Electric potential — definition V = −∫E·dl: Visual walkthrough

Step 0 — The two pictures we start from

Before any formula, let's lock down the two objects in play, each anchored to a picture.

(a) A vector. An arrow with a length (how big) and a direction (which way). We draw the electric field $E$ as a set of arrows filling space. Bold letters like $E$ mean "this is a vector — it points somewhere."

(b) A tiny test charge $q_{0}$ . A small positive ball we drop into the field. "Positive" means: the field's arrows tell it which way it gets pushed — along the arrow.

Figure — Electric potential — definition V = −∫E·dl

This is the whole reason $E$ is called "force per unit charge" — see Electric field E — definition and Coulomb's law.

Step 1 — We want a "height," so we must move the charge

WHAT. We ask: how much energy is stored when the ball sits at a chosen point? Energy only appears when you do work — when you push something over a distance. So to measure the "electrical height" of that point, we carry the ball to it from a starting point and add up the work.

WHY this and not something simpler? You can't read energy off a stationary ball. Energy is a difference between two places. So the natural quantity is not "the height of one point" but "the height of the end point compared to a start point." Movement is unavoidable. We'll name the start $a$ and the end $b$ — and later, when we send $a$ off to infinity, the end point $b$ is exactly the evaluation point we call $P$ .

PICTURE. We choose a path — any wiggly curve — from a start point $a$ to an end point $b$ , and we'll walk the ball along it in tiny steps.

Each little step is a tiny displacement vector $d l$ : a short arrow pointing the way we just moved.

Step 2 — To keep energy honest, the agent pushes against the field

WHAT. We (the "external agent") apply our own force $F_{ext}$ to move the ball. We choose it to exactly cancel the field's push: $F_{ext} = - F_{elec} = - q_{0} E .$

Term by term:

$F_{ext}$ ::: our hand's force on the ball.
$- q_{0} E$ ::: same size as the field's push, but the minus flips it to point the opposite way.

WHY exactly opposite? If our force perfectly cancels the field's, the net force is zero, so by Newton's second law the ball never speeds up. No speed change means no kinetic energy is created or destroyed — so every joule of our work becomes stored (potential) energy, nothing leaks into motion. This is the "quasi-static" (slow, gentle) move. See Work–energy theorem in electrostatics.

PICTURE. At the ball we draw two arrows of equal length pointing opposite ways: the blue field push and our yellow counter-push.

Step 3 — Why a dot product: only motion-aligned force counts

WHAT. For one tiny step, the work we do is $d W_{ext} = F_{ext} \cdot d l .$

WHY the dot, not plain multiplication? Pushing sideways to your motion does no work — think of carrying a bag horizontally: gravity pulls down, you move sideways, gravity does nothing. Only the part of the force lined up along the step matters. The dot product is the exact tool that extracts "how much of this arrow lies along that arrow": $F \cdot d l = ∣ F ∣ ∣ d l ∣ cos θ .$

$∣ F ∣$ ::: length of the force arrow.

$∣ d l ∣$ ::: length of the step.

$θ$ ::: the angle between them.

$cos θ$ ::: the "alignment dial." $cos 0^{\circ} = 1$ (full effect, pushing straight along), $cos 9 0^{\circ} = 0$ (sideways, no work), $cos 18 0^{\circ} = - 1$ (pushing against motion, negative work).

PICTURE. We split the step arrow into a piece along the force and a piece perpendicular. Only the along-piece survives the dot product.

Step 4 — Add up all the tiny steps: the integral appears

WHAT. The total work over the whole path from $a$ to $b$ is the sum of every tiny $d W_{ext}$ . A sum of infinitely many infinitely small pieces is written with the integral sign: $W_{ext} = \int_{a}^{b} F_{ext} \cdot d l .$

WHY an integral and not a plain sum $Σ$ ? Because the field can change from step to step — near a charge it's strong, far away it's weak. The steps must be vanishingly small so $E$ is essentially constant across each one. "Sum of infinitely many vanishing pieces" is the definition of the integral $\int$ .

$\int_{a}^{b}$ ::: "add up over the path from start $a$ to end $b$ ."
$F_{ext} \cdot d l$ ::: the tiny bit of work at each point.

PICTURE. The path is chopped into many small arrows; each contributes a sliver of work; stacking the slivers gives the total.

Now substitute Step 2's $F_{ext} = - q_{0} E$ : $W_{ext} = \int_{a}^{b} (- q_{0} E) \cdot d l = - q_{0} \int_{a}^{b} E \cdot d l .$ The constant $- q_{0}$ slides out of the sum (every term shares it).

Step 5 — Divide out the charge to get a property of space alone

WHAT. Potential difference is defined as work per unit charge: $V_{b} - V_{a} \equiv \frac{W _{ext}}{q _{0}} = - \int_{a}^{b} E \cdot d l .$

WHY divide by $q_{0}$ ? Every quantity above carries a factor of $q_{0}$ — a bigger ball costs proportionally more work. Dividing it out leaves a number that depends only on the field and the two locations, not on the ball you happened to use. That reusable map is the potential. (This is the same "per unit charge" logic that made $E$ itself.)

$V_{b} - V_{a}$ ::: how much higher the electrical "hill" is at $b$ than at $a$ (volts = J/C).
The $q_{0}$ has vanished from the right side — exactly what we wanted.

Step 6 — Reading the minus sign off the picture (all cases)

WHAT. Why the minus? Consider walking with the field vs against it.

PICTURE — three cases, one figure:

Case A — walk along $E$ (downhill): $E$ and $d l$ point the same way, so $E \cdot d l > 0$ . The integral is positive, and with the minus, $V_{b} - V_{a} < 0$ — potential dropped. Correct: the field points toward lower potential, like gravity points downhill.

Case B — walk against $E$ (uphill): $E$ and $d l$ oppose, $E \cdot d l < 0$ , so $V_{b} - V_{a} > 0$ — potential rose. You climbed the hill; makes sense.

Case C — walk perpendicular to $E$ (sideways): $θ = 9 0^{\circ}$ , $cos 9 0^{\circ} = 0$ , so $E \cdot d l = 0$ everywhere on the walk — potential unchanged. Those constant-height paths are equipotential surfaces: $E$ is always perpendicular to them, which is why moving along them costs no potential change.

Step 7 — The degenerate/limiting cases you must not trip on

WHAT. Three edge scenarios the smooth derivation hides.

(i) $E = 0$ (no field). Then $E \cdot d l = 0$ for every step, so $V_{b} - V_{a} = 0$ : the potential is flat everywhere. A field-free region is one big equipotential.

(ii) Path shape doesn't matter. For the wiggly path and a straight shortcut between the same $a$ and $b$ , the integral gives the same answer — because $E$ is a conservative field (it's a gradient, next section). If it weren't, "potential" wouldn't even be well-defined.

(iii) Reference at infinity can fail. For a localized blob of charge, $E$ dies off fast enough that $\int_{\infty}^{P}$ is finite — infinity is a safe "sea level." But for an infinite line or plane of charge, $E$ doesn't die off fast enough and $\int_{\infty}^{P}$ diverges. Then you must pin $V = 0$ at a finite point instead. Only differences $V_{b} - V_{a}$ are ever physical, so this is always allowed.

The inverse direction (why $V$ is worth all this)

For a single tiny step, dividing the one-step work $d W_{ext} = F_{ext} \cdot d l$ by $q_{0}$ gives the tiny change in potential across that step. Call that change $d V$ :

Running that one-step relation backwards recovers the field from the potential: $E = - \nabla V .$ $V$ is a single number at each point (a scalar) — far easier to compute than three vector components. Compute the easy scalar, then take its slope (see Gradient operator ∇ and directional derivatives) to get $E$ back. That's the payoff of the whole construction.

The one-picture summary

Reading left to right: the field pushes the ball ( $F = q_{0} E$ ) → we counter-push ( $- q_{0} E$ ) → each step's useful work is a dot product ( $- q_{0} E \cdot d l$ ) → sum the steps ( $- q_{0} \int E \cdot d l$ ) → divide out $q_{0}$ → the potential difference. Every symbol in $V = - \int E \cdot d l$ is one arrow in this chain.

Recall Feynman: the whole walkthrough in plain words

You want to know how "high up the electric hill" a spot is. You can't just look — you have to carry a tiny charged ball there and count how hard you pushed. The field is already pushing the ball (that's $q_{0} E$ ), so to move it gently you push back exactly as hard ( $- q_{0} E$ ), which means all your effort turns into stored energy, none into speed. As you walk, only the pushing that lines up with your steps counts — that's the dot product, and it quietly ignores any sideways walking. Add up every little push over the whole trip (the integral), then divide by the size of your ball so the answer describes the place, not the ball. The minus sign is bookkeeping: walking the way the field shoves you is walking downhill, so potential goes down. Walk sideways and nothing changes — those flat paths are the equipotentials.

Connections

1.8.08 Electric potential — definition V = −∫E·dl (Hinglish)
Electric field E — definition and Coulomb's law
Potential energy of a charge configuration
Equipotential surfaces and why E ⟂ them
Gradient operator ∇ and directional derivatives
Work–energy theorem in electrostatics
Capacitance and parallel-plate fields (E = ΔV/d)

Visual walkthrough — Electric potential — definition V = −∫E·dl