A potential drop is not an energy drop for every charge
You will be able to: Convert signed potential differences into kinetic-energy and speed changes.
Why can a negative charge gain speed while moving to higher potential?
A positive charge moving to lower potential loses electric potential energy. A negative charge can lose potential energy by moving to higher potential instead. In either case, that lost energy can become kinetic energy.
A useful starting point: An infinite line needs a finite potential reference →
Words and symbols before equations
- Kinetic energy K
- Nonnegative motion energy ½mv² for a nonrelativistic particle.
- Energy balance
- For motion with only electrostatic work, ΔK = −ΔU = −qΔV.
- Reachable energy
- A candidate state must have K ≥ 0; a negative calculated value is not a real speed.
- Fixed-source approximation
- Source positions and potentials stay unchanged; any supports do no work on stationary sources.
What this picture assumes
Nonrelativistic particle in a fixed electrostatic source field, with no other work. A monotonic one-dimensional potential path between endpoints is assumed. Speed is a magnitude; negative algebraic K rejects the candidate endpoint. A zero-force particle initially at rest does not spontaneously travel.
Read the picture in three steps
- Locate the labeled sources, system boundary or graph axes. Read the units before comparing values.
- ΔU=-100 μJ; candidate K_f=100 μJ. Speed magnitude 0.3162 m/s.
- Check what the picture assumes below. Use the Explore task to predict one change before moving a control.
Connect the picture to the physics
Start from K_f + U_f = K_i + U_i when the electrostatic interaction is the only changing potential energy and no other work occurs. Substitute ΔU = q(V_f − V_i) to obtain K_f = K_i − qΔV.
Only after checking K_f ≥ 0 should you calculate speed √(2K_f/m). This gives a speed magnitude, not a direction. In a specified one-dimensional monotonic potential path it can be used to decide if an endpoint is accessible; in a more complicated landscape one must also check intervening barriers.
An electron has q = −e. A positive ΔV gives it negative ΔU and positive ΔK. A positive charge released from rest accelerates toward lower V; a negative one toward higher V. Existing velocity can initially point in either direction.
A worked example, step by step
A 2 g bead with q = +2 μC starts at rest and moves from 50 V to 0 V, with only electric work. Find its speed.
- ΔV = 0 − 50 = −50 V.
- ΔU = (2 μC)(−50 V) = −100 μJ, so K_f = 100 μJ = 1.0 × 10⁻⁴ J.
- Convert mass: 2 g = 0.002 kg.
- v = √[2(1.0 × 10⁻⁴)/0.002] = 0.316 m/s. The assumed path must be energetically accessible.
Use signed qΔV. Reject negative K; do not take a square root and present an imaginary speed as motion.
Must every charge speed up when V decreases?
Compare with an explanation
No. A negative charge gains potential energy when V decreases and can slow down.
Predict. Change one thing. Explain.
Reverse the charge while holding the potential difference fixed. Check the energy balance before interpreting the speed. This model assumes a monotonic one-dimensional path between endpoints.
On narrow screens, swipe or scroll diagrams sideways to read all labels.
ΔU=-100 μJ; candidate K_f=100 μJ. Speed magnitude 0.3162 m/s.
Nonrelativistic particle in a fixed electrostatic source field, with no other work. A monotonic one-dimensional potential path between endpoints is assumed. Speed is a magnitude; negative algebraic K rejects the candidate endpoint. A zero-force particle initially at rest does not spontaneously travel.
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 −1 μC bead of mass 1 g starts at rest and reaches a point 20 V higher in potential. Find ΔU, K_f and speed, assuming an accessible path with only electric work.
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Compare with the answer and four-point rubric
- 1 point: ΔU = (−1 μC)(+20 V) = −20 μJ.
- 1 point: K_f = +20 μJ = 2.0 × 10⁻⁵ J.
- 1 point: m = 0.001 kg.
- 1 point: v = √[2(2.0 × 10⁻⁵)/0.001] = 0.20 m/s.
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 sign belongs in ΔK = −qΔV?
The actual signed charge q.
RECALL 2Can speed be inferred if K_f is negative?
No. The candidate state is not accessible in that model.
RECALL 3What additional information is needed for a trajectory?
Initial velocity direction and the spatial field, plus any constraints.
Revisit these tomorrow and a week later. Try a fresh problem and explain why the method applies.
A potential drop is not an energy drop for every charge
- K_f = K_i − q(V_f−V_i) when only electric work changes K.
- v_f = √(2K_f/m) for K_f ≥ 0.
- Energy determines speed, not the full trajectory.
Remember: Use signed qΔV. Reject negative K; do not take a square root and present an imaginary speed as motion.
Conditions: Nonrelativistic particle in a fixed electrostatic source field, with no other work. A monotonic one-dimensional potential path between endpoints is assumed. Speed is a magnitude; negative algebraic K rejects the candidate endpoint. A zero-force particle initially at rest does not spontaneously travel.
Refresh Kid · AP Physics C: Electricity and Magnetism Unit 2 (official Unit 9) · Objectives 9.3.A · Review edition
Framework, scope and review status
Mapped to College Board CED, Topic 9.3, objectives 9.3.A. 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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