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Resolve interface orientation before combining dielectric regions

This supplied piecewise-continuum extension follows the fully filled dielectric core case.

Learning outcome: Predict both geometries for f = 0, 0.5 and 1 before checking. Draw the interface and indicate whether voltage or normal D is shared. Integrate the two layer energies and compare with the final-state capacitor expression.

Starting recap

Declare conductor boundaries, integrate potential or charging work, and distinguish controlled charge from voltage.

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Try

For A = 0.01 m², d = 2 mm, V = 4 V, kappa = 3 and f = 0.5, C0 ≈ 44.2709 pF and layered C ≈ 66.4064 pF. Vacuum and dielectric fields are 3000 and 1000 N/C, with voltage drops 3 and 1 V. The side-by-side comparison instead gives 88.5419 pF.

Open the experiment

Predict

For layers stacked across the gap with no free interface charge, which normal quantity is continuous?

Your prediction and reason

Experiment

Predict both geometries for f = 0, 0.5 and 1 before checking. Draw the interface and indicate whether voltage or normal D is shared. Integrate the two layer energies and compare with the final-state capacitor expression.

Resolve a layered dielectric interface

Worked teaching model · synthetic values · Plate area 0.01 m², gap 0.002 m, maintained 4 V. Dielectric slab normal to the field, remainder vacuum; no free interface charge or fringe effects. Piecewise ideal-linear continuum extension; comparison side-by-side area fill has a different interface orientation.

Dielectric relative permittivity: 3 ratio Fraction of gap thickness filled: 0.5 ratio

Reset keeps your prediction, captures and practice history.

Layer orientation determines the voltage sum

The field crosses the layers in sequence. The horizontal fraction is scaled; plate height is schematic and does not encode area. With no free interface charge, normal D is shared, while E and the potential slope change. Zero-width regions are absent, not hidden slabs.

Piecewise potential drop from first plate

Axes rescale independently. Curves join sampled coordinates; a joining segment is not additional measured evidence. Read physical units rather than screen angles or apparent slopes.

Piecewise potential drop from first plate: numerical coordinates

distance (mm)	accumulated drop (V)
0	0
0.0333333	0.1
0.0666667	0.2
0.1	0.3
0.133333	0.4
0.166667	0.5
0.2	0.6
0.233333	0.7
0.266667	0.8
0.3	0.9
0.333333	1
0.366667	1.1
0.4	1.2
0.433333	1.3
0.466667	1.4
0.5	1.5
0.533333	1.6
0.566667	1.7
0.6	1.8
0.633333	1.9
0.666667	2
0.7	2.1
0.733333	2.2
0.766667	2.3
0.8	2.4
0.833333	2.5
0.866667	2.6
0.9	2.7
0.933333	2.8
0.966667	2.9
1	3
1.03333	3.03333
1.06667	3.06667
1.1	3.1
1.13333	3.13333
1.16667	3.16667
1.2	3.2
1.23333	3.23333
1.26667	3.26667
1.3	3.3
1.33333	3.33333
1.36667	3.36667
1.4	3.4
1.43333	3.43333
1.46667	3.46667
1.5	3.5
1.53333	3.53333
1.56667	3.56667
1.6	3.6
1.63333	3.63333
1.66667	3.66667
1.7	3.7
1.73333	3.73333
1.76667	3.76667
1.8	3.8
1.83333	3.83333
1.86667	3.86667
1.9	3.9
1.93333	3.93333
1.96667	3.96667
2	4

Calculated values under the declared model
Quantity	Value	Unit
Series-layer capacitance	66.4064	pF
Side-by-side comparison capacitance	88.5419	pF
Free displacement magnitude	26.5626	nC/m²
Vacuum-layer field	3000	N/C
Dielectric-layer field	1000	N/C
Vacuum voltage drop	3	V
Dielectric voltage drop	1	V
Integrated layer energy	0.531251	nJ
Final-state energy	0.531251	nJ

Normal displacement continuity and the sum of layer voltage drops produce a harmonic, not arithmetic, permittivity combination.

With no free charge at the internal interface, normal D is equal in both layers. E is D/epsilon in each material. At zero layer thickness the displayed field for that absent material is a hypothetical limiting value, not a field in an existing region. Side-by-side area fill shares voltage instead and gives the arithmetic area-weighted capacitance.

Derive before checking. Retain contrasting captures and explain one assumption that could fail; calculated evidence is not actual laboratory work.

Normal displacement continuity and the sum of layer voltage drops produce a harmonic, not arithmetic, permittivity combination.

Observation to attach to a capture

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

Understand

State the extension and the interface

This supplied piecewise-continuum extension follows the fully filled dielectric core case. A slab occupies fraction f of the gap thickness, with vacuum filling the remainder. Interfaces are parallel to the plates, so the field crosses layers in sequence. There is no free charge at the internal interface; normal electric displacement D is the same in both regions.

Add voltage drops, not permittivities

With D = Q/A, the vacuum field is D/epsilon0 and dielectric field D/(kappa epsilon0). The total voltage is Dd[(1 − f) + f/kappa]/epsilon0. Thus C = C0/[(1 − f) + f/kappa], where C0 = epsilon0 A/d. The potential slope changes at the interface, while the potential itself is continuous.

Contrast a different geometry

If instead a dielectric fills fraction f of plate area across the entire gap, the regions share voltage and their charges add. Neglecting fringe effects, C = C0[(1 − f) + kappa f]. These arithmetic and harmonic combinations answer different boundary problems. At f = 0 or 1 they agree; between those limits they generally do not.

Check energy and absent regions

Integrate epsilon E²/2 over each region and add; the result is CV²/2. At f = 0 the dielectric region is absent, so its displayed field is only a hypothetical material-side limit. At f = 1 the same caveat applies to vacuum. Do not claim a measured field exists inside a zero-volume layer.

Worked original case

Investigation

Changed-case practice

Try a new practice case

For kappa = 4 and f = 0.5 in the thickness-layer geometry, what is C/C0?

Numerical result in the requested units

This public, unsaved check is not a secure assessment. Explain the source integral, symmetry or charge boundary in the retained course evidence workspace.

Sources and boundaries

Original cases checked against the official framework on 30 October 2026, with OpenStax verification on 31 October 2026 against the current official framework, Unit 10 and OpenStax capacitance. These are virtual supplied models, not instructions to charge, ground, wire or handle real equipment. Public answers are practice; retained plans and self-reports do not establish independent mastery, actual laboratory participation or AP authorization.

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

Which conductor boundary, geometric approximation or independent observation could change this conclusion?

Your explanation and limitations

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