Area and separation set parallel-plate capacitance
You will be able to: Derive C = ε₀A/d from Gauss’s law and potential difference.
Why do wider plates store more charge at the same voltage?
Imagine two square metal plates facing each other across a small gap. Spread a fixed amount of charge over a larger plate and its surface density decreases. A weaker field then produces less voltage for the same charge.
A useful starting point: Capacitance is charge per potential difference →
Words and symbols before equations
- Plate area A
- Facing area of one plate, in m².
- Gap d
- Perpendicular separation of the plates, in metres.
- Permittivity ε₀
- Vacuum constant; use 8.85 × 10⁻¹² F/m.
- Fringing
- Edge fields that depart from the uniform central-field model.
What this picture assumes
Vacuum square plates, negligible fringing, held at fixed voltage. Gap is much smaller than plate width. Optional 3D uses separate labeled lateral and gap display scales to make the small gap visible; it is a geometry view, not a field simulation.
Read the picture in three steps
- Locate the labeled sources, system boundary or graph axes. Read the units before comparing values.
- C = 177 pF; Q = 1770 pC; E = 10000 N/C; square-plate width = 0.1414 m. Schematic gap enlarged; voltage fixed.
- Check what the picture assumes below. Use the Explore task to predict one change before moving a control.
Connect the picture to the physics
An ideal infinite sheet produces |E| = σ/(2ε₀) on each side. Two equal opposite sheets add between the plates, giving E = σ/ε₀ = Q/(ε₀A), and cancel outside. Finite plates approximate this away from edges when d is small compared with their width.
A uniform field across gap d gives ΔV = Ed. Substitute E to obtain ΔV = Qd/(ε₀A); then C = Q/ΔV = ε₀A/d. A larger A increases C and a larger gap decreases it.
At fixed battery voltage, the battery supplies whatever plate charge Q = CΔV is required. The optional 3D geometry view shows facing area and perpendicular gap with independent labeled display scales; rotating the camera changes neither quantity.
A worked example, step by step
Vacuum plates have A = 0.020 m² and d = 1.0 mm. Find C and Q at 10 V.
- Convert d = 0.0010 m.
- C = ε₀A/d = 8.85×10⁻¹² × 0.020 / 0.0010 = 177 pF.
- Q = CΔV = 1770 pC = 1.77 nC.
- E = ΔV/d = 10,000 N/C, directed from positive to negative plate.
Convert millimetres to metres and use the area of one facing plate. Fringing prevents exact cancellation for real finite plates.
What happens when both area and gap double?
Compare with an explanation
Their ratio is unchanged, so ideal capacitance is unchanged.
Predict. Change one thing. Explain.
Change area and gap separately at fixed voltage. Predict C and Q before reading the result. Rotate the optional geometry view and confirm the readouts do not change.
On narrow screens, swipe or scroll diagrams sideways to read all labels.
C = 177 pF; Q = 1770 pC; E = 10000 N/C; square-plate width = 0.1414 m. Schematic gap enlarged; voltage fixed.
Optional 3D view: facing area and perpendicular gap
Each plate is a square in 3D. The gap is enlarged relative to the lateral dimensions and both display scales are stated. Rotation changes no physical quantity.
Vacuum square plates, negligible fringing, held at fixed voltage. Gap is much smaller than plate width. Optional 3D uses separate labeled lateral and gap display scales to make the small gap visible; it is a geometry view, not a field simulation.
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 questionStarting from the field of an isolated charged sheet, derive the capacitance of a narrow-gap vacuum parallel-plate capacitor. State the approximation.
This response is not submitted or saved. Copy it before leaving.
Compare with the answer and four-point rubric
- 1 point: Each sheet contributes σ/(2ε₀).
- 1 point: Between opposite sheets the fields add: E = Q/(ε₀A).
- 1 point: The uniform-field voltage is ΔV = Ed = Qd/(ε₀A).
- 1 point: Therefore C = ε₀A/d, neglecting edges when gap is small compared with plate dimensions.
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 1Which area appears in ε₀A/d?
The facing area of one plate.
RECALL 2Why do the two sheet fields add inside?
Both point from the positive plate toward the negative plate.
RECALL 3What does camera rotation change?
Only the displayed viewpoint, not area, gap or capacitance.
Revisit these tomorrow and a week later. Try a fresh problem and explain why the method applies.
Area and separation set parallel-plate capacitance
- E_between ≈ Q/(ε₀A).
- ΔV ≈ Ed.
- C ≈ ε₀A/d for a narrow vacuum gap.
Remember: Convert millimetres to metres and use the area of one facing plate. Fringing prevents exact cancellation for real finite plates.
Conditions: Vacuum square plates, negligible fringing, held at fixed voltage. Gap is much smaller than plate width. Optional 3D uses separate labeled lateral and gap display scales to make the small gap visible; it is a geometry view, not a field simulation.
Refresh Kid · AP Physics C: Electricity and Magnetism Unit 3 (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. 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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