01The Starting Point: Work
Everything in this story begins with one question: what does it cost to move something? That cost is work.
Primitive Definition
When a constant force \(\vec{F}\) acts on an object that moves through displacement \(\vec{d}\), the work done is:
$$W = \vec{F} \cdot \vec{d} = Fd\cos\theta$$
where \(\theta\) is the angle between the force and the direction of motion.
\(W\)WorkJ (joules)The energy transferred to (or from) the object by the force. Positive = force adds energy; negative = force removes it.
\(\vec{F}\)Force vectorN (newtons)The push or pull acting on the object — has both magnitude and direction.
\(\vec{d}\)Displacement vectorm (meters)How far the object moves and in what direction. Not total path length — the straight-line vector from start to end.
\(\cdot\)Dot product—Extracts the component of \(\vec{F}\) along \(\vec{d}\). Geometrically: \(|\vec{F}||\vec{d}|\cos\theta\). Only the part of the force in the direction of motion does work.
\(\theta\)Angle between \(\vec{F}\) and \(\vec{d}\)rad or °At \(\theta=0°\): full work. At \(\theta=90°\): zero work. At \(\theta=180°\): negative work (force opposes motion).
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Feynman intuition: Work is the force "helping" the motion. Push sideways on something moving forward — zero work. Push forward — full work. The dot product extracts exactly the part of the force that's in the direction of travel.
For a variable force — which is what we always have with electric fields — we add up infinitely many tiny pieces:
General work integral
$$W = \int_{\mathbf{a}}^{\mathbf{b}} \vec{F} \cdot d\vec{\ell}$$
\(W\)Total workJThe total energy delivered along the entire path from \(\mathbf{a}\) to \(\mathbf{b}\).
\(\vec{F}\)Force (variable)NThe force at each point — it can change in magnitude and direction as the object moves. This is why we need an integral instead of simple multiplication.
\(d\vec{\ell}\)Infinitesimal path elementmA tiny vector step along the path. We slice the path into infinitely many such steps and sum them up.
\(\mathbf{a},\,\mathbf{b}\)Start and end points—For conservative forces (like electric), the result depends only on these endpoints — not on which path you took. This is crucial for defining potential.
A variable force \(\vec{F}(x)\) acting along a path — work is the area under the \(F_\parallel\) vs. \(x\) curve.
This integral is the engine. Everything we derive later is just applying this to specific situations in electrostatics.
02Work and Potential Energy
Now here is the key connection: when a conservative force does work on an object, that work is exchanged with potential energy.
Conservative Force & Potential Energy
A force is conservative if the work done is path-independent. For any such force:
$$W_{\text{by force}} = -\Delta U = -(U_f - U_i) = U_i - U_f$$
\(W_{\text{by force}}\)Work by the fieldJEnergy delivered by the electric (or gravitational) force itself — not by you pushing it.
\(\Delta U\)Change in potential energyJ\(U_f - U_i\). If the force does positive work, PE decreases — energy moved from "stored" to "kinetic."
\(U_i,\,U_f\)Initial, final PEJPE is always defined relative to a reference point (usually infinity = 0).
\(-\)The crucial minus sign—Force does positive work → PE goes down. Ball falling under gravity does positive work, but loses gravitational PE. This enforces energy conservation.
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Why the minus sign? Think of gravity. If a ball falls down, gravity does positive work (force and motion align), but gravitational PE decreases. Work done by force = decrease in PE. Electric force is conservative — Coulomb's law is radial, like gravity.
External work = stored potential energy
$$W_{\text{external}} = +\Delta U = U_f - U_i$$
The energy you "spend" assembling a charge configuration is stored as potential energy in the system.
03Electric Potential \(V\)
The electric force on a test charge \(q\) is \(\vec{F} = q\vec{E}\). Let's compute the work done by the electric field to move \(q\) from \(\mathbf{a}\) to \(\mathbf{b}\):
Step 1 — Write the work integral for electric force
$$W_E = \int_{\mathbf{a}}^{\mathbf{b}} \vec{F} \cdot d\vec{\ell} = \int_{\mathbf{a}}^{\mathbf{b}} q\vec{E} \cdot d\vec{\ell}$$
Step 2 — Pull out q (it's a constant)
$$W_E = q\int_{\mathbf{a}}^{\mathbf{b}} \vec{E} \cdot d\vec{\ell}$$
That integral \(\displaystyle\int_{\mathbf{a}}^{\mathbf{b}} \vec{E} \cdot d\vec{\ell}\) appears so often we give it a name — the potential difference:
Electric Potential Difference (Voltage)
$$V(\mathbf{b}) - V(\mathbf{a}) = -\int_{\mathbf{a}}^{\mathbf{b}} \vec{E} \cdot d\vec{\ell}$$
\(V(\mathbf{b})-V(\mathbf{a})\)Potential difference (voltage)V = J/CEnergy per unit charge gained or lost moving from \(\mathbf{a}\) to \(\mathbf{b}\). This is what a voltmeter measures.
\(\vec{E}\)Electric fieldV/m = N/CForce-per-unit-charge at each point in space. It's the "slope" of the potential landscape — always points from high \(V\) to low \(V\).
\(d\vec{\ell}\)Path elementmInfinitesimal displacement along the path. The dot product \(\vec{E}\cdot d\vec{\ell}\) picks out how much of \(\vec{E}\) is "along" each step.
\(-\int\)Negative line integral—Walking with \(\vec{E}\) (downhill) makes \(\vec{E}\cdot d\vec{\ell}>0\), so \(V\) decreases. The minus enforces that \(\vec{E}\) points from high to low \(V\).
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Geometric picture: Think of electric potential like elevation on a map. \(\vec{E}\) points "downhill" — from high \(V\) to low \(V\), exactly like gravity. A positive charge released from rest rolls downhill; a negative charge rolls uphill.
Equipotential surfaces (concentric rings) around a positive charge. \(\vec{E}\) field lines always point perpendicular to equipotentials — always "downhill."
Potential from a Point Charge
Let's derive \(V\) for a single point charge \(Q\). We move a test charge from infinity (where \(V=0\)) to radius \(r\):
Step 1 — Coulomb field
$$\vec{E} = \frac{1}{4\pi\epsilon_0}\frac{Q}{r^2}\hat{r}$$
Step 2 — Radial path element (conservative, so pick the easy path)
$$d\vec{\ell} = dr\,\hat{r}$$
Step 3 — Plug into the definition
$$V(r) = -\int_{\infty}^{r} \vec{E}\cdot d\vec{\ell} = -\int_{\infty}^{r} \frac{Q}{4\pi\epsilon_0 r'^2}\,dr'$$
Step 4 — Evaluate: \(\int r'^{-2}\,dr' = -r'^{-1}\)
$$V(r) = -\frac{Q}{4\pi\epsilon_0}\left[-r'^{-1}\right]_{\infty}^{r} = -\frac{Q}{4\pi\epsilon_0}\left(-\frac{1}{r} + 0\right)$$
Result
$$\boxed{V(r) = \frac{1}{4\pi\epsilon_0}\frac{Q}{r}}$$
\(V(r)\)Potential at radius \(r\)VEnergy cost per unit charge to arrive at distance \(r\) from \(Q\). Positive for \(+Q\), negative for \(-Q\).
\(\frac{1}{4\pi\epsilon_0}\)Coulomb's constant \(k\)N·m²/C²\(\approx 8.99\times10^9\) N·m²/C². The \(4\pi\) comes from the surface area of a sphere — it's a 3D geometry constant.
\(Q\)Source chargeCThe charge creating the field. Positive \(Q\) → hill-like potential. Negative \(Q\) → pit-like potential.
\(r\)Distance from \(Q\)mPotential falls as \(1/r\) — slower than the field (\(1/r^2\)), because potential is the accumulated integral of the field.
Potential Energy of a Charge in a Field
Potential Energy
$$U = qV$$
\(U\)Potential energyJThe actual energy stored — this is what gets converted to kinetic energy if the charge is released.
\(q\)Test chargeCThe charge whose energy we're calculating. Can be positive or negative — this flips the sign of \(U\).
\(V\)Electric potentialV = J/CA property of the field alone, independent of \(q\). Think of it as the "price per unit charge" to be at that location.
04Potential Energy of Two Charges
Let's build our first real energy calculation. Imagine empty space. We bring two charges \(q_1\) and \(q_2\) from infinity to positions separated by \(r_{12}\).
Step A — Bring in \(q_1\) alone
Space is empty. No field. No force. No work needed.
$$W_1 = 0$$
Step B — Bring in \(q_2\) against the field of \(q_1\)
Potential at r₂ due to q₁
$$V_1(\mathbf{r}_2) = \frac{1}{4\pi\epsilon_0}\frac{q_1}{r_{12}}$$
Work done = energy stored
$$W_2 = q_2 \cdot V_1(\mathbf{r}_2) = \frac{1}{4\pi\epsilon_0}\frac{q_1 q_2}{r_{12}}$$
Two-Charge Potential Energy
$$\boxed{U = \frac{1}{4\pi\epsilon_0}\frac{q_1 q_2}{r_{12}}}$$
\(U\)Interaction potential energyJBelongs to the pair, not either charge individually. Release both and this becomes kinetic energy.
\(q_1,\,q_2\)The two chargesCSame signs → positive \(U\) (repulsive). Opposite signs → negative \(U\) (attractive, bound).
\(r_{12}\)Separation distancemAs \(r_{12}\to\infty\): \(U\to0\). As \(r_{12}\to0\): \(|U|\to\infty\) — point charges infinitely close store infinite energy.
\(\frac{1}{4\pi\epsilon_0}\)Coulomb's constantN·m²/C²\(\approx 8.99\times10^9\). Sets the scale — electric forces are strong, so small charge separations store significant energy.
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Sign meaning:
• Same-sign charges: \(U > 0\) — it cost energy to bring them together. They'll fly apart if released.
• Opposite-sign charges: \(U < 0\) — they released energy coming together. You must work to pull them apart.
Potential energy \(U\) vs. separation \(r\) for same-sign (repulsive, gold) and opposite-sign (attractive, teal) charges.
⚡ Real Life — Chemical Bonds
The negative potential energy of opposite-sign charges is what holds atoms together in ionic bonds (like NaCl). The Na⁺ and Cl⁻ ions sit at a separation where the attractive electrical PE is minimized — that's the bond. Breaking it = supplying energy equal to \(|U|\).
05Energy Stored in a Capacitor
A capacitor is two conducting plates holding \(+Q\) and \(-Q\), separated by distance \(d\).
Capacitance Definition
$$Q = CV$$
\(C\) is capacitance (Farads): how much charge per volt. Bigger plates, smaller gap → bigger \(C\).
Derivation: Work to Charge a Capacitor
Step 1 — Tiny work to move dq against current voltage V
$$dW = V\,dq = \frac{q}{C}\,dq$$
Step 2 — Integrate from 0 to final charge Q
$$W = \int_0^Q \frac{q}{C}\,dq = \frac{1}{C}\int_0^Q q\,dq$$
Step 3 — Evaluate: \(\int_0^Q q\,dq = Q^2/2\)
$$W = \frac{1}{C}\cdot\frac{Q^2}{2} = \frac{Q^2}{2C}$$
Step 4 — Use Q = CV to write all equivalent forms
$$\boxed{U = \frac{Q^2}{2C} = \frac{1}{2}CV^2 = \frac{1}{2}QV}$$
\(U\)Stored energyJTotal work to charge the capacitor from 0 to \(Q\). Available to be released into a circuit, flash bulb, or defibrillator pulse.
\(Q\)Total charge storedCCharge on the positive plate. More charge = higher voltage = more energy stored quadratically.
\(C\)CapacitanceF (farads)Geometric property: charge per volt. Bigger plates, smaller gap, dielectric material all increase \(C\).
\(V\)Voltage across platesVEnergy scales as \(V^2\) — doubling voltage quadruples stored energy. This is why high-voltage capacitors are dangerous.
\(\frac{1}{2}\)The half factor—From integrating a linearly growing voltage — you pay the average voltage \(V/2\), not the full final voltage. Same math as spring energy \(\frac{1}{2}kx^2\).
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Why ½? The first charge moves when voltage is 0 — easy. The last charge moves against the full final voltage — hard. You pay the average voltage, which is \(V/2\). Hence the ½ — it's the area of a triangle, not a rectangle.
Charging a capacitor: area under the \(V\) vs. \(Q\) line = stored energy \(U = \frac{1}{2}QV\).
⚡ Real Life — Camera Flash & Defibrillators
A camera flash charges a ~100 μF capacitor to ~330 V. Energy stored: \(U = \frac{1}{2}(100\times10^{-6})(330)^2 \approx 5.4\text{ J}\). That energy discharges in ~1 ms — releasing ~5400 W instantaneously. A defibrillator uses the same principle: ~32 J released in a pulse to restart a heart.
06Energy Density of the Electric Field
We've been saying "the capacitor stores energy." But where exactly? The answer: the energy lives in the electric field itself.
Derivation: Energy Density from a Parallel-Plate Capacitor
Known — Capacitance of parallel plates
$$C = \frac{\epsilon_0 A}{d}$$
Known — Uniform field between plates
$$E = \frac{V}{d} \implies V = Ed$$
Step 1 — Substitute into U = ½CV²
$$U = \frac{1}{2}\cdot\frac{\epsilon_0 A}{d}\cdot(Ed)^2 = \frac{1}{2}\cdot\frac{\epsilon_0 A}{d}\cdot E^2 d^2$$
Step 2 — Cancel one d
$$U = \frac{1}{2}\epsilon_0 E^2 \cdot Ad$$
Step 3 — Recognize Ad = volume \(\mathcal{V}\), then divide both sides by it
$$\boxed{u = \frac{U}{\mathcal{V}} = \frac{1}{2}\epsilon_0 E^2}$$
\(u\)Energy densityJ/m³Energy per unit volume. Every cubic meter where \(\vec{E}\neq0\) holds this much energy. The field itself is the storage medium.
\(\epsilon_0\)Permittivity of free spaceC²/(N·m²)\(\approx 8.85\times10^{-12}\). The "stretchiness" of the vacuum to electric fields — how much energy is stored per unit field squared.
\(E^2\)Field magnitude squaredV²/m²Energy density grows as the square of the field. Double the field → four times the energy density. Same quadratic as kinetic energy \(\frac{1}{2}mv^2\).
\(\frac{1}{2}\)The half factor—Traces back to the same ½ in \(\frac{1}{2}CV^2\). A universal signature of energy stored in linear systems.
Universal Energy Density
$$u_E = \frac{1}{2}\epsilon_0 E^2 \quad\left[\frac{\text{J}}{\text{m}^3}\right]$$
Not just a capacitor trick — every point in space where \(\vec{E} \neq 0\) contains energy. The field is the energy.
\(u_E\)Electric energy densityJ/m³A local quantity — integrate over any volume to get total electric energy there.
\(\frac{1}{2}\epsilon_0\)PrefactorC²/(N·m²) × ½In a material medium, \(\epsilon_0\) → \(\epsilon = \epsilon_r\epsilon_0\) where \(\epsilon_r\) is the dielectric constant — more polarizable material stores more energy per unit field.
\(E\)Local field magnitudeV/mValid for any field configuration — uniform, curved, wave fields. Wherever you can measure \(E\), you can compute local energy density.
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A profound shift in worldview. Energy belongs to the field, not just the charges. When you separate opposite charges, you create field lines — and those lines carry the energy. Light (EM waves) carries energy exactly this way, through oscillating \(E\) and \(B\) fields, even through empty space.
Energy density \(u \propto E^2\). Brightness shows where energy is concentrated — near the charges, where \(E\) is strongest.
⚡ Real Life — Lightning & MRI
Lightning: Storm clouds build fields up to ~3×10⁶ V/m (air breakdown). Energy density: \(u = \frac{1}{2}(8.85\times10^{-12})(3\times10^6)^2 \approx 40\text{ J/m}^3\). Large storm systems store terajoules this way before discharge.
MRI machines: The \(\frac{1}{2}\epsilon_0 E^2\) and analogous \(\frac{B^2}{2\mu_0}\) formulas are used to precisely budget energy in the pulse sequences that create medical images.
Total Energy from Energy Density
Total Field Energy
$$U = \int_{\text{all space}} \frac{1}{2}\epsilon_0 E^2\, d\mathcal{V}$$
\(U\)Total field energyJFound by summing local energy density over every point in space where \(E\neq0\).
\(\int_{\text{all space}}\)Volume integral—For a capacitor: just the gap volume \(Ad\). For a point charge: all of space, and the integral diverges — hinting at the deep puzzle of point-particle self-energy.
\(d\mathcal{V}\)Volume elementm³In Cartesian: \(dx\,dy\,dz\). In spherical: \(r^2\sin\theta\,dr\,d\theta\,d\phi\). Multiply local density by each tiny volume, add them all up.
08Why \(V \sim \tfrac{1}{r}\) but \(E \sim \tfrac{1}{r^2}\)?
Look at these two results side by side and you'll feel a nagging suspicion — they're almost the same equation, but one has \(r\) and the other \(r^2\). That's not a coincidence or an arbitrary choice. It is one of the most beautiful relationships in electrostatics.
The Two Equations, Side by Side
$$E = \frac{1}{4\pi\epsilon_0}\frac{Q}{r^2} \qquad\qquad V = \frac{1}{4\pi\epsilon_0}\frac{Q}{r}$$
Same source charge \(Q\). Same constant out front. Field falls as \(1/r^2\), potential falls as \(1/r\). Why?
View 1: \(V\) is the Accumulated History of \(E\)
Potential is the integral of the field — we defined it that way in Section 3:
Definition (from Section 3)
$$V(r) = -\int_{\infty}^{r} \vec{E}\cdot d\vec{\ell}$$
\(V\) at radius \(r\) is the accumulated "push" from the field along the entire path from infinity. It's a running total.
So the question "why does \(V\) fall slower than \(E\)?" is really asking: why does the running total of \(1/r^2\) give you \(1/r\)?
Step 1 — Start with the Coulomb field
$$E(r) = \frac{kQ}{r^2} = kQ \cdot r^{-2}$$
Step 2 — Integrate from \(\infty\) to \(r\)
$$V(r) = -kQ\int_{\infty}^{r} r'^{-2}\,dr'$$
Step 3 — Power rule: \(\int r^{-2}\,dr = -r^{-1}\)
$$V(r) = -kQ\left[-r'^{-1}\right]_{\infty}^{r} = -kQ\left(-\frac{1}{r} + 0\right)$$
Step 4 — Result: exponent went from −2 to −1
$$\boxed{V(r) = \frac{kQ}{r}}$$
Integration raised the power by 1. That is the entire reason \(V\) falls slower than \(E\).
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The calculus is the physics. Integration always "softens" decay. \(E\) is steep (\(1/r^2\)). \(V\) is the total accumulated effect of that field from infinity — distant weak contributions add up, producing the gentler \(1/r\) falloff. The potential "remembers" all the field that came before.
View 2: Geometric — Spreading Over a Sphere
There's a deeper geometric reason, independent of calculus. The total "electric flux" leaving charge \(Q\) is fixed by Gauss's law at \(Q/\epsilon_0\) — regardless of distance. But that fixed flux spreads over a sphere whose area grows:
Surface area of sphere at radius r
$$A_{\text{sphere}} = 4\pi r^2$$
Field = (fixed total flux) / (growing area)
$$E = \frac{Q/\epsilon_0}{4\pi r^2} = \frac{Q}{4\pi\epsilon_0 r^2} \sim \frac{1}{r^2}$$
The \(r^2\) in the denominator is literally the sphere's surface area growing in 3D. A purely geometric fact — nothing specific to electricity.
Same total flux, spread over ever-larger spheres. Area grows as \(r^2\), so field strength falls as \(1/r^2\).
\(V\) is the integral of those thinning field slices from infinity. Each slice contributes \(E(r')\,dr' \sim \frac{1}{r'^2}\,dr'\). Summing them from \(\infty\) to \(r\) gives \(\sim 1/r\) — the spreading dilutes the field quadratically, but the integral of quadratic dilution only gives linear dilution.
View 3: The Derivative — \(E\) is the Slope of \(V\)
Fundamental relationship (radial)
$$E = -\frac{dV}{dr}$$
Step 1 — Differentiate V = kQ/r = kQr⁻¹
$$\frac{dV}{dr} = kQ \cdot (-1) \cdot r^{-2} = -\frac{kQ}{r^2}$$
Step 2 — Apply E = −dV/dr
$$E = -\left(-\frac{kQ}{r^2}\right) = \frac{kQ}{r^2} \checkmark$$
We recover the Coulomb field exactly. Differentiation drops the power by 1: \(r^{-1} \to r^{-2}\).
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Three views, one truth:
1. Calculus: Integration raises the power — \(\int r^{-2}\,dr = -r^{-1}\).
2. Geometry: \(E\) is flux ÷ sphere area (\(\sim 1/r^2\)). \(V\) is the cumulative integral of that flux (\(\sim 1/r\)).
3. Derivative: \(E\) is the slope of \(V\). Slope of \(1/r\) is \(-1/r^2\). Same function, two perspectives.
The One-Line Summary
\(V \sim 1/r\) because it is the integral of \(E\), and \(\int r^{-2}\,dr = -r^{-1}\). Equivalently, \(E \sim 1/r^2\) because it is the derivative of \(V\), and \(\frac{d}{dr}(r^{-1}) = -r^{-2}\). They are the same function — one accumulated, one instantaneous.
⚡ Real Life — Antennas & Gravity
Antenna design: In the "near field" (close to antenna), potential-like \(1/r\) terms dominate. In the "far field" (radiation zone), energy-carrying wave fields fall as \(1/r\) — not \(1/r^2\). The \(1/r\) vs \(1/r^2\) distinction is the exact dividing line between reactive near-field energy and radiated far-field energy.
Gravity (same math): Gravitational potential \(\Phi \sim 1/r\) and field strength \(g \sim 1/r^2\) obey the same relationship for the same geometric reason — both are inverse-square forces in 3D. The \(1/r\) vs \(1/r^2\) pattern is a fingerprint of three spatial dimensions.