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LESSON 06 / 20 · TOPIC 8.3

The field belongs to the sources; force also depends on the test charge

You will be able to: Calculate a point-charge field and the force on a signed test charge.

Calculus-based electrostaticsFree study resourceReview editionTeacher review pending

Why does a negative charge accelerate opposite to an electric-field arrow?

At one location a field points right with strength 100 N/C. A +2 μC test charge feels a force of 0.00020 N right. A −2 μC test charge feels the same magnitude left, while the source field stays the same.

A useful starting point: Separate charge first; disconnect ground at the right time →

Words and symbols before equations

Electric field E
Force per unit test charge, measured in N/C; a vector.
Test charge q
A small probe assumed not to rearrange the source charges appreciably.
Source charge Q
The charge producing the field; distinguish it from the probe q.
Radial direction
Away from a source position; inward is the opposite direction.
Separate field direction from test-charge forceSource Q = 2 nC at x = 0; probe at x = 0.3 mField E200 N/CForce F-0.0006 NArrows show direction only; E and F have different units.
Read this model snapshot. Eₓ = 200 N/C; probe q = -3 μC feels Fₓ = -0.0006 N. Changing q leaves the source field unchanged.
What this picture assumes

A fixed point source at x = 0, observation point on +x, vacuum. Test charges do not perturb the source. Field and force arrows are directional with separate units; compare numbers, not their lengths.

Read the picture in three steps

  1. Locate the labeled sources, system boundary or graph axes. Read the units before comparing values.
  2. Eₓ = 200 N/C; probe q = -3 μC feels Fₓ = -0.0006 N. Changing q leaves the source field unchanged.
  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

Define E = F/q using the signed test charge. Equivalently F = qE. The standard field direction is the force direction for a positive test charge. A negative test charge reverses the force arrow.

For a stationary point source in vacuum, |E| = k|Q|/r². It points away from positive Q and toward negative Q. The field exists whether or not a test charge is present.

A field map samples this vector at many positions. An actual particle trajectory also depends on its initial velocity and inertia; a field arrow is not automatically a path. For an unconstrained particle with only electric force, a = qE/m.

A worked example, step by step

A +2 nC source is at x = 0. Find E at x = 0.30 m and the force on a −3 μC test charge there.

  1. The observation point is to the right of a positive source, so E points +x.
  2. E = (9 × 10⁹)(2 × 10⁻⁹)/(0.30)² = +200 N/C.
  3. Fₓ = qEₓ = (−3 × 10⁻⁶)(200) = −6.0 × 10⁻⁴ N.
  4. The negative test charge is pulled left; reversing the test charge would not reverse the source field.
Common mix-up

Changing the test-charge sign changes its force, not the field of fixed sources.

CHECK THE IDEA

If q doubles in a fixed field, does E double?

Compare with an explanation

No. The force doubles, so the ratio F/q remains the same.

Now investigate one change Explore →

Predict. Change one thing. Explain.

Keep Q and r fixed. Reverse the probe q. Observe the separate field and force readouts, which have different units and separate drawing scales.

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

Separate field direction from test-charge forceSource Q = 2 nC at x = 0; probe at x = 0.3 mField E200 N/CForce F-0.0006 NArrows show direction only; E and F have different units.

Eₓ = 200 N/C; probe q = -3 μC feels Fₓ = -0.0006 N. Changing q leaves the source field unchanged.

The source field at points on +xEₓ (N/C)distance r (m)0.1-2700.3253150.559000.775148512070

A fixed point source at x = 0, observation point on +x, vacuum. Test charges do not perturb the source. Field and force arrows are directional with separate units; compare numbers, not their lengths.

Explain what you noticed: Which quantity changed? Which stayed fixed? Use the relevant charge, vector superposition, electric field, flux or symmetry 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. An electric field points up. A negative test charge feels electric force…

Show answer and reasoning

down. F = qE reverses direction for q < 0.

2. The field 0.20 m from a +1 nC point charge has magnitude…

Show answer and reasoning

225 N/C. E = 9/(0.20)² = 225 N/C.

Original written challenge

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

At a point E = −300 N/C along x. Find the electric force on +2 μC and −2 μC probes, then explain whether removing both probes removes E.

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

Compare with the answer and four-point rubric
  1. 1 point: For +2 μC, Fₓ = −6.0 × 10⁻⁴ N.
  2. 1 point: For −2 μC, Fₓ = +6.0 × 10⁻⁴ N.
  3. 1 point: The force reverses with probe sign.
  4. 1 point: Fixed sources still produce E when the probes are removed.

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 sets field direction?

The force direction for a positive test charge.

RECALL 2Why must a probe be small?

To avoid appreciably rearranging the sources.

RECALL 3Does a field arrow specify a trajectory?

No. Motion also depends on initial velocity and mass.

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

The field belongs to the sources; force also depends on the test charge

  • E = F/q; F = qE.
  • Point source: E = kQ r̂/r².
  • E unit: N/C; test charge small enough not to disturb sources.

Remember: Changing the test-charge sign changes its force, not the field of fixed sources.

Conditions: A fixed point source at x = 0, observation point on +x, vacuum. Test charges do not perturb the source. Field and force arrows are directional with separate units; compare numbers, not their lengths.

Refresh Kid · AP Physics C: Electricity and Magnetism Unit 1 (official Unit 8) · Objectives 8.3.A · Review edition

Framework, scope and review status

Mapped to College Board CED, Topic 8.3, objectives 8.3.A. CED effective Fall 2024, current PDF ©2026; checked September 16, 2026. This is E&M Unit 1: Electric Charges, Fields, and Gauss’s Law, numbered Unit 8 in the official combined Physics C sequence. Topics 8.1–8.6 retain their official identifiers. Quantitative force examples use at most four point charges. Field integrals use the specified rods, ring, arc and infinite wire; Gauss-law field calculations use spherical, cylindrical or planar symmetry. Optional projected 3D views clarify area normals and geometry; camera rotation never changes the physics. 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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