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SCIENCE · INVESTIGATE AND EXPLAIN

Explain why dielectric insertion needs a boundary condition

A dielectric is an insulating material that polarizes in an electric field. Bound positive and negative charge distributions shift slightly relative to one another, producing an opposing field. A linear dielectric that fills an ideal capacitor increases its capacitance by a factor κ, the relative permittivity. The material response must be distinguished from free charge supplied to conducting plates.

Learning outcome: Capture κ=3 in both boundary modes. Reconstruct ΔU=source work+external mechanical work. Repeat κ=1 to verify that no change yields zero work. State which quantities are constrained rather than discovered.

Starting recap

Use signed charge, vector components, inverse-square ratios and energy conservation. Distinguish potential from potential energy and name the system boundary.

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Try

Initially C₀=100 pF and V₀=10 V, so Q₀=1 nC and U₀=5 nJ. For κ=3, disconnected: C=300 pF, Q=1 nC, V=10/3 V, U=5/3 nJ. Connected: Q=3 nC, V=10 V, U=15 nJ; the source supplies 20 nJ and external mechanical work is −10 nJ.

Open the experiment

Predict

If the final voltage stays at 10 V after insertion, can you still assume the free plate charge stayed at 1 nC?

Experiment

Capture κ=3 in both boundary modes. Reconstruct ΔU=source work+external mechanical work. Repeat κ=1 to verify that no change yields zero work. State which quantities are constrained rather than discovered.

Distinguish fixed charge from fixed voltage

Worked teaching model · synthetic values · Initial vacuum capacitor C0=100 pF at 10 V, separation 2 mm. A linear dielectric completely fills the gap. Disconnected retains free plate charge; connected retains voltage. Quasistatic insertion, no dissipation. The energy balance includes electrical source work and external mechanical work.

Reset keeps your prediction, captures and practice history.

0550relative permittivitystored energy (nJ)

Axes rescale with settings; compare numerical values, not screen angles or apparent lengths. Solid/points show the selected model relationship or observations; the dashed comparison is explained below.

Calculated evidence ledger — synthetic teaching model
QuantityValueUnit
Final capacitance300pF
Final free plate charge1nC
Final potential difference3.3333V
Final electric field1666.7V/m
Final stored energy1.6667nJ
Change in field energy-3.3333nJ
Electrical source work0nJ
External mechanical work-3.3333nJ
Accessible plot coordinates
Primary plotted points
relative permittivitystored energy (nJ)
15
22.5
31.6667
41.25
51

Disconnected: free charge stays fixed while voltage and energy fall.

Polarization produces a field opposing the original free-charge field. With fixed voltage the source adds charge, maintaining the net field instead of letting it fall. The signed ledger obeys ΔU = source work + external mechanical work. Negative mechanical work means the external agent removes energy during controlled insertion.

Disconnected: free charge stays fixed while voltage and energy fall.

Polarization produces a field opposing the original free-charge field. With fixed voltage the source adds charge, maintaining the net field instead of letting it fall. The signed ledger obeys ΔU = source work + external mechanical work. Negative mechanical work means the external agent removes energy during controlled insertion.

Captures preserve recipes and replay their calculated results. They are synthetic model evidence, not physical measurements.

Understand

Establish the model and evidence

A dielectric is an insulating material that polarizes in an electric field. Bound positive and negative charge distributions shift slightly relative to one another, producing an opposing field. A linear dielectric that fills an ideal capacitor increases its capacitance by a factor κ, the relative permittivity. The material response must be distinguished from free charge supplied to conducting plates.

Before predicting voltage or energy, ask whether the capacitor remains connected to its source. A disconnected ideal capacitor cannot exchange free plate charge, so Q stays fixed. Because C grows, V=Q/C falls; its field and stored energy also decrease. Polarization changes internal distribution without conducting free charge across the insulating gap.

A connected ideal capacitor has a fixed voltage. Its source supplies additional free plate charge as C increases, leaving V and the gap field unchanged. The dielectric still produces an opposing contribution; the extra free charge compensates it. Saying that a dielectric always lowers the net field silently assumes fixed free charge and would be wrong for this connected boundary.

Energy accounting includes more than the field. In quasistatic insertion the material is attracted into the gap, and an external agent controlling its motion removes mechanical energy. With a connected fixed source, source work is VΔQ and field energy rises by only half that amount in this ideal setup; the other half leaves through mechanical work. The activity explicitly assumes no dissipation. Real capacitor experiments and high-voltage handling are outside this virtual task.

Worked evidence

Initially C₀=100 pF and V₀=10 V, so Q₀=1 nC and U₀=5 nJ. For κ=3, disconnected: C=300 pF, Q=1 nC, V=10/3 V, U=5/3 nJ. Connected: Q=3 nC, V=10 V, U=15 nJ; the source supplies 20 nJ and external mechanical work is −10 nJ.

Predict and investigate

If the final voltage stays at 10 V after insertion, can you still assume the free plate charge stayed at 1 nC?

Capture κ=3 in both boundary modes. Reconstruct ΔU=source work+external mechanical work. Repeat κ=1 to verify that no change yields zero work. State which quantities are constrained rather than discovered.

Transfer to changed conditions

A 200 pF capacitor begins at 6 V. A κ=2 dielectric fills its gap while the source stays connected. Find ΔQ, ΔU, source work and external mechanical work.

Compare the worked transfer

Initial Q=1.2 nC and U=3.6 nJ. Final C=400 pF, Q=2.4 nC and U=7.2 nJ. ΔQ=1.2 nC, ΔU=3.6 nJ, source work=6×1.2=7.2 nJ, external mechanical work=−3.6 nJ.

Explain the boundary

Explain how polarization and fixed voltage can coexist with an unchanged net field in the gap.

Permittivity identifies a material response

For this linear isotropic dielectric, absolute permittivity is ε=κε₀. Vacuum ε₀ is the reference in the force and capacitor equations; κ is not another name for free plate charge. Composition and the ease of local electronic displacement affect polarization, while a conductor’s freely mobile charge requires a different equilibrium description.

Predict and check: compare fully filled capacitors with κ=2 and κ=4, identical plate geometry and the same fixed free charge. The second has twice the capacitance, half the voltage and half the net gap field. Repeat at fixed voltage instead: field stays the same and free plate charge doubles. Explain why these two comparisons do not contradict the material’s induced opposing field. Do not infer a universal constant permittivity for every nonlinear, frequency-dependent or conducting material. (Current clauses 10.1.C and 10.6.A.6.)

Sources and authorship

Physical relationship reference. Current AP Physics 2 framework, checked 30 October 2026. These cases, diagrams and numerical records are original. All experiments are bounded browser calculations or synthetic data, not measured laboratory evidence. Do not construct or handle charged capacitors, high-voltage sources or particle-beam equipment for these activities.

Clause-level source/action reconciliation checked 31 October 2026 against the linked current framework and Fall 2026 clarification. Added packets are original virtual comparisons; no new practical or learner-completion claim.

Apply

A disconnected ideal capacitor is fully filled with κ=4 dielectric. Which prediction follows?

Public practice checks a specific claim. It does not establish independent mastery.

Explain how polarization and fixed voltage can coexist with an unchanged net field in the gap.

Continue

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