When both charges move, share the released energy
You will be able to: Combine energy and momentum conservation for two freely separating charges.
Does each released charge get all of the lost potential energy?
Release two positive charged beads from rest in isolation. They push apart in opposite directions. Their pair potential energy decreases and the two kinetic energies share the released amount; it cannot all be assigned to both beads.
A useful starting point: Check the barrier before accepting an endpoint →
Words and symbols before equations
- Isolated pair
- Both charges and their interaction form the system, with negligible external forces and work.
- Pair separation r
- Distance between the two particles, not the displacement of either one.
- Zero total momentum
- Starting from rest gives m₁v₁ + m₂v₂ = 0 in the chosen inertial frame.
- Nonrelativistic model
- Speeds are far below light speed; radiation losses and magnetic corrections are neglected.
What this picture assumes
Two like-signed point charges released from rest in isolation. Classical nonrelativistic energy and momentum model; radiation, magnetic corrections and external forces are neglected. State comparison at two separations, not a time simulation. Positive x is right.
Read the picture in three steps
- Locate the labeled sources, system boundary or graph axes. Read the units before comparing values.
- Potential-energy release 4.5 mJ. K₁=3.375 mJ; K₂=1.125 mJ. Velocities -2.598 and 0.866 m/s; total momentum 4.337e-19 kg·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
For two like-signed charges initially at separation r_i, U_i = kq₁q₂/r_i. At a larger r_f, the released energy is ΔK_total = kq₁q₂(1/r_i − 1/r_f). If both start at rest, this is K₁ + K₂.
Momentum conservation gives equal opposite momentum magnitudes p. Since K = p²/(2m), the lighter particle receives more kinetic energy. Specifically K₁ = ΔK_total m₂/(m₁ + m₂) and K₂ = ΔK_total m₁/(m₁ + m₂).
If one source were held fixed, the support would absorb momentum, and this isolated two-body momentum partition would no longer apply. Choose the system and constraints before using either model. The model compares two separations and does not animate a time history.
A worked example, step by step
Two +1 μC beads of masses 1 g and 3 g start at rest 1 m apart. Find their kinetic energies and speeds when 2 m apart.
- Released energy is 0.009(1 − 1/2) = 0.0045 J = 4.5 mJ.
- K₁ = (3/4)(4.5) = 3.375 mJ and K₂ = (1/4)(4.5) = 1.125 mJ.
- Speeds are √(2×0.003375/0.001) = 2.598 m/s and √(2×0.001125/0.003) = 0.866 m/s.
- Their momenta have equal magnitude 0.002598 kg·m/s and opposite directions; total kinetic gain is 4.5 mJ.
Do not give each particle the total released energy. Fixed-source and isolated-pair models use different momentum accounting.
Would the lighter bead have the larger momentum?
Compare with an explanation
No. Their momentum magnitudes are equal; its smaller mass gives it larger speed and kinetic energy.
Predict. Change one thing. Explain.
Keep charges and initial/final separation fixed while changing one mass. Predict whether total energy release changes and which particle receives the larger share.
On narrow screens, swipe or scroll diagrams sideways to read all labels.
Potential-energy release 4.5 mJ. K₁=3.375 mJ; K₂=1.125 mJ. Velocities -2.598 and 0.866 m/s; total momentum 4.337e-19 kg·m/s.
Two like-signed point charges released from rest in isolation. Classical nonrelativistic energy and momentum model; radiation, magnetic corrections and external forces are neglected. State comparison at two separations, not a time simulation. Positive x is right.
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 questionTwo equal-mass +1 μC beads, each 2 g, start at rest 0.5 m apart. Find each kinetic energy and speed when the separation reaches 1 m.
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Compare with the answer and four-point rubric
- 1 point: Released energy = 0.009(1/0.5 − 1/1) = 0.009 J.
- 1 point: Equal masses receive equal kinetic energies: 0.0045 J each.
- 1 point: Each speed is √[2(0.0045)/0.002] = 2.121 m/s.
- 1 point: Velocities point oppositely; both momentum and total energy balances hold.
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 remains fixed while the masses change in this example?
The potential-energy drop for specified charges and separations.
RECALL 2Why does the lighter bead get more kinetic energy?
Equal momentum magnitudes give K = p²/(2m).
RECALL 3Why can’t the same partition be used for a clamped source?
The clamp exchanges momentum with the system.
Revisit these tomorrow and a week later. Try a fresh problem and explain why the method applies.
When both charges move, share the released energy
- K₁ + K₂ = kq₁q₂(1/r_i − 1/r_f) from rest.
- m₁v₁ + m₂v₂ = 0 for the isolated pair.
- K₁/K₂ = m₂/m₁ when their momentum magnitudes are equal.
Remember: Do not give each particle the total released energy. Fixed-source and isolated-pair models use different momentum accounting.
Conditions: Two like-signed point charges released from rest in isolation. Classical nonrelativistic energy and momentum model; radiation, magnetic corrections and external forces are neglected. State comparison at two separations, not a time simulation. Positive x is right.
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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