6)^2 \approx 4.4$ Pa — about the pressure of a gentle breeze. At $E = 3\times10^6$ V/m (air breakdown): pressure ≈ 40 Pa. The field is quite literally pushing the plates together. When you pull them apart, you're doing work against that pressure — and that work goes into the field energy.
Why $E^2$ — the field energy is quadratic, like kinetic energy: $\frac{1}{2}\varepsilon_0 E^2$ mirrors $\frac{1}{2}mv^2$. Double the field: four times the stored energy per cubic meter. This is why electrical breakdowns are explosively violent — you can charge a region to near-breakdown field strength, storing enormous energy density, and when it discharges, all that $\frac{1}{2}\varepsilon_0 E^2 \times \mathrm{volume}$ releases in microseconds. The $\varepsilon_0$ is playing the role of "inertia" — it tells you how much energy it costs the vacuum to maintain a field of strength $E$.
N/m² = Pa — the field pushes on surfaces with a real mechanical pressure: J/m³ = N·m/m³ = N/m². The energy density has the same dimensions as pressure. And it is pressure: the field between two oppositely charged plates pushes the plates toward each other with a force per unit area equal to $\frac{1}{2}\varepsilon_0 E^2$ Pascals. This is electrostatic pressure. If $E = 10^6$ V/m between the plates: pressure = $\frac{1}{2}(8.85\times10^{-12})(10^6)^2 \approx 4.4$ Pa — about the pressure of a gentle breeze. At $E = 3\times10^6$ V/m (air breakdown): pressure ≈ 40 Pa. The field is quite literally pushing the plates together. When you pull them apart, you're doing work against that pressure — and that work goes into the field energy.
Why $E^2$ — the field energy is quadratic, like kinetic energy: $\frac{1}{2}\varepsilon_0 E^2$ mirrors $\frac{1}{2}mv^2$. Double the field: four times the stored energy per cubic meter. This is why electrical breakdowns are explosively violent — you can charge a region to near-breakdown field strength, storing enormous energy density, and when it discharges, all that $\frac{1}{2}\varepsilon_0 E^2 \times \mathrm{volume}$ releases in microseconds. The $\varepsilon_0$ is playing the role of "inertia" — it tells you how much energy it costs the vacuum to maintain a field of strength $E$.
Field energy = capacitor energy — same answer two ways
Parallel plates: $A = 0.01$ m², $d = 1$ mm, $V = 100$ V. $C = \varepsilon_0 A/d = (8.85\times10^{-12})(0.01/10^{-3}) = 88.5$ pF. Circuit formula: $U = \frac{1}{2}CV^2 = \frac{1}{2}(88.5\times10^{-12})(100)^2 = $ 442 nJ. Field formula: $E = V/d = 10^5$ V/m. $u = \frac{1}{2}\varepsilon_0 E^2 = \frac{1}{2}(8.85\times10^{-12})(10^5)^2 = 44.25$ mJ/m³. Volume = $Ad = 10^{-5}$ m³. $U = u\cdot\text{Vol} = 44.25\times10^{-3}\times10^{-5} = $ 442 nJ ✓. Identical — because the energy really is in the field, and the capacitor formula is just the compact version of the field integral.
26
Chapter 26
Current, Resistance & Ohm's Law
26.1Current — $I = dQ/dt$ and $\vec{J} = nqv_d$
$$I = \frac{dQ}{dt}, \qquad \vec{J} = nqv_d, \qquad I = \int\vec{J}\cdot d\vec{A}$$
I UNITS:
A = C/s=V/Ω=W/V
J UNITS:
A/m²=C/(s·m²)
C/s — you're counting how much charge crosses an imaginary plane per second: slice a copper wire with an imaginary plane perpendicular to the wire. Count the net Coulombs of charge that cross that plane every second — that's the current in C/s = A. At 1 A, $6.24\times10^{18}$ electrons cross per second. But each electron drifts at only ~0.02 mm/s (drift velocity). The reason 1 A is enormous in terms of electrons, yet the electrons barely move, is that copper has $8.5\times10^{28}$ free electrons per m³ — the density is astronomical. The C/s unit is a pure throughput measurement: Coulombs per second flowing past a cross-section, regardless of how fast the individual carriers move.
V/Ω — you're measuring the flow that results from a voltage "pressure" fighting a resistance: apply 12 V across a 4 Ω resistor. The voltage pushes; the resistance fights back. The resulting current: $I = V/R = 12/4 = 3$ A. Units: V/Ω = (J/C)/(J·s/C²) = C/s = A ✓. The Ohm was defined so that this ratio gives Amperes cleanly. Every time you compute $V/R$, you're finding the equilibrium flow rate where the voltage's "pressure" exactly balances the resistor's opposition.
W/V — you're reading current from power consumption and voltage simultaneously: a 1200 W hair dryer running on 120 V. The power tells you energy per second; the voltage tells you energy per Coulomb. Divide: $I = P/V = 1200/120 = 10$ A = 10 C/s. Units: W/V = (J/s)/(J/C) = C/s = A ✓. The Joules cancel. This form is how you size a fuse: what current does this device draw? You don't need to know the resistance — just the rated wattage and operating voltage.
A/m² for $\vec{J}$ — you're measuring current per unit of cross-sectional area, not total current: two wires, both carrying 10 A. Wire 1: 2 mm radius. Wire 2: 0.5 mm radius. Wire 2 has $(2/0.5)^2 = 16\times$ more current density $J = I/A$ (A/m²). Joule heating power per unit volume = $J^2/\sigma$ (W/m³) — quadratic in $J$. Wire 2 generates 256× more heat per cubic centimeter. This is why wire gauge matters for safety: the current in Amps is the same, but the A/m² is what determines whether the wire melts. The unit A/m² is physically more fundamental than A because it's a material-level quantity.
V/Ω — you're measuring the flow that results from a voltage "pressure" fighting a resistance: apply 12 V across a 4 Ω resistor. The voltage pushes; the resistance fights back. The resulting current: $I = V/R = 12/4 = 3$ A. Units: V/Ω = (J/C)/(J·s/C²) = C/s = A ✓. The Ohm was defined so that this ratio gives Amperes cleanly. Every time you compute $V/R$, you're finding the equilibrium flow rate where the voltage's "pressure" exactly balances the resistor's opposition.
W/V — you're reading current from power consumption and voltage simultaneously: a 1200 W hair dryer running on 120 V. The power tells you energy per second; the voltage tells you energy per Coulomb. Divide: $I = P/V = 1200/120 = 10$ A = 10 C/s. Units: W/V = (J/s)/(J/C) = C/s = A ✓. The Joules cancel. This form is how you size a fuse: what current does this device draw? You don't need to know the resistance — just the rated wattage and operating voltage.
A/m² for $\vec{J}$ — you're measuring current per unit of cross-sectional area, not total current: two wires, both carrying 10 A. Wire 1: 2 mm radius. Wire 2: 0.5 mm radius. Wire 2 has $(2/0.5)^2 = 16\times$ more current density $J = I/A$ (A/m²). Joule heating power per unit volume = $J^2/\sigma$ (W/m³) — quadratic in $J$. Wire 2 generates 256× more heat per cubic centimeter. This is why wire gauge matters for safety: the current in Amps is the same, but the A/m² is what determines whether the wire melts. The unit A/m² is physically more fundamental than A because it's a material-level quantity.
Drift velocity calculation — C/s with incredibly slow electrons
Copper wire, 1 mm radius, $I = 1$ A. $n = 8.5\times10^{28}$ m⁻³, $A = \pi(10^{-3})^2 = 3.14\times10^{-6}$ m². $v_d = I/(nqA) = 1/(8.5\times10^{28}\times1.6\times10^{-19}\times3.14\times10^{-6})$. Units: A/[(m⁻³)(C)(m²)] = (C/s)/(C/m·s... ) — let's just compute: denominator = $4.27\times10^4$ C/(m·s). $v_d = 1/4.27\times10^4 \approx $ $2.3\times10^{-5}$ m/s = 0.023 mm/s. The electrons crawl at 0.023 mm/s. Yet when you flip a switch, the light responds in nanoseconds — because the electric field signal propagates at near $c$, not at $v_d$. The C/s unit is about charge flow; the signal speed is electromagnetic.
equivA — Ampere (current)
Form
Derived from
Physical meaning
A = C/s
$I=dQ/dt$
Charge per second — the definition. $Q=It$: current × time = charge.
A = V/Ω
$I=V/R$, Ohm's law
Circuit analysis: given voltage drop and resistance, find current
A = W/V
$P=IV$, so $I=P/V$
Sizing: given power demand and voltage, find current to size wire/fuse
A = √(W/Ω)
$P=I^2R$, so $I=\sqrt{P/R}$
Thermal: current from power dissipation and known resistance
26.2Ohm's Law — $V=IR$, $\vec{J}=\sigma\vec{E}$, $R=\rho L/A$
$$V = IR, \qquad \vec{J} = \sigma\vec{E}, \qquad R = \rho\frac{L}{A}$$
R UNITS:
Ω = V/A=
J·s/C²=
W/A²=
s/F
V/A — you're measuring how many Volts of pressure it takes to push one Ampere through: connect an ohmmeter across a resistor. It applies a known voltage, measures the resulting current, divides. A 100 Ω resistor needs 100 V to drive 1 A, or 10 V to drive 0.1 A, or 1 V to drive 10 mA — always the same ratio. That ratio is what resistance means: Volts of driving pressure per Ampere of resulting flow. A low resistance is like a wide pipe — a small pressure drives a large flow. A high resistance is a narrow pipe — you need lots of pressure for even a trickle. The V/A unit is a literal measurement of that pipe's narrowness.
J·s/C² — you're measuring energy dissipated per unit of charge-flow-squared: $\Omega = J\cdot s/C^2$. This looks cryptic until you plug it into $P = I^2 R$: $(C/s)^2 \times (J\cdot s/C^2) = C^2/s^2 \times J\cdot s/C^2 = J/s = W$. The $C^2$ from $I^2$ cancels the $C^2$ in the denominator of the Ohm; the seconds nearly cancel except one $s$ in J·s divided by $s^2$ gives $s^{-1}$, leaving J/s = W. What it's saying physically: the resistance sets how many Joules of heat are generated per second, per (Coulombs-per-second)² of current. A 10 Ω resistor with 3 A generates $9 \times 10 = 90$ W of heat — and the J·s/C² unit is what makes the dimensional bookkeeping of $I^2R$ produce Watts.
W/A² — you're reading resistance from a heat measurement and a current measurement: you have a heating element. You run 10 A through it, put your hand near it, and measure 1200 W of radiated heat (by calorimetry or a power meter). $R = P/I^2 = 1200/100 = 12\ \Omega$. Units: W/A² = (J/s)/(C/s)² = (J/s)·s²/C² = J·s/C² = Ω ✓. This is the thermal engineer's definition of resistance: "how many Watts of heat does this thing generate per square Ampere of current?" The quadratic $I^2$ matters — double the current and the heating quadruples.
s/F — you're reading resistance as a time constant per unit capacitance: $\tau = RC$ must come out in seconds. So $R = \tau/C$ must be in s/F. Check: s/F = s/(s/\Omega) = \Omega ✓. The physical meaning: connect this resistor to a 1 F capacitor. The time it takes for the voltage to reach 63% of its final value is exactly $R$ seconds. Every Ohm of resistance "costs" 1 second of charging time per Farad. A 10 kΩ resistor charging a 100 µF capacitor: $\tau = 10^4 \times 10^{-4} = 1$ s — and the s/F unit is why that works dimensionally without any conversion factor.
$\vec{J} = \sigma\vec{E}$ — Ohm's Law stated as a material property, without any wire geometry: apply a field $E$ (V/m) inside a block of material. Electrons feel that force and drift. The resulting current density $J$ (A/m²) is proportional to the field: $J = \sigma E$. The conductivity $\sigma$ has units A/(V·m) — it tells you how many Amps flow per square meter of cross-section, per V/m of applied field. Copper: $\sigma \approx 6\times10^7$ A/(V·m) — 1 V/m drives $6\times10^7$ A through every square meter. Rubber: $\sigma \approx 10^{-15}$ A/(V·m) — 22 orders of magnitude less. That's the entire conductor-insulator spectrum, encoded in a single number with units A/(V·m).
J·s/C² — you're measuring energy dissipated per unit of charge-flow-squared: $\Omega = J\cdot s/C^2$. This looks cryptic until you plug it into $P = I^2 R$: $(C/s)^2 \times (J\cdot s/C^2) = C^2/s^2 \times J\cdot s/C^2 = J/s = W$. The $C^2$ from $I^2$ cancels the $C^2$ in the denominator of the Ohm; the seconds nearly cancel except one $s$ in J·s divided by $s^2$ gives $s^{-1}$, leaving J/s = W. What it's saying physically: the resistance sets how many Joules of heat are generated per second, per (Coulombs-per-second)² of current. A 10 Ω resistor with 3 A generates $9 \times 10 = 90$ W of heat — and the J·s/C² unit is what makes the dimensional bookkeeping of $I^2R$ produce Watts.
W/A² — you're reading resistance from a heat measurement and a current measurement: you have a heating element. You run 10 A through it, put your hand near it, and measure 1200 W of radiated heat (by calorimetry or a power meter). $R = P/I^2 = 1200/100 = 12\ \Omega$. Units: W/A² = (J/s)/(C/s)² = (J/s)·s²/C² = J·s/C² = Ω ✓. This is the thermal engineer's definition of resistance: "how many Watts of heat does this thing generate per square Ampere of current?" The quadratic $I^2$ matters — double the current and the heating quadruples.
s/F — you're reading resistance as a time constant per unit capacitance: $\tau = RC$ must come out in seconds. So $R = \tau/C$ must be in s/F. Check: s/F = s/(s/\Omega) = \Omega ✓. The physical meaning: connect this resistor to a 1 F capacitor. The time it takes for the voltage to reach 63% of its final value is exactly $R$ seconds. Every Ohm of resistance "costs" 1 second of charging time per Farad. A 10 kΩ resistor charging a 100 µF capacitor: $\tau = 10^4 \times 10^{-4} = 1$ s — and the s/F unit is why that works dimensionally without any conversion factor.
$\vec{J} = \sigma\vec{E}$ — Ohm's Law stated as a material property, without any wire geometry: apply a field $E$ (V/m) inside a block of material. Electrons feel that force and drift. The resulting current density $J$ (A/m²) is proportional to the field: $J = \sigma E$. The conductivity $\sigma$ has units A/(V·m) — it tells you how many Amps flow per square meter of cross-section, per V/m of applied field. Copper: $\sigma \approx 6\times10^7$ A/(V·m) — 1 V/m drives $6\times10^7$ A through every square meter. Rubber: $\sigma \approx 10^{-15}$ A/(V·m) — 22 orders of magnitude less. That's the entire conductor-insulator spectrum, encoded in a single number with units A/(V·m).
R = ρL/A — extension cord fire risk in units
Copper $\rho = 1.68\times10^{-8}\ \Omega\cdot\mathrm{m}$. Extension cord: $L = 30$ m, $A = 0.5\ \mathrm{mm}^2 = 5\times10^{-7}\ \mathrm{m}^2$. $R = (1.68\times10^{-8})(30)/(5\times10^{-7})$. Units: (Ω·m)(m)/m² = Ω. $R = 1.0\ \Omega$. Running a 1200 W vacuum at 120 V: $I = P/V = 10$ A. Cord heating: $P_\text{cord} = I^2R = 100\times1.0 = $ 100 W wasted as heat in the cord — that's why cheap extension cords with thin wire get hot. A 14 AWG cord ($A = 2.08\ \mathrm{mm}^2$) would have $R = 0.24\ \Omega$ and waste only 24 W. The unit chain Ω·m × m/m² = Ω is exactly what's protecting (or failing to protect) your house.
equivΩ — Ohm (resistance)
Form
Derived from
Physical meaning
Ω = V/A
$R=V/I$
Volts per Ampere — the definition of resistance
Ω = J·s/C²
V/A = (J/C)/(C/s)
Energy dissipation: $P=I^2R=(C/s)^2\cdot(J\cdot s/C^2)=J/s=W$ ✓
Ω = W/A²
$P=I^2R$, so $R=P/I^2$
Thermal engineering: resistance from measured power and current
Ω = s/F
$\tau=RC$ must be seconds
RC time constants: R×C=s. Know this cold for Ch. 28+.
Ω = kg·m²/(A²·s³)
Full SI
Dimensional proofs from base units
26.3Power — $P=IV=I^2R=V^2/R$
$$P = IV = I^2R = \frac{V^2}{R}$$
ALL THREE:
A·V=(C/s)·(J/C)=J/s=W ✓
A²·Ω=(C/s)²·(J·s/C²)=J/s=W ✓
V²/Ω=V·(V/Ω)=V·A=W ✓
$P = IV$ — you're watching two quantities cancel to reveal energy per second: current brings Coulombs per second past a cross-section; voltage brings Joules per Coulomb stored at that potential. Multiply: (C/s) × (J/C). The Coulombs cancel. You're left with J/s — Joules being delivered every second. This cancellation is exact and physically meaningful: the Coulombs are the carrier that transports energy from the voltage source to the load. They pick up $V$ joules per Coulomb at the source, carry them through the wire at a rate of $I$ Coulombs per second, and deposit them at the load. The Coulombs themselves don't accumulate — they're a conveyor belt. The W = V·A unit is the throughput of that belt.
$P = I^2R$ — you're measuring heat generation, and it's quadratic in current — this is the dangerous one: substitute $V = IR$ into $P = IV$. Current appears twice: $P = I(IR) = I^2R$. Units: $(C/s)^2 \times (J\cdot s/C^2) = J/s = W$ ✓. The squaring is what makes overcurrent situations dangerous. At 10 A, a 1 Ω wire dissipates 100 W. At 20 A — double the current — it dissipates 400 W, four times as much heat, in the same wire. Wire insulation melts, and fires start, not because the voltage got too high, but because $I^2R$ grew out of control. Every fuse, every circuit breaker, every current rating on a wire or device exists to keep $I^2R$ from reaching ignition temperatures.
$P = V^2/R$ — you're computing power at a fixed voltage, and resistance now reduces it: on a 120 V household circuit, voltage is essentially held constant by the grid. Power drawn by a device is $P = V^2/R = 14400/R$. A larger resistance draws less power — it restricts current more. A 100 W bulb: $R = 14400/100 = 144\ \Omega$. A 60 W bulb: $R = 240\ \Omega$. The brighter bulb has lower resistance, not higher, because it needs to draw more current. Unit: V²/Ω = V × (V/Ω) = V × A = W ✓. This form also explains why a short circuit (R → 0) is catastrophic: $P = V^2/R$ → ∞ as R → 0. The protective element (fuse or breaker) must open the circuit before the wire reaches that condition.
Why high-voltage transmission is efficient — all three forms working together: you need to send power $P$ from a plant to a city over a wire of resistance $R_\text{wire}$. The current needed: $I = P/V$ (from $P=IV$). The line loss: $P_\text{loss} = I^2 R_\text{wire} = (P/V)^2 R_\text{wire} = P^2 R_\text{wire}/V^2$. Double the transmission voltage $V$: loss drops by a factor of 4. The $V^2$ in the denominator of $P_\text{loss}$ is built from combining $P=IV$ and $P=I^2R$. That's why transformers were the key invention of electrical power distribution — they let you transmit at 500,000 V (tiny current, tiny $I^2R$ loss) and then step down to 120 V at the house. The unit structure of W = V·A and W = A²·Ω together are what force this engineering solution.
$P = I^2R$ — you're measuring heat generation, and it's quadratic in current — this is the dangerous one: substitute $V = IR$ into $P = IV$. Current appears twice: $P = I(IR) = I^2R$. Units: $(C/s)^2 \times (J\cdot s/C^2) = J/s = W$ ✓. The squaring is what makes overcurrent situations dangerous. At 10 A, a 1 Ω wire dissipates 100 W. At 20 A — double the current — it dissipates 400 W, four times as much heat, in the same wire. Wire insulation melts, and fires start, not because the voltage got too high, but because $I^2R$ grew out of control. Every fuse, every circuit breaker, every current rating on a wire or device exists to keep $I^2R$ from reaching ignition temperatures.
$P = V^2/R$ — you're computing power at a fixed voltage, and resistance now reduces it: on a 120 V household circuit, voltage is essentially held constant by the grid. Power drawn by a device is $P = V^2/R = 14400/R$. A larger resistance draws less power — it restricts current more. A 100 W bulb: $R = 14400/100 = 144\ \Omega$. A 60 W bulb: $R = 240\ \Omega$. The brighter bulb has lower resistance, not higher, because it needs to draw more current. Unit: V²/Ω = V × (V/Ω) = V × A = W ✓. This form also explains why a short circuit (R → 0) is catastrophic: $P = V^2/R$ → ∞ as R → 0. The protective element (fuse or breaker) must open the circuit before the wire reaches that condition.
Why high-voltage transmission is efficient — all three forms working together: you need to send power $P$ from a plant to a city over a wire of resistance $R_\text{wire}$. The current needed: $I = P/V$ (from $P=IV$). The line loss: $P_\text{loss} = I^2 R_\text{wire} = (P/V)^2 R_\text{wire} = P^2 R_\text{wire}/V^2$. Double the transmission voltage $V$: loss drops by a factor of 4. The $V^2$ in the denominator of $P_\text{loss}$ is built from combining $P=IV$ and $P=I^2R$. That's why transformers were the key invention of electrical power distribution — they let you transmit at 500,000 V (tiny current, tiny $I^2R$ loss) and then step down to 120 V at the house. The unit structure of W = V·A and W = A²·Ω together are what force this engineering solution.
All three power forms — toaster in complete units
120 V toaster, rated 1200 W. From $P=IV$: $I = 1200/120 = $ 10 A. From $R = V^2/P = 14400/1200 = $ 12 Ω. Check: $P=I^2R = 100\times12 = $ 1200 W ✓. Check: $P=V^2/R = 14400/12 = $ 1200 W ✓. Verify Ohm: $V=IR = 10\times12 = $ 120 V ✓. All four relationships close — the heating elements are 12 Ω wires, the $I^2R = 1200$ W of Joule heating is what toasts the bread, and the unit (C/s)·(J/C) = J/s = W is doing the bookkeeping every step.
equivW — Watt (power)
Form
Derived from
Physical meaning
W = J/s
$P=dU/dt$
Energy per second — the definition. 100 W = 100 J every second.
W = V·A
$P=IV$
Voltage × current — Coulombs cancel: (C/s)·(J/C)=J/s
W = A²·Ω
$P=I^2R$
Joule heating — quadratic in current. Double I → 4× heat.
W = V²/Ω
$P=V^2/R$
Fixed voltage — higher R means less power drawn
W = N·m/s
$P=Fv$ (mechanical)
Same unit as mechanical power — energy is energy, regardless of source
W = kg·m²/s³
Full SI
Proves electrical and mechanical Watts are identical physically
Why W = N·m/s = V·A is profound
Mechanical: $P=Fv$ → (N)(m/s) → kg·m²/s³. Electrical: $P=IV$ → (C/s)(J/C) → J/s → kg·m²/s³. Identical. This is why an electric motor can be rated in Watts or horsepower interchangeably, why generators convert mechanical Watts to electrical Watts, and why energy is conserved across the mechanical-electrical boundary. The unit equivalence is the mathematical statement that mechanical and electrical energy are the same physical thing.
Complete Unit Equivalence Reference
Every named unit in all its equivalent forms. Base SI: kg · m · s · A. Each equivalence is proven by the formula listed.
C — CoulombElectric charge
Form
Proven by
Physical meaning
C = A·s
$I=dQ/dt$
Current × time = charge moved. A 2 A wire in 3 s moves 6 C.
C = J/V
$V=J/C$
Energy per volt. $U=QV$: C·V=J ✓
C = F·V
$Q=CV$
Charge on a capacitor = capacitance × voltage
C = W·s/V
$P=IV$, $Q=It$
Power × time ÷ voltage — charge moved in terms of energy delivered
V — VoltElectric potential
Form
Proven by
Physical meaning
V = J/C
$V=U/q$
Energy per charge. "Electrical altitude." The definition.
V = N·m/C
J=N·m
Work per charge. Force × distance per Coulomb.
V = W/A
$P=IV$
Power per Ampere. 120 V at 0.5 A = 60 W ✓
V = Ω·A
$V=IR$
Resistance × current. The circuit analysis form.
V = kg·m²/(A·s³)
Full SI
Base unit expansion for dimensional proofs
F — FaradCapacitance
Form
Proven by
Physical meaning
F = C/V
$C=Q/V$
Coulombs per Volt. Charge stored per volt of applied pressure.
F = C²/J
C/V=C²/J
Energy form: $U=Q^2/(2C)$, C²/F=J ✓
F = s/Ω
$\tau=RC$ must be seconds
Ω×F=s. RC time constants. The most important circuit equivalence.
F = A·s/V
C/V with C=A·s
Current form: $I=C\,dV/dt$. Capacitor resists rapid voltage change.
F = A²·s⁴/(kg·m²)
Full SI
In LC resonance: $\omega=1/\sqrt{LC}$, units rad/s
Ω — OhmResistance
Form
Proven by
Physical meaning
Ω = V/A
$R=V/I$
Volts per Ampere. Voltage needed to drive 1 A through.
Ω = J·s/C²
V/A=(J/C)/(C/s)
$P=I^2R$: (C/s)²·(J·s/C²)=J/s=W ✓
Ω = W/A²
$R=P/I^2$
Resistance from measured heating power and current
Ω = s/F
$\tau=RC$
RC circuits: R=τ/C. Ω×F=s — the time constant identity.
Ω = kg·m²/(A²·s³)
Full SI
Dimensional proofs from base units
W — WattPower
Form
Proven by
Physical meaning
W = J/s
$P=dU/dt$
Energy per second. The definition.
W = V·A
$P=IV$
(C/s)·(J/C)=J/s. Coulombs cancel.
W = A²·Ω
$P=I^2R$
Joule heating. Quadratic in current — double I → 4× heat.
W = V²/Ω
$P=V^2/R$
Fixed voltage. Higher R → less power drawn.
W = N·m/s
$P=Fv$
Mechanical power. Same unit — energy is energy.
W = kg·m²/s³
Full SI
Proves mechanical W = electrical W at base unit level.
ε₀ — PermittivityPermittivity of free space
Form
Proven by
Physical meaning
C²/(N·m²)
$k=1/(4\pi\varepsilon_0)$
Fundamental form. "How much flux per Coulomb does vacuum permit."
F/m
$C=\varepsilon_0 A/d$
"Farads per meter of plate geometry." Most useful for capacitors.
A²·s⁴/(kg·m³)
Full SI
$c=1/\sqrt{\mu_0\varepsilon_0}$: speed of light is encoded in $\varepsilon_0$.
The speed of light hidden in ε₀
$c = 1/\sqrt{\mu_0\varepsilon_0}$. Units: $1/\sqrt{(\mathrm{H/m})(\mathrm{F/m})} = 1/\sqrt{\mathrm{s^2/m^2}} = \mathrm{m/s}$. Maxwell measured $\varepsilon_0$ from electrostatics and $\mu_0$ from magnetostatics in 1865, computed $c = 1/\sqrt{\mu_0\varepsilon_0}$, and got exactly the known speed of light. That's when humanity realized light is an electromagnetic wave. The F/m unit of $\varepsilon_0$ you're already using in your capacitor formulas contains that discovery.
Cross-reference: formula → unit equivalence
XREFWhich formula proves which equivalence
| Equivalence | Proven by | Algebraic path |
|---|---|---|
| N/C = V/m | $\vec{E}=-dV/dx$ | V/m=(N·m/C)/m=N/C ✓ |
| F = s/Ω | $\tau=RC$ must be seconds | Ω·F=s → F=s/Ω ✓ |
| V = Ω·A | $V=IR$ | Direct from Ohm's law ✓ |
| W = V·A | $P=IV$ | (C/s)·(J/C)=J/s=W ✓ |
| J = C·V | $U=qV$ | C·(J/C)=J ✓ |
| F/m = C²/(N·m²) | $C=\varepsilon_0 A/d$ | F/m=(C/V)/m=C/(N·m²/C·m)=C²/(N·m²) ✓ |
| Ω = W/A² | $P=I^2R$ | R=P/I², W/A²=Ω ✓ |
| J = W·s | $P=dU/dt$ | (J/s)·s=J ✓ |
| eV = 1.602×10⁻¹⁹ J | $U=qV$, $q=e$, $V=1$ V | $e\times1\ \mathrm{V}=1.6\times10^{-19}\ \mathrm{C}\times\mathrm{J/C}=1.6\times10^{-19}\ \mathrm{J}$ ✓ |
| N/m² = J/m³ | $u=\frac{1}{2}\varepsilon_0E^2$, energy density = pressure | J/m³=N·m/m³=N/m²=Pa ✓ |