United StatesPhysics C: Electricity and MagnetismSyllabus dot point

How is electric potential related to the field through integration, and how do you find the field from the potential?

Topic 9.2 Electric Potential: relate potential to the field by line integral, find potential by superposition, and recover the field as the gradient of the potential.

A calculus-based answer to AP Physics C E&M Topic 9.2, covering electric potential as potential energy per charge, the line-integral relation to the field, potential of point and continuous distributions, equipotentials, and recovering the field as a gradient.

Generated by Claude Opus 4.812 min answerUpdated 2026-06-04

Reviewed by: AI editorial process; not yet individually human-reviewed

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Quick answer

Electric potential is the electric potential energy per unit charge: $V = \dfrac{U}{q}$ , in volts (J/C). It relates to the field by a line integral: $V_B - V_A = -\displaystyle\int_A^B \vec{E}\cdot d\vec{r}$ , or for a point charge $V = \dfrac{kq}{r}$ (taking $V = 0$ at infinity). Because potential is a scalar, the potential of several charges is the simple algebraic sum $V = \sum_i \dfrac{kq_i}{r_i}$ , with no components. The field is recovered as the negative gradient of the potential: $E_x = -\dfrac{\partial V}{\partial x}$ (and similarly for $y$ , $z$ ), so the field points toward decreasing potential. Equipotential surfaces are everywhere perpendicular to the field, and no work is done moving a charge along one.

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What this topic is asking
Defining potential
Potential and the field
Superposition: a scalar sum
The field as a gradient
Try this

What this topic is asking

The College Board (Topic 9.2) wants you to define electric potential as potential energy per unit charge, relate it to the field through a line integral, compute it by superposition for point and continuous distributions, and recover the field from the potential as a gradient. Potential is a scalar, which makes it far easier to compute than the vector field, and the gradient gives the field back.

Defining potential

A charge $q$ at potential $V$ has potential energy $U = qV$ , and moving it through a potential difference $\Delta V$ changes its energy by $\Delta U = q\Delta V$ .

Potential and the field

Potential is the line integral of the field:

V_B - V_A = -\int_A^B \vec{E}\cdot d\vec{r}

A charge moving in the direction of the field moves to lower potential. For a single point charge, taking $V = 0$ at infinity, this integrates to

V = \frac{1}{4\pi\varepsilon_0}\frac{q}{r} = k\frac{q}{r}

a scalar that keeps the sign of $q$ (positive near a positive charge, negative near a negative charge).

Superposition: a scalar sum

Because potential is a scalar, superposition is just addition, with no direction bookkeeping:

V = \sum_i k\frac{q_i}{r_i} \qquad\text{(point charges)}, \qquad V = \int \frac{k\,dq}{r} \qquad\text{(continuous)}

This is the great advantage of potential: where the field needs vector components, the potential needs only signed magnitudes. Often you find $V$ first, then differentiate to get the field.

The field as a gradient

The field is recovered from the potential by taking the negative gradient:

E_x = -\frac{\partial V}{\partial x}, \qquad E_y = -\frac{\partial V}{\partial y}, \qquad E_z = -\frac{\partial V}{\partial z}

In one dimension this is simply $E = -\dfrac{dV}{dx}$ . The minus sign encodes that the field points from high to low potential. Equipotentials (surfaces of constant $V$ ) are perpendicular to the field everywhere; moving along one does no work because $\vec{E}\cdot d\vec{r} = 0$ .

Potential and field of two charges on a line

Charges $+q$ at $x = -a$ and $-q$ at $x = +a$ (a dipole). Find the potential at a point on the x-axis at $x > a$ , then the field there.

step 1 Add the potentials (scalars)

Distances to the point at $x$ : from $+q$ it is $x + a$ , from $-q$ it is $x - a$ . $V = k\dfrac{+q}{x+a} + k\dfrac{-q}{x-a} = kq\left(\dfrac{1}{x+a} - \dfrac{1}{x-a}\right)$ .

step 2 Combine over a common denominator

$V = kq\dfrac{(x-a) - (x+a)}{(x+a)(x-a)} = kq\dfrac{-2a}{x^2 - a^2} = -\dfrac{2kqa}{x^2 - a^2}$ .

step 3 Differentiate for the field

$E_x = -\dfrac{dV}{dx} = 2kqa\dfrac{d}{dx}\left(\dfrac{1}{x^2-a^2}\right) = 2kqa\cdot\dfrac{-2x}{(x^2-a^2)^2} = -\dfrac{4kqax}{(x^2-a^2)^2}$ .

step 4 Check the far field

For $x \gg a$ , $V \approx -\dfrac{2kqa}{x^2}$ , the dipole potential ( $\propto 1/x^2$ ), and $E_x \approx -\dfrac{4kqa}{x^3}$ , falling as $1/x^3$ , both the signatures of a dipole.

Try this

Q1. A $+6.0$ nC charge sits at the origin. Find the potential $0.30$ m away ( $k = 8.99\times10^9$ ). [2 points]

Cue. $V = k\dfrac{q}{r} = (8.99\times10^9)\dfrac{6.0\times10^{-9}}{0.30} = 180$ V.

Q2. The potential in a region is $V = 5x$ (volts, $x$ in meters). Find $E_x$ . [1 point]

Cue. $E_x = -\dfrac{dV}{dx} = -5$ V/m.

Exam-style practice questions

Practice questions written in the style of College Board exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

AP 2023 (style)1 marksSection I (multiple choice). In a region the potential is

V(x) = 6x^2

(in volts, with

x

in meters). The x-component of the electric field at

x = 2

m is (A)

-24

V/m (B)

+24

V/m (C)

-12

V/m (D)

+12

V/m. Justify your reasoning.

Show worked answer →

A 1-point MCQ on the field-potential gradient. The answer is (A).

The field is minus the gradient: $E_x = -\dfrac{dV}{dx} = -\dfrac{d}{dx}(6x^2) = -12x$ . At $x = 2$ , $E_x = -24$ V/m. The field points toward lower potential, here the $-x$ direction. The trap is forgetting the minus sign or not differentiating.

AP 2024 (style)6 marksSection II (FRQ, derivation). A thin ring of radius

R

carries total charge

Q

spread uniformly. (a) Derive the potential at a point on the axis a distance

x

from the center. (b) Use the result to find the on-axis field. (c) State the potential at the center.

Show worked answer →

A 6-point FRQ linking potential and field by integration and differentiation.

(a) Potential (3 points): every element $dq$ on the ring is the same distance $r = \sqrt{x^2 + R^2}$ from the axial point, so $dV = \dfrac{k\,dq}{r}$ and $V = \dfrac{k}{\sqrt{x^2+R^2}}\displaystyle\int dq = \dfrac{kQ}{\sqrt{x^2+R^2}}$ . Potential is a scalar, so no components are needed.
(b) Field from gradient (2 points): $E_x = -\dfrac{dV}{dx} = -kQ\dfrac{d}{dx}(x^2+R^2)^{-1/2} = \dfrac{kQx}{(x^2+R^2)^{3/2}}$ , matching the ring result.
(c) Center (1 point): set $x = 0$ : $V = \dfrac{kQ}{R}$ (non-zero, even though the field there is zero).

Markers reward the constant $r$ in the scalar integral, the differentiation for the field, and the center value.

Related dot points

Sources & how we know this

AP Physics C: Electricity and Magnetism Course and Exam Description — College Board (2024)