Equipotential surfaces — perpendicular to field: Worked examples

This page is a workout. The parent note built the single big fact — the electric field $E$ is always perpendicular to an equipotential surface, and $E = - \nabla V$ . Here we drag that fact through every kind of situation it can meet, so that when an exam or a real problem throws a case at you, you have already seen its twin.

Before anything, three symbols we will reuse constantly, in plain words:

The scenario matrix

Every problem this topic can throw is one of these cells. The examples below are labelled with the cell(s) they cover, and together they hit all of them.

#	Case class	What is special about it	Covered by
A	Positive point charge	Spheres, $E$ points outward	Ex 1
B	Negative point charge	Spheres, $E$ points inward (sign flip)	Ex 2
C	Uniform field (parallel plates)	Flat, equally spaced planes	Ex 3
D	Two equal charges (dipole-ish geometry)	Curved surfaces, a zero-field saddle point	Ex 4
E	Degenerate: on the surface itself	Motion along — must give $W = 0$	Ex 5
F	Limiting behaviour: $r \to \infty$ and $r \to 0$	Spacing spreads / crowds without bound	Ex 6
G	Conductor surface	Whole body one equipotential, $E ⊥$ surface	Ex 7
H	Real-world word problem	Numbers with units, capacitor-style	Ex 8
I	Exam twist: sign / direction trap	Where the naive answer is wrong	Ex 9

Cells A and B split the sign of the source. C, D cover the shapes. E, F cover the degenerate/limiting inputs. G is the killer application. H, I are the word-problem and trap cells.

Ex 1 — Positive point charge (Cell A)

Worked example A charge

$+ Q = + 2 nC$ sits at the origin. Find $V$ and $E$ at $r = 0.30 m$ , and check they are perpendicular to the equipotential there.

Forecast: Guess the shape of the equipotential and whether $E$ points toward or away from the charge before reading on.

Step 1 — Write the potential. For a point charge, $V (r) = k \frac{Q}{r}, k = 8.99 \times 1 0^{9} N\cdotm^{2} / C^{2} .$ Why this step? $V$ depends only on $r$ (distance from the charge). " $V = const$ " therefore means " $r = const$ " — a sphere. That already tells us the equipotential shape.

Step 2 — Plug the numbers. $V = (8.99 \times 1 0^{9}) \frac{2 \times 1 0 ^{- 9}}{0.30} = 59.9 V .$ Why this step? We want a concrete height on our "landscape" to anchor the picture.

Step 3 — Field magnitude. $E = k \frac{Q}{r ^{2}} = (8.99 \times 1 0^{9}) \frac{2 \times 1 0 ^{- 9}}{0.3 0 ^{2}} = 199.8 V/m .$ Why this step? $E = - d V / d n$ and here the normal step $d n$ is just the radial step $d r$ (the spheres' normal is radial); differentiating $Q / r$ gives $- Q / r^{2}$ , so the field falls off faster than $V$ .

Step 4 — Direction & perpendicularity. $E = E \overset{r}{^}$ points radially outward (away from $+ Q$ ). The outward radial direction $\overset{r}{^}$ is exactly the normal to the sphere $r = const$ . So $E ⊥$ equipotential. ✓

The figure below draws three of these spheres (blue) and the field arrows (orange) shooting straight out through them. Look at the small green right-angle mark where an arrow crosses a sphere: that $9 0^{\circ}$ crossing is the whole point of Cell A — and notice the arrows near the charge are packed tighter, foreshadowing the stronger field there.

Verify: Units: $\frac{N\cdotm ^{2} / C ^{2} \cdot C}{m ^{2}} = N/C = V/m$ ✓. Sanity: $E / V = 1/ r = 1/0.30 = 3.33 m^{- 1}$ , and indeed $199.8/59.9 = 3.34$ ✓.

Figure — Equipotential surfaces — perpendicular to field

Ex 2 — Negative point charge (Cell B — the sign flip)

Worked example Replace the source with

$- Q = - 2 nC$ at the origin. What changes at $r = 0.30 m$ ?

Forecast: Which way does $E$ point now — and does the perpendicularity fact survive?

Step 1 — Potential goes negative. $V = (8.99 \times 1 0^{9}) \frac{- 2 \times 1 0 ^{- 9}}{0.30} = - 59.9 V .$ Why this step? The sign of $Q$ carries straight into $V$ . The "landscape" is now a valley, not a hill.

Step 2 — Equipotentials are still spheres. " $V = const$ " still forces " $r = const$ ." Shape unchanged. Why this step? The geometry depends on the dependence on $r$ , not on the sign — this is the point of the cell.

Step 3 — Field flips direction. $E = k \frac{( - Q )}{r ^{2}} \overset{r}{^} = - 199.8 \overset{r}{^} V/m,$ i.e. magnitude $199.8 V/m$ pointing inward (toward the charge). Why this step? $E = - \nabla V$ : since $V$ increases as you move away from the negative charge (from $- 59.9$ toward $0$ ), the field points the opposite way — inward.

Step 4 — Perpendicularity. Inward radial is still along $\overset{r}{^}$ (just $- \overset{r}{^}$ ), still the sphere's normal. $E ⊥$ equipotential holds. ✓

Verify: $∣ E ∣$ identical to Ex 1 ( $199.8 V/m$ ) because $E \propto ∣ Q ∣/ r^{2}$ . Direction reversed. Consistent with " $E$ points from higher $V$ ( $0$ far away) to lower $V$ ( $- 59.9$ near the charge)" — see Work and Conservative Forces for why downhill is toward lower $V$ .

Ex 3 — Uniform field / parallel plates (Cell C)

Worked example Two plates

$d = 5.0 mm$ apart hold a potential difference $Δ V = 100 V$ . Find $E$ , the equipotential spacing for $Δ V = 20 V$ steps, and confirm the planes are flat and evenly spaced. See Parallel Plate Capacitor.

Forecast: Will the $20 V$ sheets be evenly or unevenly spaced?

Step 1 — Field from spacing. $E = \frac{Δ V}{d} = \frac{100}{5.0 \times 1 0 ^{- 3}} = 2.0 \times 1 0^{4} V/m .$ Why this step? Inside the plates $E$ is uniform, so $E = - d V / d n$ becomes a plain ratio (no calculus needed): total drop over total gap, since the normal step $d n$ runs straight across from plate to plate.

Step 2 — $V$ is linear across the gap. With $E = E \overset{x}{^}$ , integrating gives $V = - E x + C$ . So " $V = const$ " $\Rightarrow$ " $x = const$ ": flat planes perpendicular to $\overset{x}{^}$ . Why this step? Linear $V$ is what makes the sheets flat and (next step) evenly spaced.

Step 3 — Spacing per $20 V$ step. Each equal $Δ V$ needs an equal $Δ x$ : $Δ x = \frac{Δ V _{step}}{E} = \frac{20}{2.0 \times 1 0 ^{4}} = 1.0 \times 1 0^{- 3} m = 1.0 mm .$ Five equal sheets across the $5.0 mm$ gap. ✓ Why this step? Equal $Δ V$ and equal spacing $\Rightarrow$ uniform field — the signature of the parallel-plate cell.

The figure below shows the two plates (red $+ 100 V$ , blue $0 V$ ), the four green dashed equipotential planes at $20 V$ steps, and the orange field arrows running from high to low. Notice the arrows are all the same length and the dashed planes are all the same distance apart — that even spacing is the visual signature of a uniform field.

Verify: $5$ steps $\times 1.0 mm = 5.0 mm = d$ ✓. Units: $V / m$ from $V / m$ ✓.

Ex 4 — Two equal positive charges: the saddle / zero-field point (Cell D)

Worked example Two charges

$+ Q = + 2 nC$ sit on the $x$ -axis at $x = \pm a$ with $a = 0.10 m$ . Show that at the midpoint (origin) the field is zero, yet the point still sits on a perfectly good equipotential.

Forecast: If $E = 0$ at a point, what does "perpendicular to the equipotential" even mean there?

Step 1 — Field from each charge at the origin. Each charge is a distance $a$ away: $E_{each} = k \frac{Q}{a ^{2}} = (8.99 \times 1 0^{9}) \frac{2 \times 1 0 ^{- 9}}{0.1 0 ^{2}} = 1798 V/m .$ Why this step? We build the total field by superposition — add the two arrows.

Step 2 — They cancel. The left charge pushes a test $+$ charge to the right ( $+ \overset{x}{^}$ ); the right charge pushes it to the left ( $- \overset{x}{^}$ ). Equal magnitudes, opposite directions: $E_{total} = 1798 \overset{x}{^} - 1798 \overset{x}{^} = 0.$ Why this step? This is the degenerate zero-field input the matrix demands.

Step 3 — Potential is not zero there. Potential is a scalar — no cancellation: $V_{origin} = 2 \times k \frac{Q}{a} = 2 \times (8.99 \times 1 0^{9}) \frac{2 \times 1 0 ^{- 9}}{0.10} = 359.6 V .$ Why this step? It shows $V$ and $E$ are different animals: adding numbers vs. adding arrows.

Step 4 — Meaning of " $⊥$ " when $E = 0$ . Where $E = 0$ , the "downhill direction" is undefined — there is no direction of steepest descent, so perpendicularity is vacuously satisfied. The origin is a saddle of the potential landscape: down along the $x$ -axis toward each charge is uphill, but stepping off the axis it curves the other way. Equipotentials near a saddle look like the crossed hyperbolas of a mountain pass — but the two branches belong to the same $V$ value, so this is not two different equipotentials crossing (that is still forbidden).

Verify: $E_{total} = 0$ exactly by symmetry ✓. $V_{origin} = 2 \times 179.8 = 359.6 V$ ✓ (twice the single-charge value at $0.10 m$ ).

Ex 5 — Degenerate move: sliding along an equipotential (Cell E)

Worked example Take

$+ Q = + 2 nC$ at the origin (Ex 1). A charge $q = + 5 nC$ is dragged from point $A = (0.30, 0) m$ to point $B = (0, 0.30) m$ along the circular arc $r = 0.30 m$ . Find the work done by the electric force.

Forecast: Guess the work before computing. Does the curvy path matter?

Step 1 — Are $A$ and $B$ on the same equipotential? Both are at $r = 0.30 m$ , so both have $V = 59.9 V$ (from Ex 1). Yes — same sphere. Why this step? The whole point of the equipotential is that "same $V$ " is all that matters.

Step 2 — Apply the work formula. $W_{A B} = q (V_{A} - V_{B}) = q (59.9 - 59.9) = 0 J .$ Why this step? The electric force is conservative (Work and Conservative Forces): work depends only on endpoints, and here the endpoints tie.

Step 3 — Geometric echo. Along the arc, every step $d l$ is tangent to the sphere, and $E$ is radial (normal). Their dot product $E \cdot d l = 0$ at every point, so no work accumulates — consistent with Step 2. Why this step? It connects the algebra ( $V_{A} = V_{B}$ ) back to the geometry (perpendicularity) — the two are the same fact.

Verify: $W = 5 \times 1 0^{- 9} \times (59.9 - 59.9) = 0$ ✓. Any other path on the same sphere gives $0$ too — path length is irrelevant.

Ex 6 — Limiting behaviour of the spacing (Cell F)

Worked example For

$+ Q = + 2 nC$ , draw equipotentials at $V = 60, 30, 15 V$ (each half the previous). Find their radii and comment on the spacing as $r \to \infty$ and $r \to 0$ .

Forecast: As you halve the voltage, does the gap between spheres grow, shrink, or stay equal?

Step 1 — Invert $V (r)$ to get radii. From $V = \frac{k Q}{r}$ (with $k$ Coulomb's constant, so $k Q = 17.98 V\cdotm$ ): $r = \frac{k Q}{V} .$ Why this step? We want the positions of the surfaces to measure their spacing.

Step 2 — Compute. $r_{60} = \frac{17.98}{60} = 0.300 m, r_{30} = \frac{17.98}{30} = 0.599 m, r_{15} = \frac{17.98}{15} = 1.199 m .$ Gaps: $r_{30} - r_{60} = 0.299 m$ , then $r_{15} - r_{30} = 0.599 m$ — doubling each time. Why this step? This makes the spreading concrete.

Step 3 — The two limits.

As $r \to \infty$ ( $V \to 0^{+}$ ): spheres spread farther and farther apart for equal $Δ V$ → field weakens toward $0$ ( $E \sim 1/ r^{2}$ ).
As $r \to 0$ ( $V \to \infty$ ): spheres crowd infinitely close → field blows up. The point charge is a singular ("infinitely steep") spot.

Why this step? These are the boundary inputs the matrix insists on — the reader now knows what happens at the extremes, not just in the comfortable middle.

Verify: Ratio check — halving $V$ doubles $r$ : $r_{30} / r_{60} = 0.599/0.300 = 2.00$ ✓, $r_{15} / r_{30} = 1.199/0.599 = 2.00$ ✓.

Ex 7 — Conductor surface (Cell G — the killer application)

Worked example A solid metal sphere of radius

$R = 0.05 m$ carries $Q = + 2 nC$ . Find $V$ everywhere inside and the surface value, and explain the field direction just outside. See Conductors in Electrostatic Equilibrium.

Forecast: Is the inside of the metal at a lower, higher, or equal potential compared with its surface?

Step 1 — Inside a conductor $E = 0$ . Free electrons rearrange until no field remains inside (electrostatic equilibrium). Why this step? $E = 0$ is the starting axiom for conductors.

Step 2 — Zero field ⇒ constant $V$ . Since $E = - d V / d n = 0$ everywhere inside, $V$ cannot change: the entire sphere, surface included, is one equipotential. $V_{surface} = \frac{k Q}{R} = \frac{17.98}{0.05} = 359.6 V .$ Why this step? Outside a charged sphere the field is that of a point charge, so $V = k Q / r$ ; evaluating at $r = R$ gives the constant interior value.

Step 3 — Field just outside is perpendicular. The surface is an equipotential, so $E$ (which is $⊥$ every equipotential) must meet the metal at exactly $9 0^{\circ}$ . Any tangential component would shove the surface charges until it cancelled. Why this step? This is why field lines strike metal at right angles — the whole reason this topic matters in real devices.

Verify: $V_{surface} = 17.98/0.05 = 359.6 V$ ✓. Note it matches Ex 4's origin value coincidentally because both used $k Q$ and a $0.10$ / $0.05$ -type ratio; here $r = R = 0.05$ .

Ex 8 — Real-world word problem (Cell H)

Worked example An air-gap capacitor in a smoke detector has plates

$2.0 mm$ apart across $9.0 V$ . (a) What is the field between the plates? (b) A dust particle carrying $q = + 1.6 \times 1 0^{- 19} C$ is moved fully from the low plate to the high plate — how much work does the external agent do against the field? (c) How far apart are the $3.0 V$ equipotential sheets?

Forecast: Will the work be positive or negative for an external agent pushing a $+$ charge toward the high plate?

Step 1 — Field (uniform). $E = \frac{V}{d} = \frac{9.0}{2.0 \times 1 0 ^{- 3}} = 4500 V/m .$ Why this step? Parallel plates ⇒ uniform field ⇒ the simple ratio $E = - d V / d n$ becomes total drop over the gap.

Step 2 — Work by external agent. Moving $+ q$ from the low plate ( $V = 0$ ) to the high plate ( $V = 9.0 V$ ), the field does $W_{field} = q (V_{A} - V_{B}) = q (0 - 9.0)$ . The agent does the opposite: $W_{agent} = q (V_{B} - V_{A}) = (1.6 \times 1 0^{- 19}) (9.0) = 1.44 \times 1 0^{- 18} J .$ Positive — the agent must push a $+$ charge up the potential hill. Why this step? Sign discipline: agent work $= + q Δ V$ against the field.

Step 3 — Spacing of $3.0 V$ sheets. $Δ x = \frac{3.0}{4500} = 6.67 \times 1 0^{- 4} m = 0.667 mm .$ Three sheets fill the $2.0 mm$ gap. Why this step? Equal- $Δ V$ spacing again — the everyday capacitor picture.

Verify: $3 \times 0.667 mm = 2.0 mm = d$ ✓. $W_{agent} = 1.6 \times 1 0^{- 19} \times 9.0 = 1.44 \times 1 0^{- 18} J$ ✓. Units: $C \cdot V = J$ ✓.

Ex 9 — Exam twist: the direction trap (Cell I)

Worked example Between two plates, the left plate is at

$V = + 40 V$ and the right at $V = + 10 V$ , separated by $d = 0.020 m$ . A student writes " $E$ points from left to right because left is more positive... wait, or is it the other way?" Find $E$ 's magnitude and correct direction.

Forecast: Left plate is at higher potential. Does $E$ point away from it or toward it?

Step 1 — Magnitude. $E = \frac{∣ V _{left} - V _{right} ∣}{d} = \frac{∣40 - 10∣}{0.020} = \frac{30}{0.020} = 1500 V/m .$ Why this step? Only the potential difference and the gap set the strength; the absolute values ( $40$ vs $10$ ) don't matter.

Step 2 — Direction from $E = - \nabla V$ . The field points downhill in $V$ , i.e. from the higher-potential left plate ( $40 V$ ) toward the lower-potential right plate ( $10 V$ ) — so left → right. See Gradient and Directional Derivative: $- \nabla V$ points toward decreasing $V$ . Why this step? The trap is thinking " $+$ plate repels, so field goes... ?" The clean rule is always "field points from high $V$ to low $V$ ", no repulsion story needed.

Step 3 — Perpendicularity check. Equipotentials here are vertical planes ( $x = const$ ); $E$ is horizontal (along $x$ ) — meets them at $9 0^{\circ}$ . ✓

Verify: $E = 30/0.020 = 1500 V/m$ ✓. Direction: high $\to$ low = left $\to$ right ✓. (Had both plates been negative, say $- 40$ and $- 10$ , the field would still point from $- 40$ toward $- 10$ , i.e. toward the less negative plate — the absolute sign never enters, only the difference.)

Recall Quick self-test across the matrix

Sign flip changes what about a point charge's equipotentials? ::: Only the field direction (and the sign of $V$ ); the spherical shape is unchanged. At a zero-field saddle point, is $E ⊥$ equipotential? ::: Vacuously yes — there is no field direction, so no violation; and the crossing hyperbolas are the same $V$ , not two different surfaces. Work sliding a charge along one equipotential? ::: Zero, for any path — the force is conservative and $V_{A} = V_{B}$ . As $r \to \infty$ for a point charge, equal- $Δ V$ spheres get... ::: farther apart (field weakens as $1/ r^{2}$ ). Direction of $E$ between plates at $+ 40 V$ and $+ 10 V$ ? ::: From the $40 V$ plate toward the $10 V$ plate (high $V$ → low $V$ ).

Connections

Parent topic
Electric Potential · Electric Field Lines · Gradient and Directional Derivative
Conductors in Electrostatic Equilibrium · Parallel Plate Capacitor · Work and Conservative Forces