Define voltage (potential difference) and its units: Visual walkthrough

Step 1 — A charge makes an invisible "hill"

WHAT. We place one positive charge in empty space. It fills the space around it with an electric field — arrows that point away from it, telling any other charge "if you were here, this is the way I'd push you."

WHY. Before we can talk about pushing a charge, we need the thing doing the pushing. That thing is the electric field. It is the reason moving a charge costs (or releases) energy at all — without a field, moving a charge is free.

PICTURE. Look at figure s01. The red dot is our source charge. The black arrows are the field: longer near the charge (strong push), shorter far away (weak push). Think of this exactly like the slope of a hill — steep near the top, gentle far out.

Figure — Define voltage (potential difference) and its units

Step 2 — The field turns into a force on our charge: $F = q E$

WHAT. We put a small positive test charge $q$ into the field. The field $E$ at its location becomes an actual push — a force $F$ — on that charge.

WHY. The field by itself is just a map; a map doesn't move anything. The moment a real charge sits in it, the field's "force per coulomb" gets multiplied by "how many coulombs we have" to give a real force. That multiplication is the missing link the parent note glossed over.

PICTURE. Figure s02 shows the same field arrow at one point, then the charge $q$ dropped in, and the resulting force arrow — pointing the same way as the field (for a positive charge), but scaled by $q$ .

Step 3 — Moving against the force costs work (and why it's a sum, not one multiply)

WHAT. We shove the charge from a far point $B$ inward to a closer point $A$ , straight against the force. The energy we spend is the work $W_{A B}$ .

WHY. Fighting a force over a distance is the definition of work. But there's a catch the parent note hid: the force is not constant — it's strong near the source, weak far away (Step 1). You cannot just multiply one force by the whole distance.

PICTURE. Figure s03 chops the path into many tiny steps. Over each tiny step the force is almost constant, so its little bit of work is (force there) × (tiny length). We add up all the little bits. That "add up tiny force-times-length pieces" is exactly what the integral symbol $\int$ means.

Step 4 — The field is "conservative": only endpoints matter

WHAT. We move the charge from $B$ to $A$ by a different route — a big detour instead of the straight path — and measure the work again. It comes out exactly the same.

WHY. This is the deep property that lets voltage exist at all. If the work from $B$ to $A$ depended on which wiggly path you took, then "the energy of point $A$ " would have no single value — it would be a different number for every route. Because the electric field is conservative, the work depends only on the two endpoints, never on the path. That single-valued-ness is what makes a potential difference meaningful.

PICTURE. Figure s04 draws two very different paths (straight and looping) from $B$ to $A$ . A red tag on each reads the same work value. The loop that returns to its start costs zero net work.

Step 5 — The work becomes stored energy $U$ (the "height")

WHAT. The work we spent didn't vanish. It is now stored as electric potential energy. We give this stored energy its own symbol, $U$ , to keep it separate from "work I did."

WHY. Work done by us against the field and energy stored are two views of the same joules: $U_{A} - U_{B} = W_{A B}$ . Naming the stored energy $U$ prevents the confusion of using one letter for both. The stored energy is the "height" of the charge on the electric hill (remember: height = energy, not distance).

PICTURE. Figure s05 redraws the field as a hill. Point $B$ sits low ( $U_{B}$ ), point $A$ sits high ( $U_{A}$ , red). The vertical red gap is $U_{A} - U_{B}$ , which equals $W_{A B}$ .

Step 6 — Why we divide by charge

WHAT. We now carry a bigger charge — $2 q$ instead of $q$ — up the same hill, and watch the work double.

WHY. From Step 2, $F = q E$ , so double $q$ doubles the force at every point; and Step 3 sums force-times-length, so the total work doubles too. But the hill (the field $E$ ) didn't change. To describe the hill itself, independent of how much charge we haul, we divide the work by the charge. That ratio cancels the charge and leaves a pure property of the two points.

PICTURE. Figure s06 shows the same hill twice. Left: charge $q$ , work $W$ . Right: charge $2 q$ , work $2 W$ (red energy-bar twice as tall). But $W / q = 2 W /2 q$ — the red hill-height marker is identical.

Step 7 — Name the ratio: this IS voltage (with signs)

WHAT. We name that surviving ratio. The work done per unit charge to move from $B$ to $A$ is the potential difference — the voltage $V_{A B}$ .

WHY. We didn't discover a new physical thing; we named a useful ratio, exactly the way "speed" is just distance-per-time named.

PICTURE. Figure s07 is the payoff: the hill, with the vertical red gap now labelled $V_{A B}$ instead of raw energy.

Step 8 — Building the unit: what one volt means

WHAT. Put in exactly $1$ joule of work and exactly $1$ coulomb of charge. The ratio is $1$ , and we name it one volt.

WHY. A number needs a unit. Since voltage is joules divided by coulombs, its unit is joules per coulomb — we shorthand that as "volt." The volt is not fundamental; it is J/C.

PICTURE. Figure s08: a single coulomb (red) lifted through a gap that costs exactly 1 joule; the red marker reads 1 V.

Step 9 — Degenerate case and reading the formula both ways

WHAT (degenerate). If $A$ and $B$ are at the same height, no work is needed, $W = 0$ , so $V = 0/ q = 0$ .

WHY. Zero voltage does not mean "no charge around." It means "no height difference," so charge has no reason to flow — like water flat on a level floor.

PICTURE. Figure s09 shows two points at the same height on a flat stretch: the red drop-arrow shrinks to zero length.

WHAT (rearrangements). Once $V = W / q$ is built, algebra hands us its cousins: $V = \frac{W}{q} W = q V q = \frac{W}{V}$ And connecting to current $I = q / t$ and power: $P = \frac{W}{t} = \frac{q V}{t} = V \cdot \frac{q}{t} = V I$

$\frac{q}{t}$ ::: charge per second = current $I$ .
$V \cdot I$ ::: voltage times current = power (watts).

Recall Forecast: does more charge raise the voltage?

Q: Between two fixed points, you carry 5 C instead of 1 C. Does $V$ go up? A: No. From $F = q E$ and the work-sum, the work goes up 5×, but $q$ goes up 5× too. $V = W / q$ is unchanged — the hill height is a property of the points, not of what you carry.

The one-picture summary

Figure s10 compresses the whole walkthrough: the field-hill on the left, the force $F = q E$ that a coulomb feels, that coulomb lifted through height $V$ costing $W$ joules, and the boxed result $V = W / q = J/C = 1$ volt on the right.

Recall Feynman: the whole walkthrough in plain words

A charge sitting in space makes an invisible hill around it — a field $E$ that says how hard a charge would be pushed at each spot. Drop a real charge $q$ in, and that map becomes a real force, $F = q E$ : twice the charge, twice the push. To carry the charge up the hill you spend energy, but the hill's steepness keeps changing, so you add up lots of tiny force-times-step pieces — that's the integral. A lovely fact: no matter which wiggly route you take, the total energy is the same (the field is "conservative"), so each spot has a single stored energy $U$ . Naming things: the difference in stored energy divided by the charge is the voltage. Carrying twice the charge costs twice the energy but the ratio doesn't budge — so voltage belongs to the two points, not to your charge. Go uphill and it's positive; downhill flips the sign; a negative charge flips it too. When lifting one coulomb costs one joule, we call that height one volt. And remember the whole time: "height" in these pictures means energy, not literal metres in the air.

Connections

Electric field and potential energy — the field $E$ , the force $F = q E$ , and the stored energy $U$ .
Electric charge and the coulomb — the $q$ we divide by.
Electric current and the ampere — $I = q / t$ used in Step 9.
Power in electric circuits P=VI — derived in Step 9.
Ohm's Law V=IR — voltage across a resistor.
Batteries and EMF — a device that holds a fixed hill-height.
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