The battery connection decides the dielectric response
You will be able to: Compare dielectric insertion at fixed Q and fixed voltage using energy conservation.
Does a dielectric always lower the capacitor voltage?
Two identical capacitors begin at the same voltage. Disconnect one battery and leave the other attached, then insert the same dielectric into both. Both capacitances rise, but their charge and voltage changes differ.
A useful starting point: Bound charges reduce the field for fixed free charge →
Words and symbols before equations
- Initial capacitance C₀
- The vacuum capacitance before insertion.
- Full insertion
- A homogeneous dielectric fills the entire gap.
- Constraint
- The quantity held fixed by the physical connection.
- External work
- Work by the agent controlling slow insertion; negative means the agent removes energy.
What this picture assumes
Compare empty and fully filled endpoint states at fixed plate geometry. Ideal linear dielectric and slow controlled insertion. Charge stays fixed when isolated; voltage stays fixed with the battery. Work into the capacitor system is positive. Detailed fringe forces during insertion are not simulated.
Read the picture in three steps
- Locate the labeled sources, system boundary or graph axes. Read the units before comparing values.
- U: 5 → 1.25 nJ. ΔU -3.75 = battery 0 + external -3.75 nJ. Free charge stays fixed; voltage and field decrease for κ > 1.
- 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 an isolated capacitor, Q stays fixed and C becomes κC₀. Thus V becomes V₀/κ, E becomes E₀/κ and U becomes U₀/κ. The decrease in stored energy can do mechanical work as the slab is pulled inward; slow controlled insertion requires negative external work.
For a capacitor connected to an ideal fixed-voltage battery, V and E = V/d stay fixed. C, Q and stored energy increase by κ. The battery supplies W_b = VΔQ = (κ−1)C₀V², twice the stored-energy increase.
The remaining battery energy is available for mechanical work: W_ext = ΔU−W_b = −ΔU for slow ideal insertion. Both cases can attract the slab. The model compares fully empty and fully filled endpoint states, not detailed fringe-force trajectories.
| Quantity | Isolated capacitor | Ideal battery connected |
|---|---|---|
| Fixed quantity | Free plate charge Q | Voltage ΔV |
| Capacitance | Multiplied by κ | Multiplied by κ |
| Field and voltage | Divided by κ | Unchanged |
| Plate charge | Unchanged | Multiplied by κ |
| Stored energy | Divided by κ | Multiplied by κ |
A worked example, step by step
A C₀ = 100 pF capacitor begins at V₀ = 10 V. Insert a κ = 4 dielectric. Compare disconnected and connected results.
- Initially Q₀ = 1 nC and U₀ = 5 nJ. Final capacitance is 400 pF.
- Disconnected: Q = 1 nC, V = 2.5 V and U = 1.25 nJ.
- Connected: V = 10 V, Q = 4 nC and U = 20 nJ.
- Connected battery work is 10(3 nC) = 30 nJ; ΔU = 15 nJ, so W_ext = −15 nJ.
Do not say a dielectric always reduces E or V: those reductions apply when free charge stays fixed.
Why can connected stored energy rise while the slab is attracted?
Compare with an explanation
The battery supplies more energy than the capacitor stores; the difference can become mechanical work.
Predict. Change one thing. Explain.
Switch between isolated and connected modes for the same initial C₀ and V₀. Increase κ and explain the sign of battery work and external work. Set κ = 1 to recover the starting state.
On narrow screens, swipe or scroll diagrams sideways to read all labels.
U: 5 → 1.25 nJ. ΔU -3.75 = battery 0 + external -3.75 nJ. Free charge stays fixed; voltage and field decrease for κ > 1.
Compare empty and fully filled endpoint states at fixed plate geometry. Ideal linear dielectric and slow controlled insertion. Charge stays fixed when isolated; voltage stays fixed with the battery. Work into the capacitor system is positive. Detailed fringe forces during insertion are not simulated.
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 questionA 200 pF capacitor at 6 V receives a κ = 3 dielectric. Compute final Q, V and U for isolated and connected cases.
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Compare with the answer and four-point rubric
- 1 point: Initially Q₀ = 1.2 nC and U₀ = 3.6 nJ; final C = 600 pF.
- 1 point: Isolated Q = 1.2 nC and V = 2 V.
- 1 point: Isolated U = 1.2 nJ.
- 1 point: Connected V = 6 V, Q = 3.6 nC and U = 10.8 nJ.
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 changes in both cases?
Capacitance increases by κ for complete filling.
RECALL 2Why can connected Q increase?
The battery supplies additional free plate charge.
RECALL 3What happens at κ = 1?
No dielectric change: all endpoint changes and work terms are zero.
Revisit these tomorrow and a week later. Try a fresh problem and explain why the method applies.
The battery connection decides the dielectric response
- Fully filled: C = κC₀.
- Isolated: Q fixed, V and U divided by κ.
- Connected: V fixed, Q and U multiplied by κ.
- Energy balance: ΔU = W_b + W_ext.
Remember: Do not say a dielectric always reduces E or V: those reductions apply when free charge stays fixed.
Conditions: Compare empty and fully filled endpoint states at fixed plate geometry. Ideal linear dielectric and slow controlled insertion. Charge stays fixed when isolated; voltage stays fixed with the battery. Work into the capacitor system is positive. Detailed fringe forces during insertion are not simulated.
Refresh Kid · AP Physics C: Electricity and Magnetism Unit 3 (official Unit 10) · Objectives 10.4.A · Review edition
Framework, scope and review status
Mapped to College Board CED, Topic 10.4, objectives 10.4.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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