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LESSON 05 / 17 · TOPIC 9.2

A voltage compares two points using one reference

You will be able to: Calculate a potential difference and distinguish it from an arbitrary potential zero.

Calculus-based electrostaticsFree study resourceReview editionTeacher review pending

Why does a voltmeter need two leads?

A meter compares its red-lead potential with its black-lead potential. If those points are at +12 V and +5 V relative to the same reference, the reading is +7 V—not +12 V.

A useful starting point: Zero potential and zero field are different statements →

Words and symbols before equations

Potential difference ΔV
Final or red-lead potential minus initial or black-lead potential, as defined in the problem.
Common reference shift C
Adding the same constant C to every V value.
Battery
A device using chemical processes to maintain charge separation and a terminal potential difference.
Electron-volt eV
An energy unit: 1 eV = e × 1 V ≈ 1.60 × 10⁻¹⁹ J.
Both endpoints use the same shifted zeroV · same scale for all bars0V_A-2V_B4V_B − V_A6
Read this model snapshot. Meter red B minus black A: 6 V. Probe ΔU = 18 μJ. Common reference offset 0 V changes neither difference.
What this picture assumes

Endpoints use the same reference. Red lead at B, black lead at A. A common reference shift changes neither the voltage difference nor the probe energy change. No current or battery circuit is simulated.

Read the picture in three steps

  1. Locate the labeled sources, system boundary or graph axes. Read the units before comparing values.
  2. Meter red B minus black A: 6 V. Probe ΔU = 18 μJ. Common reference offset 0 V changes neither difference.
  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

Set the order first: ΔV = V_B − V_A for a move A → B. Swapping the endpoints reverses the sign. A voltmeter with red at B and black at A reads this difference.

Shift both values by any constant C. Then (V_B + C) − (V_A + C) = V_B − V_A, so forces and energy changes are unaffected. A reference change is not a physical battery adjustment.

A battery maintains a potential difference through chemical processes that separate charge; it does not create net charge. Its voltage alone does not specify a circuit current. For a probe crossing its terminal difference, ΔU = qΔV; inside an operating battery, chemical forces also act.

A worked example, step by step

A is at −2 V and B at +4 V. Find ΔV for A → B and ΔU for +3 μC. Then add +10 V to both reference values.

  1. ΔV = 4 − (−2) = +6 V.
  2. ΔU = (+3 μC)(+6 V) = +18 μJ.
  3. The shifted values are V_A = +8 V and V_B = +14 V.
  4. The difference remains +6 V, so the energy change is still +18 μJ.
Common mix-up

Use one reference for both endpoints. A battery voltage is not a current or an amount of stored charge.

CHECK THE IDEA

Would shifting only V_B be a reference change?

Compare with an explanation

No. That changes the physical potential difference relative to A.

Now investigate one change Explore →

Predict. Change one thing. Explain.

Move only the reference-offset slider. Verify that absolute readouts shift together while the meter difference and energy change stay fixed. Reverse the endpoint values next.

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

Both endpoints use the same shifted zeroV · same scale for all bars0V_A-2V_B4V_B − V_A6

Meter red B minus black A: 6 V. Probe ΔU = 18 μJ. Common reference offset 0 V changes neither difference.

Probe energy changeμJ · same scale for all bars0ΔU = qΔV18

Endpoints use the same reference. Red lead at B, black lead at A. A common reference shift changes neither the voltage difference nor the probe energy change. No current or battery circuit is simulated.

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.

1. Red lead at 3 V, black lead at 8 V gives…

Show answer and reasoning

−5 V. The meter reads red minus black: 3 − 8 = −5 V.

2. Adding 20 V to both endpoint potentials changes ΔV by…

Show answer and reasoning

0 V. A common constant cancels in the difference.

Original written challenge

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

An electron moves through ΔV = +5 V. Find ΔU in eV and joules, then state the electric work.

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

Compare with the answer and four-point rubric
  1. 1 point: Electron charge is −e.
  2. 1 point: ΔU = qΔV = −5 eV.
  3. 1 point: ΔU = −8.0 × 10⁻¹⁹ J.
  4. 1 point: W_e = −ΔU = +5 eV = +8.0 × 10⁻¹⁹ J.

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 1What does the meter compare?

Potential at its red lead minus potential at its black lead.

RECALL 2Can a common zero shift change a force?

No. The spatial derivatives remain the same.

RECALL 3Is eV a unit of potential?

No. It is an energy unit.

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

A voltage compares two points using one reference

  • ΔV = V_B − V_A.
  • ΔU = qΔV.
  • Adding the same constant to V everywhere leaves ΔV and E unchanged.

Remember: Use one reference for both endpoints. A battery voltage is not a current or an amount of stored charge.

Conditions: Endpoints use the same reference. Red lead at B, black lead at A. A common reference shift changes neither the voltage difference nor the probe energy change. No current or battery circuit is simulated.

Refresh Kid · AP Physics C: Electricity and Magnetism Unit 2 (official Unit 9) · Objectives 9.2.A · Review edition

Framework, scope and review status

Mapped to College Board CED, Topic 9.2, objectives 9.2.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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