Visual walkthrough — Define electric field and electric potential

We build them side by side so you always see how they relate.

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Step 1 — The one fact we start from: two charges push

WHAT. Before any "field" or "potential" exists, there is only one experimental fact: two point charges exert a force on each other. Put a source charge $Q$ at the origin and a tiny test charge $q_0$ a distance $r$ away. The measured force is Coulomb's law:

$\vec{F}=\underset{\text{constant }k}{\underbrace{\frac{1}{4\pi {\epsilon }_0}}}\underset{\text{source}}{\underbrace{\frac{Q}{}}}\underset{\text{visitor}}{\underbrace{\frac{q_0}{}}}\underset{\text{fades with distance}}{\underbrace{\frac{1}{r^2}}}\underset{\text{direction}}{\underbrace{\hat{r}}}$

Reading it symbol by symbol:

• $k=\frac{1}{4\pi {\epsilon }_0}\approx 8.99\times 10^9$ — a fixed number of nature; it just sets the strength.
• $Q$ — the charge that makes the influence (the source).
• $q_0$ — the charge that feels it (the test / visitor).
• $r$ — how far apart they are; $\frac{1}{r^2}$ means twice as far → one quarter the force.
• $\hat{r}$ — a "direction hat": a unit arrow pointing straight out from $Q$ toward $q_0$.

WHY start here. Everything else on this page is this one law, re-read two different ways. We are not adding new physics — only new bookkeeping.

PICTURE. Two dots, one arrow of force on the test charge, pointing away because both are positive.

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Step 2 — Divide out the visitor: the field is born

WHAT. The force above depends on both charges. But we want a number that belongs to the space alone, not to whoever wanders in. So we divide by the visitor's charge $q_0$. That is the definition of the field:

$\vec{E}\equiv \frac{\vec{F}}{q_0}$

Substituting Coulomb's law and cancelling $q_0$:

$\vec{E}=\frac{1}{q_0}\cdot \frac{kQ{q}_0}{r^2}\hat{r}=\frac{kQ}{r^2}\hat{r}$

• The $q_0$ on top and bottom cancel — that is the whole point.
• What survives, $\frac{kQ}{r^2}\hat{r}$, mentions only $Q$, $r$, $\hat{r}$ — the source and the place.

WHY divide, not multiply? Because force scales linearly with the visitor: double $q_0$, double $\vec{F}$. So $\vec{F}/q_0$ is the one combination that stays fixed no matter who visits — a genuine property of the location.

PICTURE. Same point in space, but now three different test charges ($q_0$, $2q_0$, $3q_0$) feel three different force arrows — yet dividing each arrow by its own charge gives the same purple field arrow. The space "remembers" one arrow.

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Step 3 — Paint the field everywhere (the arrow map)

WHAT. Step 2 gave the field at one place. Now sweep $r$ around and mark the arrow at every point. Two rules only:

• Length shrinks as $1/r^2$ — arrows near $Q$ are long, far arrows are stubby.
• Direction is radial — every arrow points straight out (for $+Q$) or straight in (for $-Q$).

WHY do this map. A single formula is abstract; the arrow map is what physicists actually see when they say "the field." It also sets up Step 4: to define potential we will walk a test charge through this map.

PICTURE. A starburst of blue arrows around $+Q$ (pointing out) and, beside it, around $-Q$ (pointing in). Watch the arrows get short fast as you move outward — that is the $1/r^2$ fade.

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Step 4 — A different question: how much work to walk in?

WHAT. Now change the question entirely. Instead of "how hard is the push here?" ask "how much energy did it cost to bring the test charge here from far away?" To carry $q_0$ from infinity in to radius $r$ against the field (slowly, no speeding up), the work I must supply is:

$W_{\infty \to r}=-{\int }_{\infty }^r{\vec{F}}_{\text{elec}}\cdot d\vec{l}$

Term by term:

• ${\vec{F}}_{\text{elec}}$ — the electric force the field pushes with; here $=q_{0}E$ outward.
• $d\vec{l}$ — a tiny step of my path.
• The dot product $\vec{F}\cdot d\vec{l}$ measures only the part of the force along my step — the part I actually fight.
• The minus sign is the heart of it: the work I do is the negative of the work the field does. When I push inward against an outward field, the field does negative work on me, so my applied work is positive.

WHY an integral, and why the dot product? The force is not constant along the trip — it grows as I approach. "Force × distance" only works for constant force, so we chop the path into tiny steps $d\vec{l}$, use the local force on each, and sum them up — that summation of infinitely many tiny pieces is the integral. The dot product is the right tool because only the along-the-path component costs energy; sideways motion is free.

PICTURE. A test charge crawling in from the right along the dashed path, the outward field arrow fighting it at each step; the little green step $d\vec{l}$ and red force $\vec{F}$ shown with the angle between them.

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Step 5 — Do the integral (the only calculus on this page)

WHAT. Move straight in along a radial line, so $\vec{F}$ and $d\vec{l}$ point along the same line and the dot product is just the product of magnitudes. With ${\vec{F}}_{\text{elec}}=q_0\frac{kQ}{r^2}$:

$W=-{\int }_{\infty }^r{q}_0\frac{kQ}{r^2}dr=-q_{0}kQ{\int }_{\infty }^r{r}^{-2}dr$

The only fact we need is the antiderivative: $\int r^{-2}dr=-r^{-1}=-\frac{1}{r}$

So $W=-q_{0}kQ{\left[-\frac{1}{r}\right]}_{\infty }^r=q_{0}kQ\left(\frac{1}{r}-\frac{1}{\infty }\right)=\frac{kQ{q}_0}{r}$

• $\frac{1}{\infty }=0$: starting infinitely far away costs nothing to be there — that is why infinity is our "zero" reference.
• The two minus signs cancel, leaving a clean positive $\frac{kQ{q}_0}{r}$ for like charges (you did work to shove them together — makes sense).

WHY $1/r^2$ becomes $1/r$. Integrating (accumulating) a $1/r^2$ quantity over distance raises the power by one in the denominator: the steepness fell as $1/r^2$, the accumulated height falls only as $1/r$. This is the numerical shadow of "potential is the running total of the field."

PICTURE. The area under the $F$-vs-$r$ curve from $\infty$ down to $r$ — the shaded region is the work. See how most of the area piles up close to $Q$.

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Step 6 — Divide out the visitor again: potential is born

WHAT. Just like the field, we want an energy-per-charge that belongs to the space, so divide the work by $q_0$. That is the potential energy $U=W$ turned into electric potential $V$:

$V\equiv \frac{W_{\infty \to r}}{q_0}=\frac{1}{q_0}\cdot \frac{kQ{q}_0}{r}=\frac{kQ}{r}$

• The $q_0$ cancels once more → a property of the place, not the visitor.
• No $\hat{r}$, no arrow: $V$ is a scalar, a single number (a height) at each point.

WHY it matters. This scalar is exactly what a voltmeter reads and what a battery quotes. See Voltage and Potential Difference.

PICTURE. The potential drawn as a height landscape: a tall spike over $+Q$ falling off as $1/r$, and a deep well over $-Q$. Height = how much energy per coulomb it took to get there.

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Step 7 — Sew them together: field is the slope of the hill

WHAT. The field came from dividing force by $q_0$; the potential came from dividing (integrated) force by $q_0$. Since integration built $V$ from $E$, differentiation must undo it:

$E=-\frac{dV}{dr},V(r)=-{\int }_{\infty }^{r}Edr$

Check it directly: $\frac{d}{dr}\left(\frac{kQ}{r}\right)=-\frac{kQ}{r^2}$, so $-\frac{dV}{dr}=\frac{kQ}{r^2}=E$. ✔

• $\frac{dV}{dr}$ — the slope of the height landscape.
• The minus — a positive charge rolls downhill, from high $V$ toward low $V$, so the field points down the slope.

WHY it's the punchline. Field and potential are not two separate objects; they are the slope and the height of one and the same hill. Steep hill → strong field. Flat hill → zero field (even if you're high up).

PICTURE. The $1/r$ height curve on top; directly below, its steepness (the field) as $1/r^2$. A tangent line marks the slope at one point; its steepness equals the field arrow's length.

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Step 8 — Edge and degenerate cases (never get surprised)

WHAT. Check every corner so no scenario ambushes you.

Case	Field $E=kQ/r^2$	Potential $V=kQ/r$	Picture cue
$Q>0$	points out	positive hill	starburst out / spike up
$Q<0$	points in	negative well	starburst in / dip down
$r\to \infty$	$\to 0$	$\to 0$	flat, far horizon
$r\to 0^{+}$	$\to \infty$ ($1/r^2$)	$\to \infty$ ($1/r$)	infinitely tall/deep spike
$Q=0$	$E=0$ everywhere	$V=0$ everywhere	flat empty space

Two famous "gotcha" spots (from the parent's mistakes list):

• Midpoint between $+Q$ and $-Q$: the two field arrows are equal and point the same way (don't cancel) → $E\ne 0$; but the two potentials $+kQ/r$ and $-kQ/r$ add to zero → $V=0$. So $V=0$ with $E\ne 0$.
• Inside a hollow charged sphere (via Gauss's Law): field cancels to $E=0$, yet you climbed a hill to get inside, so $V\ne 0$ (constant). So $E=0$ with $V\ne 0$.

WHY show these. They prove field and potential are genuinely independent lenses — knowing one at a point does not fix the other.

PICTURE. Left: the $+Q,-Q$ midpoint with both arrows the same way but heights cancelling. Right: hollow sphere, flat-zero field inside sitting on a raised constant-height plateau.

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The one-picture summary

WHAT. One canvas ties the whole journey together: Coulomb force → (÷$q_0$) → field arrows → (integrate over distance) → potential hill → (slope) → back to field.

• Read it clockwise from the top-left force arrow.
• $÷q_0$ strips the visitor twice — once for $E$, once for $V$.
• $\int dr$ turns $1/r^2$ into $1/r$; $d/dr$ turns it back.

Recall Feynman retelling — explain the whole walkthrough to a friend

We started with the one thing we can actually measure: two charges push on each other, and the push weakens as the square of the distance. That push depends on both charges, which is annoying — so we divided by the little visiting charge to get a number that belongs to the empty space itself. That number, with a direction, is the field: a map of push-arrows, long near the charge and shrinking fast as $1/r^2$.

Then we asked a totally different question — not "how hard is the push?" but "how much work did it take to drag the visitor in from far away?" Because the push changes along the trip, we sliced the path into tiny steps and added up force-times-step; that sum is an integral, and integrating a $1/r^2$ push over distance gives a $1/r$ energy. Dividing that energy by the visitor again gives the potential: a height-map, a hill over a positive charge and a valley over a negative one.

Finally we noticed the two maps are the same hill seen twice: the potential is the height, and the field is the steepness pointing downhill. Steep means strong; flat means zero — even if you're standing high up. And in weird spots the arrows can cancel while the heights don't, or the heights stay while the arrows cancel — which is exactly why we keep both lenses.

Recall Quick self-test

Field from force? ::: Divide by the test charge: $\vec{E}=\vec{F}/q_0$. Potential from field? ::: Integrate over distance from infinity: $V=-{\int }_{\infty }^{r}Edr$. Field back from potential? ::: Negative slope: $E=-dV/dr$. Why $E\sim 1/r^2$ but $V\sim 1/r$? ::: Integrating $r^{-2}$ over $r$ gives $r^{-1}$; potential accumulates the field. A place with $E\ne 0$ but $V=0$? ::: Midpoint between $+Q$ and $-Q$.

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Connections

• Coulomb's Law — the single measured fact Step 1 starts from.
• Electric Potential Energy — the work $W=q_{0}V$ before we divide by $q_0$ in Step 6.
• Voltage and Potential Difference — the potential of Step 6 used between two points.
• Equipotential Surfaces and Field Lines — the height-map and arrow-map drawn as contours.
• Gauss's Law — gives the hollow-sphere case in Step 8 without integration.
• Capacitance — builds on field-between-plates and potential difference.