Capacitance depends on the geometry
You will be able to: Relate capacitance to plate area, separation and dielectric material.
Why do larger, closer plates hold more separated charge per volt?
A capacitor separates equal positive and negative charges onto two conductors. A larger plate area or smaller gap can hold a larger charge magnitude at the same voltage. Capacitance describes that charge-per-volt relationship.
A useful starting point: Read field direction from a potential map →
Words and symbols before equations
- Capacitance C
- Q/|ΔV|, measured in farads (F=Coulomb/volt); here Q is magnitude on one plate.
- Plate area A
- Overlapping area of one plate, in m².
- Gap d
- Perpendicular plate separation, in m.
- Permittivity ε₀
- 8.85×10⁻¹² F/m; κ is the dielectric constant.
What this picture assumes
Vacuum parallel plates, ε₀=8.85×10⁻¹² F/m, connected to ideal 100 V source. Convert cm² to m² and mm to m. Plate gap small relative to lateral dimensions; no edge effects. Drawing is schematic, not a dimensional rendering.
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=17.7 nC on each plate; |E|=100000 V/m from + to −. Actual A=0.02 m² and d=0.001 m.
- 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 large parallel plates with a small gap and a uniform dielectric, C=κ ε₀A/d. This ideal expression neglects edge fields. Increasing A provides more charge-holding area; reducing d allows a given charge to produce a smaller voltage difference.
Capacitance is a property of this geometry and material, not independently changed by adding charge at fixed ideal geometry. Adding Q raises ΔV proportionally so Q/|ΔV| stays the same.
For opposite plates with equal charge magnitude, the whole capacitor can have net zero charge while storing energy. The approximately uniform gap field obeys |E|=|ΔV|/d; field direction is from the positive plate toward the negative plate.
A worked example, step by step
Vacuum plates have A=0.020 m² and d=0.0010 m. Find C and Q at 100 V.
- Use κ=1 and convert gap to m.
- C=(8.85×10⁻¹²)(0.020)/0.0010=1.77×10⁻¹⁰ F=177 pF.
- Q=C|ΔV|=(1.77×10⁻¹⁰)(100)=1.77×10⁻⁸ C=17.7 nC.
- Each plate carries that magnitude with opposite signs; net plate charge sums to zero.
Q means the magnitude on one plate, not the capacitor’s net charge. The symbol C can mean capacitance or the unit coulomb; read context.
If voltage doubles without changing ideal geometry, does C double?
Compare with an explanation
No. Q doubles and C=Q/|ΔV| remains fixed.
Predict. Change one thing. Explain.
Change area and separation at fixed 100 V. Compare capacitance, charge magnitude and field. The plate drawing is a schematic side view; the field readout uses the actual gap.
On narrow screens, swipe or scroll diagrams sideways to read all labels.
C=177 pF; Q=17.7 nC on each plate; |E|=100000 V/m from + to −. Actual A=0.02 m² and d=0.001 m.
Vacuum parallel plates, ε₀=8.85×10⁻¹² F/m, connected to ideal 100 V source. Convert cm² to m² and mm to m. Plate gap small relative to lateral dimensions; no edge effects. Drawing is schematic, not a dimensional rendering.
Explain what you noticed: Which quantity changed? Which stayed fixed? Use the relevant charge, field 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 questionA capacitor has C=200 pF and |ΔV|=50 V. (a) Find Q. (b) Find C after doubling A at fixed d. (c) At the same voltage find new Q. (d) State the plate-model boundary condition.
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Compare with the answer and four-point rubric
- 1 point: Q=10 nC.
- 1 point: C_new=400 pF.
- 1 point: Q_new=20 nC.
- 1 point: Plate gap much smaller than plate dimensions; neglect edge effects and use the stated dielectric.
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 sets ideal capacitance?
Geometry and dielectric material.
RECALL 2What Q appears in C=Q/V?
Magnitude of charge on one plate.
RECALL 3Why may net charge be zero but energy nonzero?
Opposite charges are separated in space, creating an electric field.
Revisit these tomorrow and a week later. Try a fresh problem and explain why the method applies.
Capacitance depends on the geometry
- C=Q/|ΔV|=κ ε₀A/d.
- Q=C|ΔV|; 1 pF=10⁻¹² F.
- Ideal uniform gap: |E|=|ΔV|/d.
Remember: Q means the magnitude on one plate, not the capacitor’s net charge. The symbol C can mean capacitance or the unit coulomb; read context.
Conditions: Vacuum parallel plates, ε₀=8.85×10⁻¹² F/m, connected to ideal 100 V source. Convert cm² to m² and mm to m. Plate gap small relative to lateral dimensions; no edge effects. Drawing is schematic, not a dimensional rendering.
Refresh Kid · AP Physics 2 Unit 2 (official Unit 10) · Objectives 10.6.A · Review edition
Framework, scope and review status
Mapped to College Board CED, Topic 10.6, objectives 10.6.A. CED effective Fall 2024, current PDF ©2026; checked September 16, 2026. Refresh Kid calls this the second AP Physics 2 unit; College Board numbers it Unit 10; the first unit in this course is official Unit 9. 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.
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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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