A field describes a place; force acts on a charge
You will be able to: Distinguish the source electric field from the force on a test charge.
Why does a negative charge move opposite the field arrow?
At one location, a +2 μC test charge feels 0.006 N to the right. Replacing it with −2 μC reverses the force, but the source field at that location has not reversed.
A useful starting point: Charge by induction: the order matters →
Words and symbols before equations
- Electric field E
- Force per positive unit test charge, a vector in N/C.
- Source charge
- Charge producing the field being studied.
- Test charge q
- Small enough not to significantly rearrange the source charges.
- Point-source field
- Magnitude E=k|Q|/r²; outward from +Q and inward toward −Q.
What this picture assumes
Uniform source field 3000 N/C rightward, unaffected by the small test charge. Force arrows show sign only; numerical force has units N. At zero test charge the source field remains nonzero.
Read the picture in three steps
- Locate the labeled sources, system boundary or graph axes. Read the units before comparing values.
- Source E stays 3000 N/C rightward. F=qE=0.006 N; rightward. Arrow lengths are schematic.
- 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 the field direction as the force direction for a positive test charge. F=qE uses signed q, so a negative test charge experiences force opposite E. Removing the test charge does not remove the source field.
A point source creates a radial field whose strength decreases with distance squared. The magnitude depends on source charge and observation point, not on the chosen test charge in the small-test-charge approximation.
Field arrows are local vector descriptions, not guaranteed particle trajectories. Motion depends on initial velocity and other forces as well as electric acceleration.
| Quantity | Defined by | Units |
|---|---|---|
| Field E | Sources and observation position | N/C |
| Force qE | Field and signed test charge | N |
A worked example, step by step
A +2 μC test charge feels 0.006 N rightward. Find E and the force on a −3 μC charge at the same point.
- Take right positive. E=F/q=0.006/(2×10⁻⁶)=3000 N/C rightward.
- Use the same source field for the new small test charge.
- F=(−3×10⁻⁶)(3000)=−0.009 N.
- The force is 0.009 N leftward; E remains rightward.
A negative test charge reverses its force, not the field produced by unchanged sources.
Does E disappear when q_test=0?
Compare with an explanation
No. The sources still create E; that particular object experiences no qE force.
Predict. Change one thing. Explain.
Keep a uniform field fixed and vary test charge through zero. Compare the unchanged field arrow with the changing force. At q=0 the field still exists but electric force on that test object is zero.
On narrow screens, swipe or scroll diagrams sideways to read all labels.
Source E stays 3000 N/C rightward. F=qE=0.006 N; rightward. Arrow lengths are schematic.
Uniform source field 3000 N/C rightward, unaffected by the small test charge. Force arrows show sign only; numerical force has units N. At zero test charge the source field remains nonzero.
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.
Original written challenge
4 points · self-check · not an official AP questionA uniform field is 400 N/C upward. (a) Find force on +5 μC. (b) Find force on −5 μC. (c) State the field after removing the test charge. (d) Explain the small-test-charge assumption.
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Compare with the answer and four-point rubric
- 1 point: 0.002 N upward.
- 1 point: 0.002 N downward.
- 1 point: Still 400 N/C upward for unchanged sources.
- 1 point: The test charge does not significantly change the source configuration.
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 1Field direction convention?
Force direction on a positive test charge.
RECALL 2Units of E?
N/C, equivalently V/m.
RECALL 3Do field arrows trace all particle paths?
No. Initial velocity and other forces matter.
Revisit these tomorrow and a week later. Try a fresh problem and explain why the method applies.
A field describes a place; force acts on a charge
- E=F/q with signed vectors.
- F=qE.
- Point source: |E|=k|Q|/r²; 1 N/C=1 V/m.
Remember: A negative test charge reverses its force, not the field produced by unchanged sources.
Conditions: Uniform source field 3000 N/C rightward, unaffected by the small test charge. Force arrows show sign only; numerical force has units N. At zero test charge the source field remains nonzero.
Refresh Kid · AP Physics 2 Unit 2 (official Unit 10) · Objectives 10.3.A · Review edition
Framework, scope and review status
Mapped to College Board CED, Topic 10.3, objectives 10.3.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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