A neutral conductor can still attract a charge
You will be able to: Distinguish redistributed surface charge from net charging and interpret a polarized sphere.
Can a conductor polarize without gaining net charge?
Bring a positive rod near a neutral metal object without touching. Electrons shift toward the rod, leaving an electron deficit farther away. The object remains neutral but its nearby negative side can produce a net attraction.
A useful starting point: A cavity is different from the surrounding metal →
Words and symbols before equations
- Polarization
- Separation of positive and negative charge within an object without necessarily changing its total charge.
- Applied field E₀
- The field that would exist without the conductor.
- Polar angle θ
- Angle from the applied-field direction, 0° at the positive-axis pole.
- Induced field
- Field created by rearranged conductor charges.
What this picture assumes
Exact surface-density result for a neutral sphere in a uniform applied field: σ = 3ε₀E₀cosθ. The diagram is a cross-section; arrow directions are schematic. A nonuniform nearby rod field is outside this quantitative model.
Read the picture in three steps
- Locate the labeled sources, system boundary or graph axes. Read the units before comparing values.
- At θ = 0°, σ = 53.1 pC/m². Internal applied field 2 plus induced field -2 = 0 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
For a neutral isolated conducting sphere in an otherwise uniform field directed along +z, equilibrium requires the induced field in the metal to be −E₀. Their vector sum is zero.
The exact spherical surface distribution is σ = 3ε₀E₀ cos θ. It is positive on the +z hemisphere, negative on the opposite hemisphere and zero at the equator. Equal signed hemispheres give zero total charge. This formula is specific to a sphere in a uniform field.
A perfectly uniform applied field exerts no net force on this neutral sphere by symmetry. Attraction to a nearby charged rod involves a nonuniform field: the closer opposite induced charge is affected more strongly. Do not use the uniform-field diagram as a quantitative rod-force model.
A worked example, step by step
A neutral sphere is placed in a uniform +z field of 2 N/C. Find the surface density at θ = 0°, 90° and 180°.
- Use σ = 3ε₀E₀ cos θ with ε₀ = 8.85 × 10⁻¹² F/m.
- The coefficient 3ε₀E₀ is 53.1 pC/m².
- The pole, equator and opposite pole have +53.1, 0 and −53.1 pC/m².
- The sphere remains neutral; induced and applied fields cancel inside.
Polarization does not imply net charge, and a uniform-field polarization model does not imply a net translational force.
Why can the rod attract a neutral conductor?
Compare with an explanation
Induced opposite charge is closer in the rod’s nonuniform field; neutrality does not require forces to cancel.
Predict. Change one thing. Explain.
Sweep θ from one pole to the other while holding E₀ fixed. Predict the sign at the equator and both poles. Then reverse E₀.
On narrow screens, swipe or scroll diagrams sideways to read all labels.
At θ = 0°, σ = 53.1 pC/m². Internal applied field 2 plus induced field -2 = 0 N/C.
Exact surface-density result for a neutral sphere in a uniform applied field: σ = 3ε₀E₀cosθ. The diagram is a cross-section; arrow directions are schematic. A nonuniform nearby rod field is outside this quantitative model.
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 questionCompare a neutral sphere in a uniform field with one near a positive rod. Explain polarization, net charge and the possibility of a net force.
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Compare with the answer and four-point rubric
- 1 point: Both situations redistribute mobile electrons.
- 1 point: Without contact or grounding, total sphere charge remains zero.
- 1 point: The symmetric uniform-field case has no net translational force.
- 1 point: The nonuniform rod field can produce attraction because the closer opposite induced charge experiences a stronger influence.
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 1Does polarization transfer charge from ground?
Not unless a grounding connection is actually present.
RECALL 2Which field vanishes in the metal?
The total field, not necessarily the applied field alone.
RECALL 3Why is the uniform-field example useful?
Its symmetry gives an exact, interpretable surface-density model.
Revisit these tomorrow and a week later. Try a fresh problem and explain why the method applies.
A neutral conductor can still attract a charge
- Inside the metal: E_induced + E₀ = 0.
- Neutral sphere in uniform E₀: σ = 3ε₀E₀ cos θ.
- Positive and negative surface contributions sum to zero.
Remember: Polarization does not imply net charge, and a uniform-field polarization model does not imply a net translational force.
Conditions: Exact surface-density result for a neutral sphere in a uniform applied field: σ = 3ε₀E₀cosθ. The diagram is a cross-section; arrow directions are schematic. A nonuniform nearby rod field is outside this quantitative model.
Refresh Kid · AP Physics C: Electricity and Magnetism Unit 3 (official Unit 10) · Objectives 10.1.A · Review edition
Framework, scope and review status
Mapped to College Board CED, Topic 10.1, objectives 10.1.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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