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LESSON 18 / 18 · TOPIC 10.7

Use voltage to predict changes in motion

You will be able to: Use ΔK=−qΔV when only electric work changes kinetic energy.

Official College Board Unit 10Free study resourceReview editionTeacher review pending

Which way does a charge gain kinetic energy?

A positive charge released from rest can speed up toward lower potential. A negative charge can speed up toward higher potential. Both move toward lower electric potential energy because U=qV includes the charge’s sign.

A useful starting point: Test capacitance with a straight-line graph →

Words and symbols before equations

Potential change ΔV
Final minus initial potential, in V.
Potential-energy change ΔU
qΔV for a fixed test charge and source configuration.
Kinetic-energy change ΔK
K_final−K_initial, in J.
System choice
Track the particle and electric interaction; include other work if gravity, drag or an external agent matters.
Endpoint energy account · signed changesμJ · same scale for all bars0Potential ΔU-120Kinetic ΔK120Candidate K_f170
Read this model snapshot. ΔU=qΔV=-120 μJ; ΔK=120 μJ; K_f=50+ΔK=170 μJ. Energy permits this endpoint; field geometry also determines actual reachability.
What this picture assumes

Endpoint energy account with initial K=50 μJ. Static source configuration; electric work only, no other work. Candidate K<0 is flagged unreachable rather than clamped to a physical state. K≥0 is necessary but alone does not guarantee a trajectory reaches that endpoint.

Read the picture in three steps

  1. Locate the labeled sources, system boundary or graph axes. Read the units before comparing values.
  2. ΔU=qΔV=-120 μJ; ΔK=120 μJ; K_f=50+ΔK=170 μJ. Energy permits this endpoint; field geometry also determines actual reachability.
  3. Check what the picture assumes below. Use the Explore task to predict one change before moving a control.

Connect the picture to the physics

If electric work alone changes kinetic energy, ΔK=W_electric=−ΔU=−qΔV. A positive ΔK indicates speeding up; a negative value means slowing down if sufficient initial kinetic energy exists.

Do not discard q’s sign. For q<0, moving to a higher V lowers U and can raise K. Conversely, a proposed endpoint with K_final<0 is unreachable under the stated energy conditions; negative kinetic energy is not a physical answer.

Electrostatic work depends on endpoints, not the chosen path. With additional work W_other, use ΔK=−qΔV+W_other. The explorer represents endpoint energy accounting, not a force map or a claim about the path.

A worked example, step by step

A +2 μC charge moves from 100 V to 40 V with electric work only. Its initial K is 50 μJ. Find final K.

  1. ΔV=40−100=−60 V.
  2. ΔU=qΔV=(2×10⁻⁶)(−60)=−120 μJ.
  3. ΔK=+120 μJ, so K_final=50+120=170 μJ.
  4. Energy is transferred from electric potential energy into kinetic energy.
Common mix-up

Negative q reverses the voltage trend. If calculated K_final is negative, reject the endpoint as inaccessible in this model.

CHECK THE IDEA

Can a negative charge gain kinetic energy while moving toward higher V?

Compare with an explanation

Yes. qΔV is then negative, so potential energy falls and kinetic energy can rise.

Now investigate one change Explore →

Predict. Change one thing. Explain.

Switch charge sign and vary the final-minus-initial potential. Compare potential-energy change and candidate final kinetic energy. Find a setting labeled unreachable and explain the missing energy.

On narrow screens, swipe or scroll diagrams sideways to read all labels.

Endpoint energy account · signed changesμJ · same scale for all bars0Potential ΔU-120Kinetic ΔK120Candidate K_f170

ΔU=qΔV=-120 μJ; ΔK=120 μJ; K_f=50+ΔK=170 μJ. Energy permits this endpoint; field geometry also determines actual reachability.

Endpoint energy account with initial K=50 μJ. Static source configuration; electric work only, no other work. Candidate K<0 is flagged unreachable rather than clamped to a physical state. K≥0 is necessary but alone does not guarantee a trajectory reaches that endpoint.

Explain what you noticed: Which quantity changed? Which stayed fixed? Use the relevant charge, field or energy 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.

1. For q<0 and ΔV>0 with only electric work, ΔK is…

Show answer and reasoning

Positive. −qΔV is positive.

2. A calculated K_final=−5 J means…

Show answer and reasoning

The endpoint is unreachable with the stated initial energy. The assumptions do not supply enough kinetic energy to reach that potential.

Original written challenge

4 points · self-check · not an official AP question

A −3 μC charge has initial K=20 μJ and moves through ΔV=+10 V with electric work only. (a) Find ΔU. (b) Find ΔK. (c) Find K_final. (d) State whether the endpoint passes the energy-accessibility check.

This response is not submitted or saved. Copy it before leaving.

Compare with the answer and four-point rubric
  1. 1 point: ΔU=−30 μJ.
  2. 1 point: ΔK=+30 μJ.
  3. 1 point: K_final=50 μJ.
  4. 1 point: Yes, K_final≥0; energy permits it, although actual motion also depends on the field geometry and initial direction.

Accept equivalent correct methods and explanations. This is a Refresh Kid teaching rubric, not an official AP scoring guideline.

Recall the ideas without notes Review →

Retrieve it before you reveal it.

RECALL 1Positive charges gain K toward which V?

Lower V, when electric work alone supplies the gain.

RECALL 2Negative charges gain K toward which V?

Higher V under the same condition.

RECALL 3Does an energy-allowed endpoint guarantee a trajectory reaches it?

No. Geometry and initial motion also matter.

Revisit these tomorrow and a week later. Try a fresh problem and explain why the method applies.

Use voltage to predict changes in motion

  • ΔU=qΔV.
  • Electric work only: ΔK=−qΔV.
  • With other work: ΔK=−qΔV+W_other; physical K≥0.

Remember: Negative q reverses the voltage trend. If calculated K_final is negative, reject the endpoint as inaccessible in this model.

Conditions: Endpoint energy account with initial K=50 μJ. Static source configuration; electric work only, no other work. Candidate K<0 is flagged unreachable rather than clamped to a physical state. K≥0 is necessary but alone does not guarantee a trajectory reaches that endpoint.

Refresh Kid · AP Physics 2 Unit 2 (official Unit 10) · Objectives 10.7.A · Review edition

Framework, scope and review status

Mapped to College Board CED, Topic 10.7, objectives 10.7.A. CED effective Fall 2024, current PDF ©2026; checked September 16, 2026. Refresh Kid calls this the second AP Physics 2 unit; College Board numbers it Unit 10; the first unit in this course is official Unit 9. 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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