Section 01
The Big Picture — Altitude for Charges
Here's the best mental model: think of a mountain landscape. At every point on the landscape you can measure altitude. The altitude doesn't depend on whether you're a hiker or a car — it's a property of the terrain. The gradient of altitude tells you which way is downhill (that's the slope — the force on something). If you carry a heavy backpack up the hill, you do work against gravity and store potential energy.
The Electric Analogy
Replace "altitude" with Electric Potential \(V\) — a property of space set up by source charges. Replace "downhill gradient" with the Electric Field \(\mathbf{E}\). Replace "carrying a heavy backpack" with Electric Potential Energy \(U\) — which depends on both the terrain and the test charge you put in it.
Feynman one-liner: Electric potential \(V\) is a property of the landscape. Potential energy \(U\) is what you get when a charge sits in that landscape. \(V\) lives alone; \(U\) needs a charge.
Both ideas come from one master principle: the work done by the electric force when moving a charge equals the decrease in potential energy. Everything else is bookkeeping.
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Section 02
Electric Potential \(V\) — The Landscape
The electric potential at a point in space is defined as the work per unit charge done by an external agent to move a positive test charge from a reference point (usually \(r = \infty\)) to that point, without accelerating the charge (quasistatically).
\[ V = \frac{W_{\text{external}}}{q_0} \]
For a single point charge \(Q\) at the origin, integrating Coulomb's law from \(\infty\) to \(r\) gives:
\[ \boxed{V = \frac{kQ}{r} = \frac{Q}{4\pi\epsilon_0 r}} \]
Critical Point
\(V\) is a scalar (just a number, no direction). You can add potentials from multiple charges by simple addition — no vector components to juggle. This is its huge computational advantage over working with \(\mathbf{E}\).
Units
\[ 1\,\text{V} = 1\,\frac{\text{J}}{\text{C}} = 1\,\frac{\text{N}\cdot\text{m}}{\text{C}} \]
"10 Volts" means "it takes 10 joules to move 1 coulomb from the reference to this point."
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Section 03
Equation Anatomy — What Each Symbol Means
\[ V = \frac{k \, Q}{r} \]
\(V\)
Electric Potential
Volts = J/C. A scalar field — just a number at every point in space. Measured with a voltmeter.
\(k\)
Coulomb's Constant
\(k = 8.99 \times 10^9 \,\text{N·m}^2/\text{C}^2\). Also written as \(\tfrac{1}{4\pi\epsilon_0}\). Sets the strength of the electric interaction.
\(Q\)
Source Charge
Creates the potential landscape. Positive \(Q\) → positive "hill". Negative \(Q\) → negative "valley". Units: Coulombs (C).
\(r\)
Distance from Source
Potential falls off as \(1/r\) — slower than the field which falls as \(1/r^2\). Units: meters (m).
Sign Convention: The sign of \(Q\) matters enormously. Positive \(Q\) → potential decreases as you move away (\(V > 0\), hill). Negative \(Q\) → potential increases as you move away (\(V < 0\), valley). A positive test charge naturally rolls from high-\(V\) to low-\(V\).
Superposition — Multiple Charges
\[ V_{\text{total}} = \sum_i \frac{k Q_i}{r_i} \]
Since \(V\) is a scalar, this is just regular addition — no vectors!
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Section 04
The Calculus Side — Where \(V\) Comes From
The electric field \(\mathbf{E}\) tells you the force per unit charge at every point. The potential is obtained by integrating that field along any path from the reference point to your destination.
The Defining Integral
\[ \Delta V = V_B - V_A = -\int_A^B \mathbf{E} \cdot d\boldsymbol{\ell} \]
\(\Delta V\)
Potential Difference
The change in potential from A to B. Often called "voltage." This is what batteries and voltmeters actually measure.
\(-\)
The Minus Sign
Critical! Moving in the direction of \(\mathbf{E}\) decreases \(V\). Fields point "downhill" on the potential landscape.
\(\mathbf{E}\)
Electric Field (vector)
Force per unit charge at each point along the path. We dot it with the path element to extract only the component doing work along the path.
\(d\boldsymbol{\ell}\)
Path Element
Infinitesimally small vector along your integration path. The dot product \(\mathbf{E}\cdot d\boldsymbol{\ell}\) picks out the component of \(\mathbf{E}\) parallel to the path.
Why the dot product? Only the component of \(\mathbf{E}\) along your direction of travel does work on the charge. If you walk perpendicular to the field lines, the field does zero work — you stay on an equipotential surface. This is why equipotential surfaces are always perpendicular to field lines.
Worked Integral — Point Charge
Deriving \(V = kQ/r\) from scratch. For a point charge \(Q\) at the origin, \(\mathbf{E} = \frac{kQ}{r^2}\hat{r}\). We integrate along a radial path from \(\infty\) to \(r\):
- Step 1
Set up: \[ V(r) - \underbrace{V(\infty)}_{=\,0} = -\int_\infty^r \frac{kQ}{r'^2}\,dr' \]
- Step 2
Integrate \(r'^{-2}\): \[ V(r) = -\left[\frac{-kQ}{r'}\right]_\infty^r = -\left(\frac{-kQ}{r} - \frac{-kQ}{\infty}\right) \]
- Step 3
Since \(kQ/\infty = 0\): \[ V(r) = \frac{kQ}{r} \quad \checkmark \]
Path Independence
The integral gives the same answer regardless of path from A to B. The electric field is conservative — it's derived from a scalar potential. Straight line, curved arc, zigzag — you always get \(\Delta V = V_B - V_A\).
Connecting Field and Potential — The Gradient
\[ \mathbf{E} = -\nabla V = -\left(\frac{\partial V}{\partial x}\hat{x} + \frac{\partial V}{\partial y}\hat{y} + \frac{\partial V}{\partial z}\hat{z}\right) \]
The gradient points toward steepest increase of \(V\). The minus sign makes \(\mathbf{E}\) point toward steepest decrease of \(V\) — downhill on the potential landscape.
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Section 05
Electric Potential Energy \(U\) — Charge in the Landscape
\[ \boxed{U = qV = \frac{kqQ}{r}} \]
\(V\) is the "altitude per unit weight" of the terrain. Multiply by the actual "weight" (the charge \(q\)) and you get the actual stored energy \(U\).
Physical Meaning: \(U\) is the work done by an external agent to assemble the configuration — to bring charge \(q\) from infinity to its current position \(r\) from \(Q\), without accelerating it. It's also the energy you'd get back if you let \(q\) fly off to infinity freely.
Work–Energy Connection
\[ W_{\text{elec}} = q(V_A - V_B) = -(U_B - U_A) = -\Delta U \]
Work equals negative change in potential energy. The electric force does positive work when potential energy decreases (charge rolls downhill).
Systems of Multiple Charges
\[ U_{12} = \frac{kq_1 q_2}{r_{12}} \qquad U_{\text{total}} = \frac{kq_1 q_2}{r_{12}} + \frac{kq_1 q_3}{r_{13}} + \frac{kq_2 q_3}{r_{23}} \]
Sign of U matters: \(U > 0\): same-sign charges pushed together — they'd fly apart if released. \(U < 0\): opposite-sign charges — you'd have to do work to pull them apart (they're bound together).
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Section 06
Equation Anatomy — \(U = qV\)
\[ U = q \, V \]
\(U\)
Potential Energy
Measured in Joules (J). Energy stored in the configuration of charge \(q\) in potential \(V\). Property of the system, not just the field.
\(q\)
Test / Probe Charge
The charge placed in the field. Sign and magnitude matter. Negative \(q\) in positive \(V\) gives negative \(U\) — the system is bound.
\(V\)
Electric Potential at that point
The potential landscape created by all other charges — evaluated at the location of \(q\). Measured in Volts.
Worked Example — Proton Near a Nucleus
A proton (\(q = +1.6\times10^{-19}\,\text{C}\)) is placed \(r = 5\times10^{-10}\,\text{m}\) from a nucleus with charge \(Q = +3.2\times10^{-19}\,\text{C}\).
- Step 1 — V
\[ V = \frac{kQ}{r} = \frac{(8.99\times10^9)(3.2\times10^{-19})}{5\times10^{-10}} = 5.75\,\text{V} \]
- Step 2 — U
\[ U = qV = (1.6\times10^{-19})(5.75) = 9.2\times10^{-19}\,\text{J} \approx 5.75\,\text{eV} \]
Positive — the two positive charges repel; energy is stored as repulsion.
- Step 3 — Read it
\(V = 5.75\,\text{V}\) is a property of the nuclear field at that point — independent of what we put there. \(U = 9.2\times10^{-19}\,\text{J}\) is the work we did to push the proton here. Release it → flies away, converting \(U\) into kinetic energy.
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Section 07
Electric Potential vs. Potential Energy — The Key Differences
| Property | Electric Potential \(V\) | Potential Energy \(U\) |
|---|
| What it describes | Property of a point in space (the landscape) | Property of a charge in a field (energy stored) |
| Formula | \(V = kQ/r\) | \(U = qV = kqQ/r\) |
| Units | Volts (V = J/C) | Joules (J) |
| Depends on test charge? | No — exists without a test charge | Yes — multiply \(V\) by \(q\) to get \(U\) |
| Gravity analogy | Altitude (meters above sea level) | Gravitational PE = \(mgh\) (needs mass \(m\)) |
| Superposition | Add potentials: \(\sum kQ_i/r_i\) | Add pairwise energies: \(\sum kq_iq_j/r_{ij}\) |
The Key Insight: \(V\) is like elevation on a topographic map — it exists whether or not any hiker is there. \(U\) is the energy a specific hiker (charge \(q\)) stores by being at that elevation.
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Section 08 · Interactive
Explore: Potential Landscape of a Point Charge
Drag the sliders to change source charge \(Q\) and the distance \(r\) to the test charge \(q\). Watch \(V\), \(U\), and \(\mathbf{E}\) update live. The rings show equipotential surfaces — lines of constant \(V\).
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Section 09
Recap — Everything in One Place
Electric Potential \(V\)
\(V = kQ/r\) — scalar landscape. Units: Volts. Independent of test charge. Related to field via \(\mathbf{E} = -\nabla V\) or \(\Delta V = -\int \mathbf{E}\cdot d\boldsymbol{\ell}\).
Potential Energy \(U\)
\(U = qV = kqQ/r\) — energy of charge \(q\) in field. Units: Joules. Positive \(U\) = repulsive. Negative \(U\) = bound. Work by field \(= -\Delta U\).
The Integral Definition
\(\Delta V = -\int_A^B \mathbf{E}\cdot d\boldsymbol{\ell}\). Path-independent (conservative field). Gradient gives back: \(\mathbf{E} = -\nabla V\).
Superposition
\(V = \sum kQ_i/r_i\) (simple addition!). Two-charge PE: \(U = kq_1q_2/r_{12}\). Scalar math throughout — no vector components.
The Mountain Analogy — Final Form
| Gravity | Altitude \(h\) | Slope \(g = -dh/dy\) | PE = \(mgh\) |
| Electricity | Potential \(V\) | Field \(\mathbf{E} = -\nabla V\) | PE = \(U = qV\) |
Addendum · Section A1
The Courier Mental Model — Charges as Energy Delivery Agents
Here is the most important conceptual shift: a charge is not the energy itself — it is the courier that carries the energy from the source to the load. The charge is the truck; the energy is the cargo.
The Delivery Truck Analogy: Imagine a fleet of trucks (electrons) driving in a loop between a warehouse (the battery) and a factory (the resistor). The warehouse loads each truck with cargo (electrical potential energy). The trucks drive to the factory, unload their cargo (doing work), and drive back empty to get reloaded. The trucks don't disappear — they cycle. The energy (cargo) is what gets transferred.
The Key Questions This Section Answers
① How exactly does a charge "carry" energy? — Via the electric potential landscape it rides through.
② How is that energy deposited into a component? — Via work done by the electric field on the charge.
③ Where does the energy go? — Into heat (resistors), light (LEDs), motion (motors), chemical bonds (chargers).
What the Battery Really Does
A battery maintains a potential difference — it keeps one terminal at a higher electric potential than the other using chemical energy. Charges fall from high potential to low potential through the external circuit, converting their potential energy into other forms along the way.
\[ \underbrace{\varepsilon}_{\text{EMF}} = \frac{W_{\text{chem}}}{q} \quad \Longrightarrow \quad \text{energy per coulomb provided by the battery} \]
A 9V battery gives 9 Joules to every coulomb that passes through it.
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Addendum · Section A2
Volts Are Joules per Coulomb — No More, No Less
\[ 1\,\text{V} = 1\,\frac{\text{J}}{\text{C}} \]
One Volt means one Joule of energy per Coulomb of charge. When the voltage across a resistor is 5 V: for every coulomb that passes through, 5 Joules of energy is deposited.
\[ \Delta V = \frac{\Delta U}{q} \quad \Longrightarrow \quad \Delta U = q \cdot \Delta V \]
\(\Delta V\)
Voltage (Potential Difference)
Joules per Coulomb across a component. The "height drop" a charge falls through. Measured with a voltmeter.
\(q\)
Charge (Coulombs)
The actual charge that passes through. "How many trucks" made the delivery run.
\(\Delta U\)
Energy Transferred (Joules)
Actual work done by the electric field on the charge — energy deposited into the component. Total cargo delivered.
Concrete Example: A 12 V car battery drives 3 coulombs through a headlight bulb:
\[\Delta U = q \cdot \Delta V = 3\,\text{C} \times 12\,\frac{\text{J}}{\text{C}} = 36\,\text{J}\]
The charge was the courier; the 36 Joules was the cargo delivered to the bulb as heat and light.
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Addendum · Section A3
Work — How the Field Actually Transfers Energy
Energy transfer happens via work: a force acts over a displacement. Inside any circuit element, \(\mathbf{E}\) exerts force \(\mathbf{F} = q\mathbf{E}\) on each charge. As the charge moves through the element, the field does work on it.
\[ W = \int_A^B \mathbf{F} \cdot d\boldsymbol{\ell} = q\int_A^B \mathbf{E} \cdot d\boldsymbol{\ell} = q(V_A - V_B) = q\,\Delta V \]
- Step 1
Force on the charge. \(\mathbf{E}\) inside the component points from high potential to low potential. Force on charge \(q\) is \(\mathbf{F} = q\mathbf{E}\) — "downhill."
- Step 2
Work = force × displacement along path. As the charge drifts from A to B, the field continuously does \(dW = q\mathbf{E}\cdot d\boldsymbol{\ell}\).
- Step 3
Integrate. \(\int_A^B \mathbf{E}\cdot d\boldsymbol{\ell} = V_A - V_B\), so \(W = q\,\Delta V\). Work done by the field equals the decrease in potential energy: \(W_{\text{field}} = -\Delta U\).
- Step 4
Energy conservation. That work came from electrical PE stored in the field configuration, originally put there by the battery (chemical → electrical PE). The field converts it to heat, light, or motion in the component.
Common Misconception — "Electricity Flows Like Water": Electrons drift at only ~mm/s! The energy is transported by the electric field at nearly the speed of light. When you flip a light switch, the field reorganizes almost instantly throughout the circuit — electrons just nudge slightly.
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Addendum · Section A4
Around the Loop — The Full Energy Journey
Let's walk a single coulomb of charge all the way around a simple circuit: battery → wire → resistor → wire → back to battery.
- Battery (−→+)
Potential rises sharply. Chemical energy lifts charges from (−) to (+). The only place potential increases. \(V\) increases by \(\varepsilon\).
- Wire (ideal)
Potential is flat. Zero resistance → no energy deposited → potential doesn't drop.
- Resistor
Potential drops. Energy deposited here. Falls by \(V = IR\).\[\Delta U_{\text{deposited}} = q \cdot IR\]
- Back to (−)
Potential returns to zero. Charge has delivered all its energy, ready for another cycle.
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Addendum · Section A5
Inside the Resistor — How Work Becomes Heat
The Drude Model — A First-Principles Picture
- 1 — Field applied
When you connect the resistor to \(\Delta V\), an electric field \(\mathbf{E} = \Delta V / L\) is established inside. This exerts force \(\mathbf{F} = eE\) on each free electron.
- 2 — Electrons accelerate
The field accelerates electrons. They gain kinetic energy: \(\Delta KE = F\cdot d\). The field is converting electrical PE into electron KE.
- 3 — Collision with lattice
After a short mean free path \(\lambda\), each electron collides with a lattice ion, transferring its KE to the ion as vigorous vibration. Vibrating ions = heat. The electron is scattered and starts accelerating again.
- 4 — Drift velocity
Accelerate + collide + repeat... Net result is a slow "drift" velocity \(v_d\). On average, electrons are not speeding up — they're at terminal velocity through the collision environment. All electrical PE converts continuously to thermal energy.
\[ \underbrace{W = q\,\Delta V}_{\text{work by field on charge}} = \underbrace{Q_{\text{heat}}}_{\text{thermal energy of lattice}} \]
Why Resistance Exists
Resistance measures how frequently electrons collide with the lattice. \(R = \rho L/A\): longer path \(L\) → more collisions; thinner \(A\) → more crowded, more collisions; higher resistivity \(\rho\) → more scattering centers.
Ohm's Law from First Principles
If drift velocity is proportional to the applied field (constant collision rate), then \(I = nqv_d A \propto E = V/L\), so \(I \propto V\):
\[ V = IR \quad \text{where} \quad R = \frac{\rho L}{A} \]
Not a fundamental law — an empirical approximation that holds when the collision rate is constant (breaks down at very high fields or temperatures).
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Addendum · Section A6
Power — Energy Per Unit Time
\[ P = \frac{dW}{dt} = \frac{d(q\,\Delta V)}{dt} = \Delta V \cdot \frac{dq}{dt} = \Delta V \cdot I \]
\[ P = I \cdot \Delta V = I^2 R = \frac{(\Delta V)^2}{R} \]
\(P\)
Power (Watts)
Joules per second. 1 W = 1 J/s. A 60 W bulb uses 60 Joules every second.
\(I\)
Current (Amperes)
Coulombs per second. \(I = dq/dt\). 1 A = 1 C/s. "Fleet size" — how many couriers pass per second.
\(\Delta V\)
Voltage Across Component
Joules per Coulomb. Energy per truck. Times \(I\) (trucks/sec) = Joules/sec = Watts.
\(I^2 R\)
Joule Heating Form
Substitute \(\Delta V = IR\) into \(P = I\Delta V\). Gives power dissipated as heat directly from current and resistance.
Dimensional Story:
\[P = \underbrace{I}_{\text{C/s}} \times \underbrace{\Delta V}_{\text{J/C}} = \frac{\text{C}}{\text{s}} \times \frac{\text{J}}{\text{C}} = \frac{\text{J}}{\text{s}} = \text{W}\]
The Coulombs cancel — they were just the courier. What remains is energy per second.
Worked Example — A Space Heater
120 V, 10 A. How much energy in 1 hour?
- Step 1 — Power
\[P = I \cdot \Delta V = 10 \times 120 = 1200\,\text{W}\]
- Step 2 — Energy
\[W = P \cdot t = 1200 \times 3600 = 4.32 \times 10^6\,\text{J} = 4.32\,\text{MJ}\]
- Step 3 — Verify
\(q = 10 \times 3600 = 36{,}000\,\text{C}\). Each coulomb carries 120 J. Total: \(36{,}000 \times 120 = 4.32\,\text{MJ}\,\checkmark\)
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Addendum · Section A7 · Interactive
Live Circuit — Energy Flow Simulator
Adjust EMF and resistance. The courier dots cycle around the loop, glowing gold when fully loaded with potential energy, dimming to blue as they deposit that energy in the resistor. The glow on the resistor shows power being dissipated as heat.
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Addendum · Section A8
The Full Mental Model — All Together
- Layer 1 — Force
The electric field \(\mathbf{E}\) exerts force \(\mathbf{F} = q\mathbf{E}\) on charges. Most fundamental level — Coulomb's law, point charges, action through the field.
- Layer 2 — Work
As charges move under that force: \(W = \int \mathbf{F}\cdot d\boldsymbol{\ell} = q\,\Delta V\). Work is the mechanism of energy transfer.
- Layer 3 — Potential
We define \(V = W/q\) — a scalar terrain map. Voltage differences drive everything.
- Layer 4 — Current
\(I = dq/dt\) in Amperes. Current is the "fleet size per second."
- Layer 5 — Power
\(P = IV\) in Watts. What your electricity bill is based on. A kilowatt-hour = \(1000 \times 3600 = 3.6 \times 10^6\,\text{J}\).
The Water Analogy — And Where It Breaks
Water Circuit
PumpBattery / EMF
Water pressureVoltage \(\Delta V\)
Flow rate (L/s)Current \(I\) (C/s)
Narrow pipeHigh resistance \(R\)
TurbineResistor / load
Electric Circuit
Chemical energy→ Electrical PE
J per C = VoltsTerrain height
C per secondCourier fleet rate
More collisionsMore heat, more \(\rho\)
Work = \(q\Delta V\)Cargo delivered
Where the analogy breaks: Water physically carries kinetic energy. Electric energy is actually carried by the electromagnetic field surrounding the wire (Poynting vector), not by the slow electrons. That's why lights turn on instantly even though electrons drift at ~mm/s.
One-Sentence Summary: The battery lifts each coulomb to a high potential (loads the courier); the charge falls through the circuit doing work \(W = q\Delta V\) on each component (delivers cargo); the rate is power \(P = IV\); and in a resistor, all that work becomes heat via electron-lattice collisions — Joule heating.