The electric work depends on endpoints, not detours
You will be able to: Calculate a uniform-field potential difference and separate electric from external work.
Does a longer path always require more electrical work?
Walk across a level room while holding a book at the same height: the gravitational energy does not change with the length of the route. In a static uniform electric field, only displacement along the field changes electric potential.
A useful starting point: A voltage compares two points using one reference →
Words and symbols before equations
- Line integral
- A sum of local E·dℓ contributions along a path.
- Dot product
- The component of a vector along a displacement, times that displacement.
- Conservative field
- A field whose work between fixed endpoints is path-independent.
- Quasistatic motion
- Slow controlled motion with negligible change in kinetic energy.
What this picture assumes
Static uniform field along x, path A(0,0) → (0,h) → (Δx,h) → B(Δx,0). The reference direct path shares A and B. External work is for slow transport with no kinetic-energy change. Coordinate paths are not free particle trajectories.
Read the picture in three steps
- Locate the labeled sources, system boundary or graph axes. Read the units before comparing values.
- Direct length 3 m; detour length 7 m. Both give ΔV = -60 V. Electric work 120 μJ; slow external work -120 μJ.
- 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 uniform E along +x, ΔV = −∫E·dℓ = −E(x_B − x_A). A perpendicular displacement contributes zero because E·dℓ = 0. Potential decreases in the field direction.
The electric force does W_e = qEΔx = −qΔV. A slow external agent balancing that force does W_ext = qΔV. A negative charge reverses both work signs relative to a positive charge.
In electrostatics, a direct path and a detour with the same endpoints have the same potential difference. Around a closed path the total is zero. This claim assumes static electric sources; induced electric fields from changing magnetism belong to a later unit.
| Quantity | Electric force | External agent |
|---|---|---|
| Work in slow transport | −qΔV | +qΔV |
| Role | Changes configuration energy into or out of motion | Supplies or removes energy while keeping K fixed |
| Condition | Electrostatic source field | No change in kinetic energy; no other work |
A worked example, step by step
E = +20 N/C along x. Move a +2 μC probe from (0,0) to (3,2) m slowly. Find ΔV and both works.
- Only the +3 m x displacement contributes; the 2 m y displacement contributes zero.
- ΔV = −20(3) = −60 V.
- Electric work is −qΔV = +120 μJ.
- For constant kinetic energy, external work is qΔV = −120 μJ. The agent removes energy.
Path length does not replace displacement in qEΔx. External work equals +ΔU only under the stated slow-motion condition.
What happens along an equipotential part of the path?
Compare with an explanation
ΔV = 0 for that segment, so the electric work on a fixed q is zero.
Predict. Change one thing. Explain.
Increase the height of the rectangular detour while keeping its endpoints’ x separation fixed. Compare path length with ΔV and electric work.
On narrow screens, swipe or scroll diagrams sideways to read all labels.
Direct length 3 m; detour length 7 m. Both give ΔV = -60 V. Electric work 120 μJ; slow external work -120 μJ.
Static uniform field along x, path A(0,0) → (0,h) → (Δx,h) → B(Δx,0). The reference direct path shares A and B. External work is for slow transport with no kinetic-energy change. Coordinate paths are not free particle trajectories.
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 questionA −3 μC probe moves from x = 0 to 2 m in Eₓ = +10 N/C. Find ΔV, ΔU and electric work; state whether a slow external agent supplies or removes energy.
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Compare with the answer and four-point rubric
- 1 point: ΔV = −10(2) = −20 V.
- 1 point: ΔU = (−3 μC)(−20 V) = +60 μJ.
- 1 point: W_e = −60 μJ.
- 1 point: The slow external agent supplies +60 μJ.
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 1Which part of displacement matters in a uniform field?
Its component along the field.
RECALL 2Why can a detour have the same work?
Electrostatic work is path-independent.
RECALL 3Does zero electric work mean every force is zero?
No. Force may be perpendicular to displacement.
Revisit these tomorrow and a week later. Try a fresh problem and explain why the method applies.
The electric work depends on endpoints, not detours
- ΔV = −EΔx for uniform E along x.
- W_e = −qΔV; slow W_ext = qΔV.
- Electrostatic closed-path integral ∮E·dℓ = 0.
Remember: Path length does not replace displacement in qEΔx. External work equals +ΔU only under the stated slow-motion condition.
Conditions: Static uniform field along x, path A(0,0) → (0,h) → (Δx,h) → B(Δx,0). The reference direct path shares A and B. External work is for slow transport with no kinetic-energy change. Coordinate paths are not free particle trajectories.
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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