Electric Potential, Energy Density, and Conservative Fields

Unit 4: Dielectrics, capacitance and stored energy

Four lessons: Polarization and electric displacement, Dielectric interfaces and field refraction, Parallel-plate, coaxial and spherical capacitance, Electrostatic energy, force and breakdown limits.

Learning outcomes

By the end of this unit, explain the physical field variables and their SI units; solve a useful analytical reference geometry; reproduce a field-level numerical calculation; identify boundary assumptions; and explain an engineering decision that would be unjustified by the ideal model alone.

Unit investigation

A fixed charge and a fixed voltage are different experiments. For a parallel-plate ideal capacitor, analyze what happens to E, Q, C and stored field energy when a dielectric slab is inserted completely while a battery remains attached. Then repeat after charging and disconnecting the battery. At fixed V, C increases and Q must enter through the circuit; at fixed Q, the voltage falls. The change in field energy cannot by itself describe mechanical work until the source’s energy exchange is considered. Explain why both models can predict dielectric attraction without a contradiction.

For a coaxial capacitor calculate C per unit length from a voltage integral of E(r). Record surface field at the inner conductor and stored energy for the same geometry. Compare increasing εr to decreasing inner radius, using one electrical constraint at a time. An engineer designing for maximum charge storage, low voltage stress and low cable mass faces competing objectives. The student report should be a Pareto comparison, not an unsupported winner based on capacitance alone.

Guided lesson path

Step 1: Polarization and electric displacement

Review the governing relationship, reproduce the numerical variant with independent unit conversions, and explicitly note one observation that would falsify the model's assumptions. See [lesson 1](../lessons/U04-L01-polarization-and-electric-displacement.html). The mathematical conclusion should follow the boundary and source assumptions, not merely the choice of familiar-looking formula.

Step 2: Dielectric interfaces and field refraction

Review the governing relationship, reproduce the numerical variant with independent unit conversions, and explicitly note one observation that would falsify the model's assumptions. See [lesson 2](../lessons/U04-L02-dielectric-interfaces-and-field-refraction.html). The mathematical conclusion should follow the boundary and source assumptions, not merely the choice of familiar-looking formula.

Step 3: Parallel-plate, coaxial and spherical capacitance

Review the governing relationship, reproduce the numerical variant with independent unit conversions, and explicitly note one observation that would falsify the model's assumptions. See [lesson 3](../lessons/U04-L03-parallel-plate-coaxial-and-spherical-capacitance.html). The mathematical conclusion should follow the boundary and source assumptions, not merely the choice of familiar-looking formula.

Step 4: Electrostatic energy, force and breakdown limits

Review the governing relationship, reproduce the numerical variant with independent unit conversions, and explicitly note one observation that would falsify the model's assumptions. See [lesson 4](../lessons/U04-L04-electrostatic-energy-force-and-breakdown-limits.html). The mathematical conclusion should follow the boundary and source assumptions, not merely the choice of familiar-looking formula.

Unit verification checklist

• Define at least one coordinate system and positive contour or normal orientation.
• State which source or boundary quantities are controlled and which are outputs.
• Compare analytical and computational predictions, including one limiting case.
• Report at least one sensitivity result rather than a single nominal value.
• Describe which material data would be needed for genuine engineering validation.
• Prepare a concise error budget identifying unit conversion, approximate geometry and material uncertainty.

Practice and project: [Unit assessment](../assessments/unit-04.html) · [Unit engineering project](../projects/unit-04.html) · [Interactive mathematical model](../simulations/model-04.html).

Unit 4, lesson 1 — Polarization and electric displacement

Book 21 — Electromagnetic Field Theory I · Prerequisites: multivariable calculus, introductory circuits and SI units. Quality note: Original educational draft; physical equations and instructional solutions require independent review before publication.

Physical idea and meaning

A dielectric does not permit freely mobile charges to neutralize an imposed static field like a metal. Instead it responds through microscopic polarization: tiny dipoles align or are induced, producing bound charge. The displacement field D is defined to separate free and bound charge bookkeeping. In a linear isotropic medium D=epsilon E; that relation is a constitutive assumption, not a universal field law. Anisotropic crystals require a tensor epsilon, and strong fields may invalidate linear response.

Governing relationships and derivation pathway

D=epsilon0 E+P and div D=rho_free. For a linear isotropic nondispersive model, P=epsilon0(εr-1)E and epsilon=epsilon0 εr. Bound charge density is rho_bound=-div P and surface bound charge sigma_bound=P·n.

Worked numerical illustration

For εr=4 and E=1000 V/m, D=4epsilon0·1000=3.542e-8 C/m² and P=3epsilon0·1000=2.656e-8 C/m². This is a macroscopic average, not the polarization of a specific molecule.

Independent numerical variant

For εr=5 at E=500 V/m, D=2.214e-8 C/m² and P=(5-1)epsilon0×500≈1.771e-8 C/m².

Engineering context: Dielectric response ledger

Measure capacitance with the same electrode geometry filled first with air and then with an insulating sample. In an ideal uniform-field situation their ratio estimates relative permittivity. However the measured ratio also includes fringe fields and parasitic leads. Repeat at several frequencies and temperatures; the apparent εr may change because polarization mechanisms are dispersive. Avoid inserting one tabulated permittivity into both a static high-voltage model and a high-frequency RF model without checking its validity range.

Sensitivity and design interpretation

At fixed E, D scales linearly with εr under the linear model; at fixed D, E varies inversely with εr.

Why the model can fail

Some texts call D “electric flux density”; it does not mean electric field lines are material wires. Frequency-dependent complex permittivity in a lossy dielectric must be handled separately from static εr.

Two resolved conceptual checkpoints

Question A. What is D in a linear dielectric of εr=3 at fixed E compared with vacuum?

Reasoned answer. It is three times the vacuum displacement value. That statement presumes the same E and a linear isotropic model.

Question B. Is all polarization charge free charge?

Reasoned answer. No. Bound charge arises from divergence of P and appears in microscopic polarization; Gauss’s law for D uses free charge.

Next steps

[Unit 4 interactive model](../simulations/model-04.html) · [Assessment](../assessments/unit-04.html) · [Project](../projects/unit-04.html) · [Shared study guidance](../STUDY-GUIDE.html).

Unit 4, lesson 2 — Dielectric interfaces and field refraction

Book 21 — Electromagnetic Field Theory I · Prerequisites: multivariable calculus, introductory circuits and SI units. Quality note: Original educational draft; physical equations and instructional solutions require independent review before publication.

Physical idea and meaning

Electrostatic boundary conditions are as important as bulk equations. The tangential component of E remains continuous at a static interface, while the normal component of D changes by any free surface charge. At an interface without free surface charge, D normal is continuous, so E normal generally differs on the two sides when the permittivities differ. The resulting direction change is sometimes called field refraction; it follows conservation and constitutive relations rather than optical Snell’s law.

Governing relationships and derivation pathway

For interface normal n pointing from medium 1 to 2, n·(D2-D1)=sigma_free and n×(E2-E1)=0 in electrostatics. When sigma_free=0, epsilon1 E1n=epsilon2 E2n. Et is the same on both sides.

Worked numerical illustration

Two planar dielectrics with εr1=2 and εr2=5 carry Dn=20 nC/m² with no free interfacial charge. E1n=20e-9/(2epsilon0)=1129 V/m while E2n=452 V/m. Assigning the same normal E to both would create a fictitious interface charge.

Independent numerical variant

For normal D=12 nC/m², εr=3 yields E≈451.8 V/m; εr=6 yields E≈225.9 V/m. Neither result requires a free interface sheet charge.

Engineering context: Two-layer insulation

Connect two dielectric sheets in series across a common electrode area. With no free interfacial charge, the normal displacement is the same, while electric-field magnitudes differ inversely with permittivity. The voltage fraction carried by a layer is its thickness times its field, not just its thickness fraction. A thinner lower-permittivity layer can experience the larger field. Record the electric stress in each layer before deciding which layer limits withstand capability.

Sensitivity and design interpretation

At zero free interface charge, doubling ε2 halves E2n for the same Dn, while Et remains equal.

Why the model can fail

At a real rough interface, local normals vary and ideal planar results approximate only the locally smooth region. Interfaces can carry trapped charge, making the zero-free-charge simplification invalid.

Two resolved conceptual checkpoints

Question A. At a charge-free interface, which normal quantity stays continuous?

Reasoned answer. Normal D stays continuous; normal E changes inversely with epsilon. Tangential E stays continuous in electrostatics.

Question B. Why does the lower-ε layer often experience higher normal electric stress?

Reasoned answer. At fixed normal D, E=D/epsilon, so smaller epsilon produces larger E even while the normal displacement matches.

Next steps

[Unit 4 interactive model](../simulations/model-04.html) · [Assessment](../assessments/unit-04.html) · [Project](../projects/unit-04.html) · [Shared study guidance](../STUDY-GUIDE.html).

Unit 4, lesson 3 — Parallel-plate, coaxial and spherical capacitance

Book 21 — Electromagnetic Field Theory I · Prerequisites: multivariable calculus, introductory circuits and SI units. Quality note: Original educational draft; physical equations and instructional solutions require independent review before publication.

Physical idea and meaning

Capacitance measures charge per potential difference for a specified geometry and dielectric configuration. Increasing electrode overlap or permittivity increases C, while increasing plate separation generally decreases it. Geometry matters: the field of a coaxial cable varies as 1/rho; treating it as a constant parallel-plate field gives a systematically wrong voltage. In a two-conductor static problem, the capacitance is positive and depends on the full conductor arrangement, not just the material label.

Governing relationships and derivation pathway

C=Q/ΔV. Ideal plates: C≈epsilon A/d when fringing is negligible. Coaxial line length L, inner radius a, outer b: C=2πepsilon L/ln(b/a). Concentric spheres radii a

Worked numerical illustration

For 1 m of coax with εr=2.25, a=1 mm and b=4 mm, C=2π(2.25epsilon0)/ln 4≈90.3 pF. Doubling length doubles C; doubling both radii leaves the ratio b/a unchanged and does not change capacitance per unit length in this ideal model.

Independent numerical variant

For 1 m coax with a=1 mm, b=5 mm, εr=2, C≈69.1 pF/m. Increasing b to 10 mm reduces C to about 48.3 pF/m.

Engineering context: Cable geometry experiment

Hold a coaxial cable’s outer radius fixed and vary its inner radius. Capacitance per metre increases as ln(b/a) decreases, but the maximum E at the inner conductor may also increase for certain constrained voltage designs. Choose what is held constant—V, Q or field limit—before comparing candidates. A trustworthy design table contains geometry, εr, C per metre, peak surface E and stored energy. No one scalar metric establishes a universally superior geometry.

Sensitivity and design interpretation

At fixed b, increasing inner radius a decreases ln(b/a), increasing capacitance per length while changing peak field.

Why the model can fail

For a cable with several dielectric layers, an effective permittivity cannot generally be obtained by arithmetic averaging. Integrate each radial electric-field segment and add the resulting voltage drops at fixed charge.

Two resolved conceptual checkpoints

Question A. For ideal coax, does doubling both radii change C per metre?

Reasoned answer. No. ln(b/a) stays fixed, and per-length C for a homogeneous medium is unchanged. However operating electric field at fixed voltage changes.

Question B. Why is C=εA/d unsafe for a coaxial cable?

Reasoned answer. Its radial field is nonuniform and cylindrical geometry requires integrating voltage from a to b; the logarithm is essential.

Next steps

[Unit 4 interactive model](../simulations/model-04.html) · [Assessment](../assessments/unit-04.html) · [Project](../projects/unit-04.html) · [Shared study guidance](../STUDY-GUIDE.html).

Unit 4, lesson 4 — Electrostatic energy, force and breakdown limits

Book 21 — Electromagnetic Field Theory I · Prerequisites: multivariable calculus, introductory circuits and SI units. Quality note: Original educational draft; physical equations and instructional solutions require independent review before publication.

Physical idea and meaning

Charging a capacitor transfers energy from the source into its field. For linear nondispersive materials, electrostatic energy density is one half of E·D. An attractive force between biased electrodes can be obtained by an appropriate energy or co-energy derivative, but the electrical constraint must be explicit: fixed voltage and fixed charge lead to different mathematical derivatives. The ideal field model cannot determine safe working voltage without material breakdown data, defects, temperature and field concentration.

Governing relationships and derivation pathway

W_e = (1/2)CV²=Q²/(2C) for a linear capacitor; u_e=(1/2)E·D. For an ideal parallel plate at fixed V, magnitude of attractive force is approximately epsilon A V²/(2d²) if fringing is negligible.

Worked numerical illustration

An ideal C=100 pF capacitor charged to 500 V stores 0.5×100e-12×500²=12.5 μJ. Increasing voltage by a factor of two quadruples stored energy, an important result for design of test circuits and discharge protections.

Independent numerical variant

A 50 pF ideal capacitor at 200 V stores 1.0 μJ. At 400 V it stores 4.0 μJ, which verifies the quadratic voltage dependence.

Engineering context: Stored charge and safe discharge

An energy calculation is a starting point for analyzing a capacitor, not a complete safety assessment. Even modest average energy can accompany dangerous high voltages or rapid currents. Model a resistive discharge with V(t)=V0 exp[-t/(RC)] and identify the period during which a measurement instrument should not be connected. Real capacitors can have dielectric absorption and residual voltage. Student simulations should use synthetic signals and should never advise building a live high-voltage test.

Sensitivity and design interpretation

At fixed C, doubling voltage multiplies energy by four; at fixed voltage, doubling C doubles energy.

Why the model can fail

Do not use vacuum electrostatic stress blindly inside nonlinear ferroelectrics. Breakdown field depends on time scale and geometry; a simulation reporting voltage alone is not an insulation qualification.

Two resolved conceptual checkpoints

Question A. If capacitor voltage triples at fixed C, what happens to stored energy?

Reasoned answer. It increases by a factor of nine because W=0.5CV², but only while C stays effectively constant.

Question B. Can an energy calculation alone specify dielectric safety?

Reasoned answer. No. Peak field, defects, duration, geometry, leakage and material dielectric strength all matter.

Next steps

[Unit 4 interactive model](../simulations/model-04.html) · [Assessment](../assessments/unit-04.html) · [Project](../projects/unit-04.html) · [Shared study guidance](../STUDY-GUIDE.html).

Unit 4 engineering project: Two-layer dielectric and cable design

Problem statement and intended decision

Compare two insulation candidates for a coaxial instrument cable using inner radius 1 mm and outer radius 5 mm. One has a uniform εr=2.5; the other uses two radial layers with εr=2 and εr=4. At specified line charge, integrate radial E to calculate voltage and capacitance per metre for each candidate. At specified voltage, compute the peak E at the inner surface. Explain why the layer that minimizes capacitance is not automatically the layer that minimizes electric stress. Produce a sensitivity chart for layer boundary radius and include assumptions about leakage and breakdown.

Detailed investigation plan

A fixed charge and a fixed voltage are different experiments. For a parallel-plate ideal capacitor, analyze what happens to E, Q, C and stored field energy when a dielectric slab is inserted completely while a battery remains attached. Then repeat after charging and disconnecting the battery. At fixed V, C increases and Q must enter through the circuit; at fixed Q, the voltage falls. The change in field energy cannot by itself describe mechanical work until the source’s energy exchange is considered. Explain why both models can predict dielectric attraction without a contradiction.

For a coaxial capacitor calculate C per unit length from a voltage integral of E(r). Record surface field at the inner conductor and stored energy for the same geometry. Compare increasing εr to decreasing inner radius, using one electrical constraint at a time. An engineer designing for maximum charge storage, low voltage stress and low cable mass faces competing objectives. The student report should be a Pareto comparison, not an unsupported winner based on capacitance alone.

Submission requirements

Follow the seven evidence deliverables and the 100-point project rubric in the [Shared study guidance](../STUDY-GUIDE.html). The unit-specific source model, investigation plan, parameter choices and physical limitations above determine whether the report is adequate.

Submission safeguard

All activities are simulation and analysis only. If a project might ordinarily be performed with mains current, charged capacitors, electromagnets, high-voltage insulation or RF power, replace hardware with the ideal mathematical model and a synthetic dataset. Do not treat this project brief as authorization or guidance to perform dangerous laboratory measurements.

Unit 4 assessment — Dielectrics, capacitance and stored energy

Instructions: Work independently before expanding answers. These twelve questions cover all four lessons. State your coordinate reference, units, sign convention and physical assumptions. The answers below explain the expected reasoning; identical numerical output alone is insufficient.

Question 1 — Polarization and electric displacement

Prompt: What is D in a linear dielectric of εr=3 at fixed E compared with vacuum?

Reveal worked reasoning

It is three times the vacuum displacement value. That statement presumes the same E and a linear isotropic model. The matching lesson and shared study guide provide additional verification questions.

Question 2 — Polarization and electric displacement

Prompt: Is all polarization charge free charge?

Reveal worked reasoning

No. Bound charge arises from divergence of P and appears in microscopic polarization; Gauss’s law for D uses free charge. The matching lesson and shared study guide provide additional verification questions.

Question 3 — Polarization and electric displacement numerical and decision challenge

Prompt: Independently derive and check this scenario: For εr=5 at E=500 V/m, D=2.214e-8 C/m² and P=(5-1)epsilon0×500≈1.771e-8 C/m². Then choose a parameter to change, predict its effect from at fixed E, D scales linearly with εr under the linear model; at fixed D, E varies inversely with εr. State which quantities remain fixed and identify one important excluded physical effect.

Reveal worked reasoning and review criteria

Reference calculation: For εr=5 at E=500 V/m, D=2.214e-8 C/m² and P=(5-1)epsilon0×500≈1.771e-8 C/m².

Sensitivity logic: At fixed E, D scales linearly with εr under the linear model; at fixed D, E varies inversely with εr.

Engineering qualification: Some texts call D “electric flux density”; it does not mean electric field lines are material wires. Frequency-dependent complex permittivity in a lossy dielectric must be handled separately from static εr. The final report must include a quantity with correct SI units, an analytical limit and a specific reason the ideal calculation is not sufficient for hardware sign-off.

Question 4 — Dielectric interfaces and field refraction

Prompt: At a charge-free interface, which normal quantity stays continuous?

Reveal worked reasoning

Normal D stays continuous; normal E changes inversely with epsilon. Tangential E stays continuous in electrostatics. The matching lesson and shared study guide provide additional verification questions.

Question 5 — Dielectric interfaces and field refraction

Prompt: Why does the lower-ε layer often experience higher normal electric stress?

Reveal worked reasoning

At fixed normal D, E=D/epsilon, so smaller epsilon produces larger E even while the normal displacement matches. The matching lesson and shared study guide provide additional verification questions.

Question 6 — Dielectric interfaces and field refraction numerical and decision challenge

Prompt: Independently derive and check this scenario: For normal D=12 nC/m², εr=3 yields E≈451.8 V/m; εr=6 yields E≈225.9 V/m. Neither result requires a free interface sheet charge. Then choose a parameter to change, predict its effect from at zero free interface charge, doubling ε2 halves E2n for the same Dn, while Et remains equal. State which quantities remain fixed and identify one important excluded physical effect.

Reveal worked reasoning and review criteria

Reference calculation: For normal D=12 nC/m², εr=3 yields E≈451.8 V/m; εr=6 yields E≈225.9 V/m. Neither result requires a free interface sheet charge.

Sensitivity logic: At zero free interface charge, doubling ε2 halves E2n for the same Dn, while Et remains equal.

Engineering qualification: At a real rough interface, local normals vary and ideal planar results approximate only the locally smooth region. Interfaces can carry trapped charge, making the zero-free-charge simplification invalid. The final report must include a quantity with correct SI units, an analytical limit and a specific reason the ideal calculation is not sufficient for hardware sign-off.

Question 7 — Parallel-plate, coaxial and spherical capacitance

Prompt: For ideal coax, does doubling both radii change C per metre?

Reveal worked reasoning

No. ln(b/a) stays fixed, and per-length C for a homogeneous medium is unchanged. However operating electric field at fixed voltage changes. The matching lesson and shared study guide provide additional verification questions.

Question 8 — Parallel-plate, coaxial and spherical capacitance

Prompt: Why is C=εA/d unsafe for a coaxial cable?

Reveal worked reasoning

Its radial field is nonuniform and cylindrical geometry requires integrating voltage from a to b; the logarithm is essential. The matching lesson and shared study guide provide additional verification questions.

Question 9 — Parallel-plate, coaxial and spherical capacitance numerical and decision challenge

Prompt: Independently derive and check this scenario: For 1 m coax with a=1 mm, b=5 mm, εr=2, C≈69.1 pF/m. Increasing b to 10 mm reduces C to about 48.3 pF/m. Then choose a parameter to change, predict its effect from at fixed b, increasing inner radius a decreases ln(b/a), increasing capacitance per length while changing peak field. State which quantities remain fixed and identify one important excluded physical effect.

Reveal worked reasoning and review criteria

Reference calculation: For 1 m coax with a=1 mm, b=5 mm, εr=2, C≈69.1 pF/m. Increasing b to 10 mm reduces C to about 48.3 pF/m.

Sensitivity logic: At fixed b, increasing inner radius a decreases ln(b/a), increasing capacitance per length while changing peak field.

Engineering qualification: For a cable with several dielectric layers, an effective permittivity cannot generally be obtained by arithmetic averaging. Integrate each radial electric-field segment and add the resulting voltage drops at fixed charge. The final report must include a quantity with correct SI units, an analytical limit and a specific reason the ideal calculation is not sufficient for hardware sign-off.

Question 10 — Electrostatic energy, force and breakdown limits

Prompt: If capacitor voltage triples at fixed C, what happens to stored energy?

Reveal worked reasoning

It increases by a factor of nine because W=0.5CV², but only while C stays effectively constant. The matching lesson and shared study guide provide additional verification questions.

Question 11 — Electrostatic energy, force and breakdown limits

Prompt: Can an energy calculation alone specify dielectric safety?

Reveal worked reasoning

No. Peak field, defects, duration, geometry, leakage and material dielectric strength all matter. The matching lesson and shared study guide provide additional verification questions.

Question 12 — Electrostatic energy, force and breakdown limits numerical and decision challenge

Prompt: Independently derive and check this scenario: A 50 pF ideal capacitor at 200 V stores 1.0 μJ. At 400 V it stores 4.0 μJ, which verifies the quadratic voltage dependence. Then choose a parameter to change, predict its effect from at fixed C, doubling voltage multiplies energy by four; at fixed voltage, doubling C doubles energy. State which quantities remain fixed and identify one important excluded physical effect.

Reveal worked reasoning and review criteria

Reference calculation: A 50 pF ideal capacitor at 200 V stores 1.0 μJ. At 400 V it stores 4.0 μJ, which verifies the quadratic voltage dependence.

Sensitivity logic: At fixed C, doubling voltage multiplies energy by four; at fixed voltage, doubling C doubles energy.

Engineering qualification: Do not use vacuum electrostatic stress blindly inside nonlinear ferroelectrics. Breakdown field depends on time scale and geometry; a simulation reporting voltage alone is not an insulation qualification. The final report must include a quantity with correct SI units, an analytical limit and a specific reason the ideal calculation is not sufficient for hardware sign-off.