Electric potential — definition V = −∫E·dl: Worked examples

Before anything, let us re-read the tools so no symbol is unearned.

The scenario matrix

Every potential problem is one of these cells. Each worked example below is tagged with the cell(s) it covers.

#	Cell (case class)	What makes it tricky	Example
A	Single positive point charge, radial path	sign of $V$ ; where $V$ is bigger	Ex 1
B	Single negative point charge	$V < 0$ everywhere; "downhill" flips	Ex 2
C	Superposition of several charges	$V$ adds as plain numbers (no vectors)	Ex 3
D	Uniform field (parallel plates), sign of walk	direction of walk vs. $E$	Ex 4
E	Recover $E$ from $V$ via gradient	which sign, which partial derivative	Ex 5
F	Work / energy of a moving charge (both signs of $q$ )	sign of $q$ times sign of $Δ V$	Ex 6
G	Degenerate / limiting: infinite line of charge	reference at $\infty$ fails (diverges)	Ex 7
H	Zero case: point on a perpendicular bisector, $V \neq = 0$ but $E = 0$ -component subtlety	$V$ big where $E$ can vanish	Ex 8
I	Real-world word problem	translate words to the formula	Ex 9
J	Exam twist: path independence — a curved path	why the shape of the path does not matter	Ex 10

Ex 1 — Positive point charge (Cell A)

Worked example Statement

A charge $q = + 3 nC$ sits at the origin. Find $V$ at $r_{1} = 0.20 m$ and at $r_{2} = 0.10 m$ . Which point is at higher potential?

Forecast: Guess before computing — as you walk toward a positive charge, does the electrical "height" go up or down? (Picture climbing a hill whose peak is the charge.)

Steps.

Use the master result from the parent note, $V (r) = \frac{k q}{r}$ . Why this step? The field of a point charge is radial, so a radial path makes $E \cdot d l = E d r$ ; the integral collapses to $k q / r$ . Reuse it, don't re-derive.
$V (r_{1}) = \frac{( 8.99 \times 1 0 ^{9} ) ( 3 \times 1 0 ^{- 9} )}{0.20} = 134.85 V$ . Why this step? Plug the far radius $r_{1}$ into $k q / r$ to get the baseline height before we move closer — we need it to compare the two points.
$V (r_{2}) = \frac{( 8.99 \times 1 0 ^{9} ) ( 3 \times 1 0 ^{- 9} )}{0.10} = 269.7 V$ . Why this step? Repeat with the nearer radius $r_{2}$ ; halving $r$ doubles $V$ , because the $1/ r$ shape means "closer = taller."

Verify: $V > 0$ everywhere (as it must be near a $+$ charge), and the closer point $r_{2}$ has the higher potential. That matches the forecast: walking toward $+ q$ is walking uphill. Units: $N\cdotm^{2} / C^{2} \cdot C / m = N\cdotm/C = J/C = V$ . ✓

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

Ex 2 — Negative point charge (Cell B)

Worked example Statement

Now $q = - 3 nC$ at the origin. Find $V$ at $r = 0.20 m$ . As you approach, does $V$ rise or fall?

Forecast: The formula $V = \frac{k q}{r}$ still holds unchanged — but now $q$ itself carries a minus sign. Guess the sign of $V$ and whether approaching makes $V$ bigger or more negative.

Steps.

Keep the single fundamental formula $V (r) = \frac{k q}{r}$ and substitute $q = - 3 \times 1 0^{- 9} C$ : $V (0.20) = \frac{( 8.99 \times 1 0 ^{9} ) ( - 3 \times 1 0 ^{- 9} )}{0.20} = - 134.85 V .$ Why this step? We do not invent a new " $- k q / r$ " formula; the same $V = k q / r$ works, and the sign of $q$ alone drives $V$ negative. That keeps one rule for every charge.
Substitute the nearer radius $r = 0.10 m$ into the same formula: $V (0.10) = \frac{( 8.99 \times 1 0 ^{9} ) ( - 3 \times 1 0 ^{- 9} )}{0.10} = - 269.7 V .$ Why this step? We compute a second point to see the trend: getting closer makes $V$ go from $- 134.85$ down to $- 269.7$ — deeper into a valley, not up a hill.

Verify: $V < 0$ everywhere near $- q$ . Field $E$ points toward the charge (inward), which is the direction of decreasing $V$ (from $- 134.85$ down to $- 269.7$ ). " $E$ points downhill" is satisfied. ✓

Ex 3 — Superposition of charges (Cell C)

Worked example Statement

Two charges: $q_{1} = + 4 nC$ at $(0, 0)$ and $q_{2} = - 2 nC$ at $(0.30, 0) m$ . Find $V$ at the point $P = (0.30, 0.40) m$ .

Forecast: Potential is a scalar, so we just add the two numbers — no arrows, no components. Guess whether the total comes out $+$ or $-$ .

Steps.

Distance from $q_{1}$ to $P$ : $r_{1} = 0.3 0^{2} + 0.4 0^{2} = 0.50 m$ . Why this step? $V$ depends only on distance, so we need each separation first.
Distance from $q_{2}$ to $P$ : it sits directly below $P$ , so $r_{2} = 0.40 m$ . Why this step? $q_{2}$ and $P$ share the same $x = 0.30$ , so their separation is just the vertical gap $0.40$ — no square root needed. We need this second distance to feed the second $k q / r$ term.
$V = \frac{k q _{1}}{r _{1}} + \frac{k q _{2}}{r _{2}} = \frac{( 8.99 \times 1 0 ^{9} ) ( 4 \times 1 0 ^{- 9} )}{0.50} + \frac{( 8.99 \times 1 0 ^{9} ) ( - 2 \times 1 0 ^{- 9} )}{0.40}$ . Why this step? Superposition: total $V$ is the plain algebraic sum of each charge's $k q / r$ . This is exactly the " $V$ is scalar" payoff — no angles to resolve.
$= 71.92 + (- 44.95) = 26.97 V$ . Why this step? Carry out the two divisions and add the signed numbers; the negative charge's term simply subtracts, no vector geometry involved.

Verify: The bigger, closer-enough positive charge wins, so $V > 0$ — matching the forecast. If the two charges were equal and opposite at equal distance, they'd cancel to $V = 0$ ; here they don't, and we get a modest positive number. ✓

Ex 4 — Uniform field, sign of the walk (Cell D)

Worked example Statement

A parallel-plate region has a uniform field $E = 500 V/m$ pointing in the $+ \overset{x}{^}$ direction. The plates are $d = 0.04 m$ apart. Find $V_{b} - V_{a}$ when you walk from $a$ (at $x = 0$ ) to $b$ (at $x = 0.04$ ).

Forecast: You are walking along $E$ . Does the potential rise or fall? (Recall $E$ points downhill.)

Steps.

$E \cdot d l = (500) (d x)$ because both point along $+ \overset{x}{^}$ ; the dot product is just the product of magnitudes. Why this step? When two arrows are parallel, the dot product loses the angle — it's plain multiplication. This is why we chose to walk straight across.
$V_{b} - V_{a} = - \int_{0}^{0.04} 500 d x = - 500 \times 0.04 = - 20 V$ . Why this step? Constant $E$ comes out of the integral, leaving just the length $d = 0.04$ to integrate over — that's why the answer is simply $- E d$ .
So $b$ is $20 V$ lower than $a$ ; equivalently $V_{a} - V_{b} = + 20 V$ , and $E = \frac{Δ V}{d} = \frac{20}{0.04} = 500 V/m$ . Why this step? Rearranging back to $E = Δ V / d$ recovers the given field, a consistency check and the reason field units are $V/m$ .

Verify: Walking along $E$ dropped the potential (negative $Δ$ ), exactly as the "downhill" rule predicts. Re-deriving $E$ from $Δ V / d$ returns the original $500 V/m$ , confirming self-consistency. See Capacitance and parallel-plate fields (E = ΔV/d). ✓

Ex 5 — Recover $E$ from $V$ (Cell E)

Worked example Statement

A region has $V (x, y, z) = 2 x^{2} - 3 y z + 4 z (volts, with x, y, z in metres)$ . Find $E$ at the point $(1, 2, 1)$ .

Forecast: $E = - \nabla V$ . Before computing, notice $V$ grows fastest in $x$ near this point — so guess which component of $E$ is largest in magnitude.

Steps.

Partial derivatives (differentiate treating the other variables as constants): $\frac{\partial V}{\partial x} = 4 x, \frac{\partial V}{\partial y} = - 3 z, \frac{\partial V}{\partial z} = - 3 y + 4.$ Why this step? $\nabla V$ is exactly this list of slopes; see Gradient operator ∇ and directional derivatives.
Flip the sign: $E = - (4 x) \overset{x}{^} - (- 3 z) \overset{y}{^} - (- 3 y + 4) \overset{z}{^} = - 4 x \overset{x}{^} + 3 z \overset{y}{^} + (3 y - 4) \overset{z}{^} .$ Why this step? $E = - \nabla V$ ; the minus turns "uphill slope" into "downhill field."
At $(1, 2, 1)$ : $E = (- 4) \overset{x}{^} + (3) \overset{y}{^} + (2) \overset{z}{^} V/m .$ Why this step? We evaluate the slope at the point of interest — a gradient is a function of position, so it only gives a definite arrow once we plug in $(1, 2, 1)$ .

Verify: Units: $V$ in volts, derivative w.r.t. metres gives $V/m$ — correct field units. The $\overset{x}{^}$ component has magnitude $4$ , largest of the three, matching the forecast that $V$ steepens fastest in $x$ here. ✓

Ex 6 — Work on a moving charge, both signs of $q$ (Cell F)

Worked example Statement

Point $a$ is at $V_{a} = 80 V$ , point $b$ at $V_{b} = 30 V$ . (i) A $+ 5 μ C$ charge moves $a \to b$ . (ii) A $- 5 μ C$ charge moves $a \to b$ . Find the work done by the field in each case.

Forecast: Use $W_{field} = q (V_{a} - V_{b})$ . Guess the sign of the work in each case before the arithmetic — one is positive, one negative.

Steps.

$V_{a} - V_{b} = 80 - 30 = 50 V$ . Why this step? The work formula needs the potential drop along the motion, $V_{a} - V_{b}$ , so we compute it once and reuse it for both charges.
(i) $W = (+ 5 \times 1 0^{- 6}) (50) = + 2.5 \times 1 0^{- 4} J$ . Why this step? A positive charge released from high $V$ to low $V$ is pushed with the field — it "rolls downhill," so the field does positive work.
(ii) $W = (- 5 \times 1 0^{- 6}) (50) = - 2.5 \times 1 0^{- 4} J$ . Why this step? A negative charge feels a force opposite to $E$ ; going the same way in space now means going uphill in energy, so the field does negative work.

Verify: Same $Δ V$ , opposite $q$ , opposite work signs — consistent with $W \propto q$ . Energy check: positive work by field = kinetic energy gained by the $+$ charge, matching the Work–energy theorem in electrostatics. ✓

Ex 7 — Degenerate case: infinite line of charge (Cell G)

Worked example Statement

An infinite straight line carries linear charge density $λ$ . Its field at distance $s$ is $E = \frac{λ}{2 π ε _{0} s}$ pointing radially out. Try $V (\infty) = 0$ ; then fix the problem.

Forecast: We integrate $1/ s$ out to infinity. Recall from calculus what $\int^{\infty} \frac{d s}{s}$ does — guess whether $V$ stays finite.

Steps.

Naive attempt: $V (s) = - \int_{\infty}^{s} \frac{λ}{2 π ε _{0} s ^{'}} d s^{'} = - \frac{λ}{2 π ε _{0}} [ln s^{'}]_{\infty}^{s} .$ Why this step? Radial path again makes $E \cdot d l = E d s$ .
$ln s^{'}$ at the upper limit $\infty$ blows up — the reference at infinity gives an infinite number. The naive choice fails. Why this step? The charge is not localized; it stretches to infinity, so "infinitely far" isn't infinitely far from the charge. The parent note's mistake box warned of exactly this.
Fix: pick a finite reference distance $s_{0}$ . Then $V (s) - V (s_{0}) = - \frac{λ}{2 π ε _{0}} ln \frac{s}{s _{0}} = \frac{λ}{2 π ε _{0}} ln \frac{s _{0}}{s} .$ Why this step? Only differences of potential are physical; a finite reference makes every number finite.

Verify: As $s \to \infty$ with fixed $s_{0}$ , $V \to - \infty$ (for $λ > 0$ ) — that divergence is real, which is why $\infty$ can't be the zero. For $s < s_{0}$ (closer to a $+$ line) we get $ln (s_{0} / s) > 0$ , so $V$ is higher near the line — correct "hill toward positive charge." ✓

Ex 8 — Zero-field spot with nonzero potential (Cell H)

Worked example Statement

Two equal charges $+ q = + 6 nC$ sit at $(\pm 0.30, 0) m$ . At the midpoint $M = (0, 0)$ , the fields cancel so $E = 0$ . Is $V = 0$ there too?

Forecast: Fields point in opposite directions at $M$ and cancel. But potentials are numbers — do numbers cancel here? Guess before computing.

Steps.

Each charge is $r = 0.30 m$ from $M$ . Why this step? $V$ needs distances, and both are equal here.
$V = \frac{k q}{r} + \frac{k q}{r} = 2 \cdot \frac{( 8.99 \times 1 0 ^{9} ) ( 6 \times 1 0 ^{- 9} )}{0.30} = 2 (179.8) = 359.6 V .$ Why this step? Both contributions are positive (same-sign charges) — they add, they do not cancel. Only the vector $E$ cancels, because its arrows point opposite ways while the two numbers $k q / r$ both point "up."

Verify: $E = 0$ but $V = 359.6 V \neq = 0$ . This is the crucial lesson: $E = - \nabla V$ measures the slope, and a hill can have a flat top (zero slope) while still being high (nonzero height). A smart way to remember it. ✓

Ex 9 — Real-world word problem (Cell I)

Worked example Statement

A cloud-to-ground lightning setup behaves roughly like a capacitor: the cloud base is $500 m$ above the ground, and the (near-uniform) field just before a strike is about $3 \times 1 0^{6} V/m$ (the breakdown field of air). Estimate the potential difference between cloud and ground.

Forecast: Uniform field over a distance $d$ : which relation connects $E$ , $Δ V$ , $d$ ? Guess the order of magnitude — millions? billions of volts?

Steps.

Model it as a parallel-plate gap: $∣Δ V ∣ = E d$ . Why this step? Over the strike channel the field is nearly constant, so $\int E d l$ becomes $E \times d$ (exactly Ex 4's logic) — the cloud and ground are our two "plates."
$∣Δ V ∣ = (3 \times 1 0^{6} V/m) (500 m) = 1.5 \times 1 0^{9} V .$ Why this step? Substitute the air-breakdown field and the $500 m$ gap into $E d$ ; multiplication is all that's left because $E$ is treated as constant.

Verify: $1.5 \times 1 0^{9} V$ (1.5 billion volts) — the correct real-world ballpark for lightning. Units: $(V/m) (m) = V$ . ✓ The huge number is why air breaks down and a spark forms.

Ex 10 — Exam twist: path independence (Cell J)

Worked example Statement

A charge $+ q = + 2 nC$ is at the origin. Compute $V_{b} - V_{a}$ between $a$ at $r_{a} = 0.10 m$ and $b$ at $r_{b} = 0.25 m$ (i) along a straight radial line, and (ii) along a weird path that first swings sideways then spirals out. Show the answer is the same.

Forecast: The electrostatic field is conservative. Guess whether the crooked path changes the answer at all.

Steps.

Radial path: $V_{b} - V_{a} = \frac{k q}{r _{b}} - \frac{k q}{r _{a}} = (8.99 \times 1 0^{9}) (2 \times 1 0^{- 9}) (\frac{1}{0.25} - \frac{1}{0.10}) .$ Why this step? $V (r) = k q / r$ already came from a radial integral, so the difference between two radii is just the difference of two $k q / r$ values.
$= 17.98 (4 - 10) = 17.98 \times (- 6) = - 107.88 V .$ Why this step? Compute $1/0.25 = 4$ and $1/0.10 = 10$ , subtract, and multiply by $k q = 17.98$ ; the result is negative because $b$ is farther out (lower on the hill).
Crooked path: split any $d l$ into a radial part $d r \overset{r}{^}$ and a sideways part. The sideways part is $⊥ E$ , so $E \cdot d l$ picks up only the radial piece $E d r$ — identical to step 1. Why this step? $E \cdot d l$ discards motion perpendicular to $E$ ; since $E$ is purely radial, only your change in radius matters, not the wandering.

Verify: Both paths give $V_{b} - V_{a} = - 107.88 V$ . Potential difference depends only on the endpoints — the hallmark of a conservative field, and the reason $V$ is well-defined at all. See Equipotential surfaces and why E ⟂ them: sideways motion along an equipotential costs no potential change. ✓

Recall

Recall Quick self-test on the matrix

Approaching a $+$ charge, does $V$ rise or fall? ::: Rise — it's a hill; $V = k q / r$ grows as $r$ shrinks. Near a $-$ charge, sign of $V$ ? ::: Negative everywhere; use the same $V = k q / r$ with $q < 0$ . How do potentials of several charges combine? ::: Plain algebraic sum of each $k q / r$ (scalar). Walking along $E$ , does $V$ go up or down? ::: Down — $E$ points toward lower $V$ . Can $E = 0$ while $V \neq = 0$ ? ::: Yes — flat hilltop between two equal charges. Why does the $\infty$ reference fail for an infinite line? ::: $\int^{\infty} d s / s = ln \infty$ diverges; use a finite reference. Does a crooked path change $V_{b} - V_{a}$ ? ::: No — the field is conservative; only endpoints matter. Field from potential in one formula? ::: $E = - \nabla V$ .

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)