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LESSON 01 / 18 · TOPIC 10.1

Charge and force: size first, direction next

You will be able to: Calculate the electric force between two point charges and explain its direction.

Official College Board Unit 10Free study resourceReview editionTeacher review pending

Why can two charged objects attract or repel?

A rubbed balloon can cling to a wall. Charge interactions underlie that effect, but a wall involves many particles. Begin with the simpler model of two small charged objects far enough apart to treat them as points.

A useful starting point: Scalars, vectors and direction →

Words and symbols before equations

Charge q
Signed electrical property, in coulombs (C). Electrons have −e, protons +e and neutrons zero charge.
Elementary charge e
Magnitude 1.60×10⁻¹⁹ C; ordinary object charge changes in integer multiples of e.
Separation r
Distance between point charges, in m; not the gap between edges of large objects.
Coulomb constant k
Use 9.0×10⁹ N·m²/C² in these examples. 1 μC=10⁻⁶ C.
Pair forces act on different objects+2 μC−3 μCr=0.5 mEach force magnitude: 0.216 N · arrows schematic
Read this model snapshot. Attraction; each object feels 0.216 N along the joining line, in opposite directions.
What this picture assumes

Vacuum point charges +2 μC and ±3 μC, k=9.0×10⁹ N·m²/C². Charge glyphs are not physical sizes. Force arrows show direction and equal pair magnitudes but do not encode a length scale.

Read the picture in three steps

  1. Locate the labeled sources, system boundary or graph axes. Read the units before comparing values.
  2. Attraction; each object feels 0.216 N along the joining line, in opposite directions.
  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

For point charges in vacuum, the force magnitude is F=k|q₁q₂|/r². First calculate a nonnegative magnitude, then use signs for direction: same signs repel; opposite signs attract along their line of separation.

The two interaction forces have equal magnitude and opposite directions, even if charges or masses differ. These forces act on different objects, so they do not cancel on either individual object’s force diagram.

Ordinary charging usually transfers electrons; it does not create charge. Losing electrons leaves an object more positive. Electric interactions among matter also underlie contact forces such as normal force and tension, though we usually model those macroscopically.

A worked example, step by step

Charges +2 μC and −3 μC are 0.50 m apart. Find the interaction force.

  1. Treat the objects as point charges in vacuum and convert μC to C.
  2. F=(9.0×10⁹)(2×10⁻⁶)(3×10⁻⁶)/(0.50)²=0.216 N.
  3. Opposite signs mean attraction: each force points toward the other object.
  4. Both experience 0.216 N; different masses could produce different accelerations.
Common mix-up

The minus sign of a charge determines interaction direction; it is not a negative force magnitude.

CHECK THE IDEA

If separation doubles, what happens to F?

Compare with an explanation

It falls to one quarter because r is squared.

Now investigate one change Explore →

Predict. Change one thing. Explain.

Change separation while keeping charges fixed, then switch the second charge’s sign. Predict the factor change and compare equal-opposite arrows. Arrow lengths are schematic; read the numerical magnitude.

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

Pair forces act on different objects+2 μC−3 μCr=0.5 mEach force magnitude: 0.216 N · arrows schematic

Attraction; each object feels 0.216 N along the joining line, in opposite directions.

Vacuum point charges +2 μC and ±3 μC, k=9.0×10⁹ N·m²/C². Charge glyphs are not physical sizes. Force arrows show direction and equal pair magnitudes but do not encode a length scale.

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. Double one charge and double separation. Force magnitude becomes…

Show answer and reasoning

Half. The factor is 2/2²=1/2.

2. An object loses electrons. Its charge becomes…

Show answer and reasoning

More positive. Removing negative charge makes its net charge more positive.

Original written challenge

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

Two point charges +1 μC and +4 μC are 0.30 m apart. (a) Calculate force magnitude. (b) State directions. (c) Compare the force on each charge. (d) Predict the force at 0.60 m.

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

Compare with the answer and four-point rubric
  1. 1 point: F=9×10⁹(4×10⁻¹²)/0.09=0.40 N.
  2. 1 point: Repulsive, away from one another along the joining line.
  3. 1 point: Equal magnitudes, opposite directions on different objects.
  4. 1 point: F=0.10 N after doubling r.

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 1Why does neutral not mean no charges?

Positive and negative charges can be present in equal amounts.

RECALL 2What usually moves during ordinary charging?

Electrons; protons remain bound in atomic nuclei.

RECALL 3Does a larger charge feel a larger pair force?

No. Pair forces are equal and opposite.

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

Charge and force: size first, direction next

  • F=k|q₁q₂|/r²; vacuum point-charge model.
  • Like signs repel; unlike signs attract.
  • 1 μC=10⁻⁶ C; ordinary net charge is an integer multiple of e.

Remember: The minus sign of a charge determines interaction direction; it is not a negative force magnitude.

Conditions: Vacuum point charges +2 μC and ±3 μC, k=9.0×10⁹ N·m²/C². Charge glyphs are not physical sizes. Force arrows show direction and equal pair magnitudes but do not encode a length scale.

Refresh Kid · AP Physics 2 Unit 2 (official Unit 10) · Objectives 10.1.A · Review edition

Framework, scope and review status

Mapped to College Board CED, Topic 10.1, objectives 10.1.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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