Connected conductors share potential, not equal charge
You will be able to: Combine charge conservation and equal potential to predict redistribution between distant spheres.
Why does the larger sphere keep more charge?
Connect a small metal sphere and a larger one with a thin wire. Charge moves until neither end has a potential advantage. Equal voltage is the stopping condition; equal charge is not.
A useful starting point: A neutral conductor can still attract a charge →
Words and symbols before equations
- Electrical contact
- A conducting path that lets charge move between objects.
- Total charge Q_total
- Signed sum on the two-sphere system; conserved if isolated.
- Common potential
- The equal final voltage of connected conductors.
- Far-separated approximation
- Sphere separation is much greater than either radius, so mutual potential corrections are neglected.
What this picture assumes
Two spheres far apart relative to their radii, connected by a thin wire with negligible charge. Mutual potential corrections are neglected. Diagram spacing is schematic; sphere radii share a display scale.
Read the picture in three steps
- Locate the labeled sources, system boundary or graph axes. Read the units before comparing values.
- Common V ≈ 179.8 V. Surface densities σ₁ = 15.92, σ₂ = 7.958 nC/m². Charge sum = 6 nC.
- 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 distant spheres, each has V ≈ kQ/R relative to infinity. A thin connecting wire brings them to V₁ = V₂, giving Q₁/R₁ = Q₂/R₂.
Use conservation as a second equation: Q₁ + Q₂ = Q_total. Therefore Q₁ = Q_total R₁/(R₁+R₂), and similarly for Q₂. Radius, not area or volume, sets the ratio in this approximation.
Surface density is Q/(4πR²), so equal-potential spheres have greater density and surface field on the smaller sphere. This illustrates a curvature effect, but does not replace solving an arbitrary irregular shape. For nearby spheres their mutual fields matter.
A worked example, step by step
Distant spheres of radii 0.10 m and 0.20 m share a total +6 nC after being connected. Find final charges and common potential.
- Equal potential gives Q₁:Q₂ = R₁:R₂ = 1:2.
- Conservation gives Q₁ = 2 nC and Q₂ = 4 nC.
- V₁ ≈ 9×10⁹ × 2×10⁻⁹/0.10 = 180 V.
- V₂ ≈ 180 V also; equal charge would not satisfy this condition.
The equal-charge shortcut works for identical symmetric conductors, not arbitrary connected conductors.
If the radii are equal, how is total charge shared?
Compare with an explanation
Equally, under the symmetric far-separated assumptions.
Predict. Change one thing. Explain.
Double one sphere radius at fixed total charge. Compare charges, common potential and surface densities. The sketch is schematic; the spheres are assumed far apart.
On narrow screens, swipe or scroll diagrams sideways to read all labels.
Common V ≈ 179.8 V. Surface densities σ₁ = 15.92, σ₂ = 7.958 nC/m². Charge sum = 6 nC.
Two spheres far apart relative to their radii, connected by a thin wire with negligible charge. Mutual potential corrections are neglected. Diagram spacing is schematic; sphere radii share a display scale.
Explain what you noticed: Which quantity changed? Which stayed fixed? Use the relevant conductor equilibrium, charge conservation, capacitance 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 questionTwo distant spheres with radii R and 2R share total charge Q. Derive both charges and compare their surface-field magnitudes.
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Compare with the answer and four-point rubric
- 1 point: Equal potential gives Q₂ = 2Q₁.
- 1 point: Conservation gives Q₁ = Q/3 and Q₂ = 2Q/3.
- 1 point: Surface field magnitudes are k|Q|/(3R²) and k|Q|/(6R²).
- 1 point: The smaller sphere has twice the field, despite having half the charge.
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 drives redistribution to stop?
Equal electric potential along the connected conductor.
RECALL 2When does the simple radius formula apply?
Distant spheres with negligible wire charge and mutual corrections.
RECALL 3What if the pair is grounded?
Its total charge need not remain fixed; ground can exchange charge.
Revisit these tomorrow and a week later. Try a fresh problem and explain why the method applies.
Connected conductors share potential, not equal charge
- V₁ = V₂ at equilibrium through a wire.
- Q₁ + Q₂ = Q_total for the isolated pair.
- Distant spheres: Q₁/R₁ = Q₂/R₂.
Remember: The equal-charge shortcut works for identical symmetric conductors, not arbitrary connected conductors.
Conditions: Two spheres far apart relative to their radii, connected by a thin wire with negligible charge. Mutual potential corrections are neglected. Diagram spacing is schematic; sphere radii share a display scale.
Refresh Kid · AP Physics C: Electricity and Magnetism Unit 3 (official Unit 10) · Objectives 10.2.A · Review edition
Framework, scope and review status
Mapped to College Board CED, Topic 10.2, objectives 10.2.A. CED effective Fall 2024, current PDF ©2026; checked September 16, 2026. This is E&M Unit 3: Conductors and Capacitors, numbered Unit 10 in the official combined Physics C sequence. Topics 10.1–10.4 retain their official identifiers. Models state the electrostatic conditions, geometry approximations and whether charge or voltage stays fixed. Capacitor geometries include parallel plates, concentric spheres and long coaxial cylinders. Dielectric comparisons assume a fully filling ideal linear material. The optional 3D plate view uses explicitly different gap and lateral scales to show the small separation. 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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