Field is negative slope, not potential height
You will be able to: Differentiate potential and interpret field signs and stationary points.
Where is the electric field strongest on a potential graph?
A high plateau can have zero slope, while a lower hillside is steep. Similarly, a large V value can have zero E, and a small V value can lie in a strong field. The change with position matters.
A useful starting point: Signed field area gives the potential change →
Words and symbols before equations
- Derivative dV/dx
- Local slope of a potential-versus-position graph, in V/m.
- Field component Eₓ
- The negative local slope, Eₓ = −dV/dx.
- Stationary point
- A point where the first derivative is zero.
- Reference offset C
- A vertical shift of the V graph that leaves its slope unchanged.
What this picture assumes
A one-dimensional prescribed potential V = C + Ax + Bx². Field is Eₓ = −A − 2Bx. C is a reference offset. A stationary point in one coordinate is not a claim of full three-dimensional stability.
Read the picture in three steps
- Locate the labeled sources, system boundary or graph axes. Read the units before comparing values.
- At x=1 m: V=7 V; slope -4 V/m; Eₓ=4 N/C. A common offset changes V but not its slope.
- Check what the picture assumes below. Use the Explore task to predict one change before moving a control.
Connect the picture to the physics
For V(x) = C + Ax + Bx², differentiate term by term: dV/dx = A + 2Bx. Thus Eₓ = −A − 2Bx. C disappears because a constant has no spatial change.
If V rises to the right, Eₓ points left. If V falls to the right, Eₓ points right. The field magnitude is larger where the V graph is steeper, not where its ordinate is larger.
In two or three dimensions, each field component is a negative partial derivative: Eₓ = −∂V/∂x and similarly for y and z. A flat slope along one direction only makes that component zero; it does not alone prove the whole field is zero.
A worked example, step by step
V(x) = 12 − 6x + x² volts with x in meters. Find Eₓ at x = 1 m and the zero-field position.
- Differentiate: dV/dx = −6 + 2x V/m.
- Eₓ = 6 − 2x N/C.
- At x = 1 m, Eₓ = +4 N/C, pointing right.
- Set Eₓ = 0 to obtain x = 3 m. V there is 3 V, not zero.
A zero or large potential value does not determine E. Read the spatial slope and its sign.
Is E zero along an entire curve on which V is constant?
Compare with an explanation
Only its tangential component is zero; a perpendicular component may remain.
Predict. Change one thing. Explain.
Move the probe along the potential parabola. Then change only the offset C; predict how the potential graph and the field graph respond.
On narrow screens, swipe or scroll diagrams sideways to read all labels.
At x=1 m: V=7 V; slope -4 V/m; Eₓ=4 N/C. A common offset changes V but not its slope.
A one-dimensional prescribed potential V = C + Ax + Bx². Field is Eₓ = −A − 2Bx. C is a reference offset. A stationary point in one coordinate is not a claim of full three-dimensional stability.
Explain what you noticed: Which quantity changed? Which stayed fixed? Use the relevant potential, energy, work, field-gradient or line-integral relationship to justify your prediction.
Apply the idea to a fresh problem Practice →Show what you understand.
Two original questions are a starting check, not proof of mastery. Explain your choice before revealing the answer.
Original written challenge
4 points · self-check · not an official AP questionFor V = 3 + 4x − x² V, determine Eₓ at x = 1 and 3 m, and locate the zero-field point.
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Compare with the answer and four-point rubric
- 1 point: dV/dx = 4 − 2x, so Eₓ = −4 + 2x.
- 1 point: At x = 1 m, Eₓ = −2 N/C.
- 1 point: At x = 3 m, Eₓ = +2 N/C.
- 1 point: Eₓ = 0 at x = 2 m, where the potential has a maximum.
Accept equivalent correct methods and explanations. This is a Refresh Kid teaching rubric, not an official AP scoring guideline.
Retrieve it before you reveal it.
RECALL 1What does graph height measure?
Potential in volts.
RECALL 2What does the negative slope measure?
The corresponding field component in N/C.
RECALL 3What does ∇ add beyond a one-dimensional derivative?
Spatial slopes in each coordinate direction.
Revisit these tomorrow and a week later. Try a fresh problem and explain why the method applies.
Field is negative slope, not potential height
- Eₓ = −dV/dx in one dimension.
- E = −∇V for electrostatic potential.
- Adding a constant to V changes no field component.
Remember: A zero or large potential value does not determine E. Read the spatial slope and its sign.
Conditions: A one-dimensional prescribed potential V = C + Ax + Bx². Field is Eₓ = −A − 2Bx. C is a reference offset. A stationary point in one coordinate is not a claim of full three-dimensional stability.
Refresh Kid · AP Physics C: Electricity and Magnetism Unit 2 (official Unit 9) · Objectives 9.2.B · Review edition
Framework, scope and review status
Mapped to College Board CED, Topic 9.2, objectives 9.2.B. CED effective Fall 2024, current PDF ©2026; checked September 16, 2026. This is E&M Unit 2: Electric Potential, numbered Unit 9 in the official combined Physics C sequence. Topics 9.1–9.3 retain their official identifiers. Models assume electrostatic fields and state whether source charges are fixed or free. Potential integrals use the specified rods, ring, arc and infinite line or cylinder. Infinite-line examples use a finite reference radius, not zero potential at infinity. The optional 3D equipotential surface uses height to represent volts, not a particle trajectory. Checked with the Fall 2026 clarifications. The lesson breakdown and questions are original Refresh Kid work, not official topic subdivisions.
Implementation and automated checks are separate from independent teacher review and observation of students. Both human review stages remain pending. This is a review edition, not a certified or validated assessment.
Optional further resource: College Board’s released questions and scoring guides. Papers can combine units; this link is an archive, not an assignment of every question to this lesson.
Our learn, explore, practice and recall sequence is informed by the IES learning guide. The exact Refresh Kid implementation has not been evaluated for learning effectiveness.
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