Match a spherical surface to spherical charge symmetry
You will be able to: Derive the field of a uniform solid sphere and a thin spherical shell.
How does Gauss’s law give the field inside and outside a charged sphere?
At a fixed radius from a spherically symmetric source, every direction is equivalent. The field has the same radial magnitude everywhere on a concentric Gaussian sphere, turning a difficult surface sum into one product.
A useful starting point: A closed surface measures net enclosed charge →
Words and symbols before equations
- Spherical symmetry
- Charge distribution depends on distance from the center, not direction.
- Gaussian radius r
- Radius of the imaginary surface where the field is wanted.
- Physical radius R
- Radius of the actual charged object; it need not equal r.
- Volume fraction
- For a uniform solid sphere, the fraction inside r is (r/R)³.
What this picture assumes
Spherically symmetric isolated source in vacuum. At r = 0 the field is zero. At a thin shell r = R, the readout reports the exterior limit; the ideal surface field has distinct one-sided limits. Solid volume charge is uniform.
Read the picture in three steps
- Locate the labeled sources, system boundary or graph axes. Read the units before comparing values.
- At r = 0.5 m: Q_enc = 0.25 nC, E_r = 9 N/C. Flux 28.27 N·m²/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
Symmetry makes E radial and constant on a concentric Gaussian sphere. Therefore ∮E·dA = E_r 4πr², and E_r = Q_enc/(4πε₀r²) = kQ_enc/r², with an outward signed component.
For a uniformly charged solid sphere, Q_enc = Q(r/R)³ when r < R. Substitution gives E_r = kQr/R³. For r ≥ R, all Q is enclosed and E_r = kQ/r².
For a uniform thin spherical shell, Q_enc = 0 inside and symmetry gives E = 0. Outside the same point-charge form applies. At an ideal charged shell the field has a discontinuity; speak of inside and outside limits rather than a unique microscopic surface value.
A worked example, step by step
A uniform solid sphere has Q = +8 nC and R = 2 m. Find E at r = 1 m and r = 4 m.
- At r = 1 m, Q_enc = 8(1/2)³ = 1 nC.
- E_r = kQ_enc/r² = 9 N/C outward.
- At r = 4 m, Q_enc = 8 nC, so E_r = 72/16 = 4.5 N/C outward.
- The interior field rises with r, but the exterior falls as 1/r². Both meet continuously at r = R for the solid sphere.
Use enclosed charge, not total charge, inside a solid distribution. Symmetry is what makes E constant on the surface.
Why can zero flux imply E = 0 inside this shell but not inside any empty box?
Compare with an explanation
Spherical symmetry makes the radial field identical over the Gaussian sphere; the empty-box case lacks that inference.
Predict. Change one thing. Explain.
Switch between a uniform solid sphere and a thin shell at the same Q and R. Move r across the physical boundary and compare the enclosed-charge and field readouts.
On narrow screens, swipe or scroll diagrams sideways to read all labels.
At r = 0.5 m: Q_enc = 0.25 nC, E_r = 9 N/C. Flux 28.27 N·m²/C.
Spherically symmetric isolated source in vacuum. At r = 0 the field is zero. At a thin shell r = R, the readout reports the exterior limit; the ideal surface field has distinct one-sided limits. Solid volume charge is uniform.
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 questionFor a solid sphere Q = +1 nC, R = 1 m, compare E at r = 0.5 m and 2 m. State what changes for a thin shell with the same Q and R.
This response is not submitted or saved. Copy it before leaving.
Compare with the answer and four-point rubric
- 1 point: Solid interior: E = 9(0.5) = 4.5 N/C outward.
- 1 point: Exterior: E = 9/4 = 2.25 N/C outward.
- 1 point: The shell interior field is zero.
- 1 point: The shell exterior field is unchanged because its total charge and spherical symmetry match.
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 1Why choose a concentric Gaussian sphere?
The field has constant magnitude and is normal everywhere on it.
RECALL 2How does interior solid-sphere field vary?
Linearly with r for uniform volume density.
RECALL 3Can r and R be used interchangeably?
No. r is the observation radius; R is the object radius.
Revisit these tomorrow and a week later. Try a fresh problem and explain why the method applies.
Match a spherical surface to spherical charge symmetry
- E_r(4πr²) = Q_enc/ε₀.
- Uniform solid: Q_enc = Q(r/R)³ for r < R.
- Uniform shell: E = 0 inside; exterior E_r = kQ/r².
Remember: Use enclosed charge, not total charge, inside a solid distribution. Symmetry is what makes E constant on the surface.
Conditions: Spherically symmetric isolated source in vacuum. At r = 0 the field is zero. At a thin shell r = R, the readout reports the exterior limit; the ideal surface field has distinct one-sided limits. Solid volume charge is uniform.
Refresh Kid · AP Physics C: Electricity and Magnetism Unit 1 (official Unit 8) · Objectives 8.6.A · Review edition
Framework, scope and review status
Mapped to College Board CED, Topic 8.6, objectives 8.6.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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