Integrate directions around a charged arc
You will be able to: Integrate angle-dependent components for a uniform arc.
Why does a semicircle of charge produce a field at its center?
A positively charged upper semicircle pushes a positive probe at its center downward. Left and right contributions cancel horizontally, while their downward components reinforce each other.
A useful starting point: A ring cancels transverse fields but keeps an axial field →
Words and symbols before equations
- Angle φ
- Position around the arc measured counterclockwise from +x, in radians.
- Arc element R dφ
- A small length along a circle of radius R.
- dq = λR dφ
- Charge in a small arc element with linear density λ.
- Arc bounds
- The start and end angles defining which source elements are present.
What this picture assumes
Uniform positive arc in the xy plane from angle 0 to β counterclockwise; field evaluated at the center. λ stays fixed as angle changes, so total charge changes. The geometry is shown with equal x/y scale.
Read the picture in three steps
- Locate the labeled sources, system boundary or graph axes. Read the units before comparing values.
- Arc charge Q = 3.142 nC. E = (0, -18) N/C; magnitude 18 N/C.
- Check what the picture assumes below. Use the Explore task to predict one change before moving a control.
Connect the picture to the physics
At the center, every element is a distance R away. For a positive element at angle φ, the field points toward the center, giving dEₓ = −(kλ/R)cosφ dφ and dEᵧ = −(kλ/R)sinφ dφ.
For an arc from 0 to β, integrate each component: Eₓ = −(kλ/R)sinβ and Eᵧ = (kλ/R)(cosβ − 1). The formula uses radians in the integral; convert a degree control by multiplying by π/180.
For a full upper semicircle β = π, Eₓ = 0 and Eᵧ = −2kλ/R. Since Q = λπR, the magnitude is 2kQ/(πR²). For a shorter arc, horizontal cancellation is generally incomplete.
A worked example, step by step
A positive upper semicircle has R = 1 m and λ = 1 nC/m. Find its center field.
- The source angles run from φ = 0 to π.
- Eₓ = −9 sinπ = 0 N/C.
- Eᵧ = 9(cosπ − 1) = −18 N/C.
- The field is 18 N/C downward; each source element repels a positive probe toward the empty lower half.
The element field direction changes around the arc. Integrating magnitudes discards cancellation.
Why is the semicircle field not zero like the full ring center?
Compare with an explanation
The missing lower half removes the contributions that would cancel the downward result.
Predict. Change one thing. Explain.
Vary the arc from 30° to 180° at fixed λ and R. Predict which component vanishes only at the semicircle. Notice that total charge also changes when arc length changes.
On narrow screens, swipe or scroll diagrams sideways to read all labels.
Arc charge Q = 3.142 nC. E = (0, -18) N/C; magnitude 18 N/C.
Uniform positive arc in the xy plane from angle 0 to β counterclockwise; field evaluated at the center. λ stays fixed as angle changes, so total charge changes. The geometry is shown with equal x/y scale.
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.
Original written challenge
4 points · self-check · not an official AP questionA quarter-circle from 0 to π/2 has R = 1 m and λ = +2 nC/m. Find both center field components and the magnitude.
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Compare with the answer and four-point rubric
- 1 point: kλ/R = 18 N/C.
- 1 point: Eₓ = −18 sin(π/2) = −18 N/C.
- 1 point: Eᵧ = 18[cos(π/2) − 1] = −18 N/C.
- 1 point: |E| = 18√2 = 25.46 N/C, toward the lower-left.
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 1What units belong in angular integration?
Radians.
RECALL 2Does extending an arc at fixed λ keep Q fixed?
No. Q = λRβ increases with arc length.
RECALL 3What cancels for an upper semicircle?
Horizontal contributions from mirrored elements.
Revisit these tomorrow and a week later. Try a fresh problem and explain why the method applies.
Integrate directions around a charged arc
- dq = λR dφ.
- Eₓ = −(kλ/R)sinβ; Eᵧ = (kλ/R)(cosβ − 1), arc 0 → β.
- Upper semicircle: Eᵧ = −2kλ/R.
Remember: The element field direction changes around the arc. Integrating magnitudes discards cancellation.
Conditions: Uniform positive arc in the xy plane from angle 0 to β counterclockwise; field evaluated at the center. λ stays fixed as angle changes, so total charge changes. The geometry is shown with equal x/y scale.
Refresh Kid · AP Physics C: Electricity and Magnetism Unit 1 (official Unit 8) · Objectives 8.4.A · Review edition
Framework, scope and review status
Mapped to College Board CED, Topic 8.4, objectives 8.4.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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