Electric potential — definition V = −∫E·dl: Question bank

Before we start, a few symbols earn their place so nothing below is a mark you haven't met:

Definition The cast of characters

Potential $V$ — a number attached to each point: the electrical "height" of that spot, in volts.
Field $E$ — an arrow at each point: it points downhill in $V$ and its length is the steepness.
Test charge $q_{0}$ — a tiny positive charge we imagine placing at a point to probe it, so small it doesn't disturb the field it's measuring. When a problem talks about a real, arbitrary charge instead we call it $q$ .
Potential difference $Δ V = V_{b} - V_{a}$ — the height gap between two points; the only part that is physically measurable.
Coulomb constant $k$ — the fixed proportionality number in $V = \frac{k q}{r}$ and Coulomb's law, $k = \frac{1}{4 π ε _{0}} \approx 9 \times 1 0^{9} N\cdotm^{2} / C^{2}$ . It just sets the scale of how strong electric effects are; treat it as a positive constant. See Electric field E — definition and Coulomb's law.
Plate separation $d$ — for two parallel charged plates, $d$ is simply the gap (distance) between them, measured along the field. It appears in $E = Δ V / d$ . See Capacitance and parallel-plate fields (E = ΔV/d).

Figure 1 below is your reference "hill" for every height-vs-slope question (True/False items 2, 3, 7 and the Why questions on scalars). Figure 2 is the reference equipotential-and-perpendicular-field picture for the equipotential items and the shell edge case — glance back at them whenever a prompt mentions "downhill", "perpendicular", or "same height".

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

A third picture settles the two "divergent reference" and "same potential, nonzero field" traps at a glance — keep it beside the Edge cases section:

True or false — justify

Every item is a claim. The reveal gives the verdict and the reason — a bare "true/false" is worth nothing.

Potential

V

is a vector because it is built from the vector

E

False.

V

is a scalar — the dot product

E \cdot d l

collapses the arrow to a number before integrating. Direction lives in

E = - \nabla V

, not in

V

E = 0

at a point, then

V = 0

there too.

False.

E

is the slope of

V

(Figure 1); slope zero means

V

is flat (a hilltop or valley), not that its height is zero. Midway between two equal

+

charges

E = 0

but

V

is large and positive.

V = 0

at a point, then

E = 0

there too.

False. Height zero says nothing about steepness (Figure 1: a hill can cross zero height while still sloping). Midway between

+ q

and

- q

V = 0

yet

E

is strong (both charges push the same way).

Moving along an equipotential surface, the field does zero work.

True. On an equipotential

Δ V = 0

, and work by field

= q (V_{a} - V_{b}) = 0

. Geometrically

E ⊥

the surface (Figure 2), so

E \cdot d l = 0

for every step — see Equipotential surfaces and why E ⟂ them.

A positive charge released from rest always moves toward higher potential.

False. The field does positive work when charge moves to lower

V

, so a free positive charge accelerates down the

V

-hill (Figure 1), toward lower potential. (A negative charge goes toward higher

V

Setting

V = 0

at infinity is always a valid choice.

False. For an infinite line or plane of charge the integral to infinity diverges (see the Edge cases for the

\int d r / r

and

\int d r

arguments); you must pick a finite reference. Only

Δ V

is physical, so the reference is a free choice — but it must exist.

Two points can have the same potential yet the field between them be nonzero.

True. Equal height at two spots doesn't forbid a hill in between (Figure 3);

\int_{a}^{b} E \cdot d l

can be zero via cancellation while

E \neq = 0

along the way.

The minus sign in

V = - \int E \cdot d l

is just a convention we could drop.

False (physically). The sign encodes that potential decreases in the direction

E

points; drop it and you'd predict positive charges climbing hills spontaneously — a violation of the Work–energy theorem in electrostatics.

Doubling the test charge

q_{0}

doubles the potential

V

at its location.

False.

V

is work per unit charge — the

q_{0}

is divided out precisely so

V

is a property of the location alone. Only the potential energy

U = q_{0} V

doubles.

Spot the error

Each line contains one flawed statement or step. The reveal names the bug and repairs it.

"Field of a plate is

E = Δ V / d

, so a bigger plate spacing gives a bigger field."

Backwards causation. For a fixed

Δ V

, larger gap

d

gives smaller

E = Δ V / d

. See Capacitance and parallel-plate fields (E = ΔV/d).

V = \frac{k q}{r}

, so at

r \to 0

the potential is zero."

Wrong limit. As

r \to 0

V = k q / r \to \infty

(blows up), not zero. It is

r \to \infty

that gives

V \to 0

(here

k

is the Coulomb constant, a positive scale factor).

"Work by the field is

W = q (V_{b} - V_{a})

Sign flipped. It is

W_{field} = q (V_{a} - V_{b})

: the field does positive work when the charge drops from high

V_{a}

to low

V_{b}

E = + \nabla V

, since field and potential rise together."

Missing minus.

E = - \nabla V

: the field points opposite to the steepest increase of

V

— downhill, not uphill. See Gradient operator ∇ and directional derivatives.

"To bring in a test charge quasi-statically the agent applies

F_{ext} = + q_{0} E

Wrong direction. The agent must oppose the field:

F_{ext} = - q_{0} E

, so the net force is zero and kinetic energy stays fixed.

"Since only the component of force along the displacement does work, we write

W = \int F d l

(no dot)."

The dot product is the statement "component along displacement"; dropping it silently assumes force is already parallel to every step, which is false on a curved path. Keep

\int F \cdot d l

"For a point charge I integrated along a curved path, so the angle matters and

E \cdot d l \neq = E d r

Overcomplication.

V

is path-independent for electrostatic fields, so choose a radial path where

E ∥ d l

and

E \cdot d l = E d r

. The result

k q / r

is the same on any path.

Why questions

Why is

V

defined as work per unit charge instead of just the work?

Because force, and hence work, scales with the test charge

q_{0}

. Dividing it out leaves a quantity that describes the field/location alone, reusable for any charge you later drop in.

Why must the test charge

q_{0}

be moved without changing kinetic energy?

So that all agent-work becomes potential energy (none leaks into speed). This makes

W_{ext} = Δ U

exact, which is what lets us equate it to

q_{0} Δ V

Why is a scalar

V

more useful than the vector

E

for calculation?

Scalars just add (no components, no angles); you sum

V

contributions, then take one gradient to recover the full vector

E = - \nabla V

. This is the "solve easy, recover hard" shortcut.

Why does

E

always point perpendicular to equipotential surfaces?

Along the surface

V

is constant, so its slope in that direction is zero. All the steepness — and thus

E

— must lie in the one remaining perpendicular direction (Figure 2). See Equipotential surfaces and why E ⟂ them.

Why can only potential differences be measured, never absolute

V

Physics depends on

E = - \nabla V

, and adding a constant to

V

everywhere leaves its gradient unchanged. The constant (the reference height) is unobservable, so only

Δ V

shows up in any experiment.

Why does potential energy

U

differ from potential

V

even though they look alike?

U = q_{0} V

: potential is the "height per kilogram", energy is the height times the actual mass. See Potential energy of a charge configuration —

U

belongs to the charge-plus-field,

V

belongs to the location.

Why is the field measured in volts per metre?

From

E = Δ V / d

, field is potential-drop per length — literally volts per metre, showing

E

is the rate at which the

V

-hill descends as you walk.

Edge cases

State what happens (and why) in each degenerate or limiting scenario.

A charge sits exactly on an equipotential and you move it around a closed loop on that surface.

Net work is zero — both because

Δ V = 0

(start=end height) and because

E ⊥

every step. Electrostatic fields are conservative, so any closed loop gives zero anyway.

The charge distribution is an infinite charged plane; where do you put the reference?

Not at infinity. A plane gives a constant field

E \sim const

, so

\int^{\infty} E d r \sim \int^{\infty} d r

diverges (grows without bound). Pick a convenient finite point (often the plane itself,

V = 0

there) and measure

Δ V

from it.

The charge distribution is an infinite line; why does infinity fail here too?

A line gives

E \sim 1/ r

, so

\int^{\infty} \frac{d r}{r} = ln r^{\infty}

diverges logarithmically. Again choose a finite reference; the physical answer is always a

Δ V = ln (r_{1} / r_{2})

-type ratio.

Inside a hollow conducting shell (no charge inside the cavity) — what is

V

V

is constant everywhere inside, equal to its surface value. Reason:

E = 0

throughout the interior, so

V_{b} - V_{a} = - \int_{a}^{b} E \cdot d l = 0

for any two interior points — no slope means uniform height, a perfectly flat plateau.

Two identical charges

+ q

sit a distance

2 a

apart; you sit at the midpoint, so each charge is a distance

r = a

away. What are

V

and

E

V = k q / a + k q / a = 2 k q / a

is large and positive (scalars add);

E = 0

because the two equal arrows point opposite ways and cancel. A hilltop: high height, zero slope.

A dipole

+ q, - q

a distance

2 a

apart; you sit at the midpoint, each charge a distance

r = a

away. What are

V

and

E

V = k q / a - k q / a = 0

(equal-and-opposite scalars cancel) but

E \neq = 0

(both arrows point the same way, from

+

-

). Zero height, nonzero slope — this is the Figure 3 situation.

The field is uniform (

E

constant) — how does

V

vary along the field?

Linearly:

V

drops by

E

per metre, giving a straight, tilted ramp. Equipotentials are equally spaced parallel planes perpendicular to

E

You bring a test charge from infinity to a point near a negative charge. Sign of

V

Negative. The field points inward (toward

- q

), doing positive work on an incoming positive charge, so its potential energy — and

V = U / q_{0}

— comes out below zero:

V = - k ∣ q ∣/ r

At the exact location of an ideal point charge (

r = 0

), what is

V

It diverges to

\pm \infty

(

V = k q / r

). This is an artefact of the idealised point; real charges have finite size, so no true infinity occurs physically.

Recall One-line self-test

If a positive charge released from rest speeds up, which way did its potential change? Which way did its potential energy change? Answer ::: Potential decreased (it rolled down the $V$ -hill toward lower $V$ ); potential energy $U = q_{0} V$ decreased too, converting into kinetic energy — exactly the Work–energy theorem in electrostatics.

Connections

Parent: Electric potential — definition
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)