Conductors, Dielectrics, Boundary Conditions, and Capacitance

Unit 5: Current flow and conduction fields

Four lessons: Conduction current density and Ohm’s law, Charge continuity and relaxation, Resistance fields and the conduction–electrostatic analogy, Field-based thermal loss and measurement.

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

Treat a uniform conductive bar as a resistor at first. Calculate R, I, J, E and dissipated power using terminal V and known σ. Verify the local power integral matches VI to rounding precision. Now divide the same bar into two serial segments of equal length but different cross-sectional areas. Because steady current must be equal in both, the narrower segment has greater J and E. Integrate voltage drop along the length, rather than attempting to assign one uniform gradient to the full device. This introduces field-based resistance as a physical consequence of boundary and conservation laws.

Extend the discussion to a rounded constriction in a two-dimensional conductive sheet. There is no simple exact one-dimensional current-density profile near the taper; show the limitations of a series-slab estimate. Explain why contacts contribute spreading resistance and how a four-terminal measurement can isolate the sensing region. The project is complete only when it states the ambient temperature used to select σ and identifies whether electrode contact resistance is included.

Guided lesson path

Step 1: Conduction current density and Ohm’s law

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/U05-L01-conduction-current-density-and-ohms-law.html). The mathematical conclusion should follow the boundary and source assumptions, not merely the choice of familiar-looking formula.

Step 2: Charge continuity and relaxation

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/U05-L02-charge-continuity-and-relaxation.html). The mathematical conclusion should follow the boundary and source assumptions, not merely the choice of familiar-looking formula.

Step 3: Resistance fields and the conduction–electrostatic analogy

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/U05-L03-resistance-fields-and-the-conduction-electrostatic-analogy.html). The mathematical conclusion should follow the boundary and source assumptions, not merely the choice of familiar-looking formula.

Step 4: Field-based thermal loss and measurement

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/U05-L04-field-based-thermal-loss-and-measurement.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-05.html) · [Unit engineering project](../projects/unit-05.html) · [Interactive mathematical model](../simulations/model-05.html).

Unit 5, lesson 1 — Conduction current density and Ohm’s law

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

In a conductor a macroscopic current is the flow of charge across a surface per unit time. The vector density J describes flow per unit area; total current is its surface integral. The pointwise Ohm law J=σE assumes a local linear isotropic material, sufficiently slow variations and a known temperature. In a resistive rod the same total current flows through each cross section at steady state, but J can be very nonuniform around bends, constrictions or heterogeneous conductors.

Governing relationships and derivation pathway

I=∫S J·dA; in homogeneous isotropic material J=σE and rho_resistivity=1/σ. For a uniform bar of length L, area A, R=L/(σA). Joule power per volume p=J·E=σ|E|².

Worked numerical illustration

A 2 m copper-like conductor with σ=5.8e7 S/m and cross-sectional area 1 mm² has R=2/(5.8e7×1e-6)=0.0345 Ω. At 2 A its ideal steady loss is I²R≈0.138 W, excluding contacts and temperature rise.

Independent numerical variant

A 1 m copper-like rod with σ=5.8e7 S/m and A=2 mm² has R≈8.62 mΩ; at 3 A its ideal loss is about 0.0776 W.

Engineering context: Current density mapping

Imagine two copper traces carrying the same total current but having different widths. A uniform cross-section estimate predicts larger average J in the narrower trace. Now add a sharp inside corner: current lines crowd toward the corner, so local J can exceed the average. To assess heating, compute an integral of J·E over the conductor rather than using peak J alone. Device-level measurements require calibrated current and temperature observations to compare with the model.

Sensitivity and design interpretation

Doubling L doubles resistance; doubling A halves resistance, for constant σ and negligible contacts.

Why the model can fail

Electrical conductivity changes with temperature and microstructure. A simulation using constant σ must not claim to predict thermal runaway; current crowding can produce local hot spots even when average J appears acceptable.

Two resolved conceptual checkpoints

Question A. If a conductor length doubles but area and conductivity stay fixed, how does R change?

Reasoned answer. It doubles under uniform-bar approximation, and at fixed current the Joule loss doubles. At fixed voltage the loss falls by half.

Question B. Why do sharp trace corners threaten a uniform-J assumption?

Reasoned answer. Current trajectories crowd near some edges, raising local J and J·E above average values.

Next steps

[Unit 5 interactive model](../simulations/model-05.html) · [Assessment](../assessments/unit-05.html) · [Project](../projects/unit-05.html) · [Shared study guidance](../STUDY-GUIDE.html).

Unit 5, lesson 2 — Charge continuity and relaxation

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

Charge cannot vanish when a wire branches. The continuity equation states that net outward conduction-current flux equals the negative rate of decrease of charge inside a region. At steady state, div J=0 even though E and J may be nonzero. The equation also reveals why conduction and capacitance phenomena are linked: during transients, conduction current may be balanced by changing electric displacement, and regions can temporarily accumulate charge.

Governing relationships and derivation pathway

Local continuity: ∂rho_free/∂t+div J=0. Integral form: dQfree/dt=-∮S J·dA. With J=σE, div E=rho/epsilon and homogeneous constant σ, charge density decays as rho(t)=rho(0)e^(-t/τ), τ=epsilon/σ.

Worked numerical illustration

For εr=3 and σ=1e-9 S/m, τ=3epsilon0/1e-9≈0.0266 s. After three relaxation times the idealized excess bulk charge has dropped to e^-3≈5.0% of its initial amount.

Independent numerical variant

If ε=4epsilon0 and σ=2e-9 S/m, τ≈17.7 ms, and after 2τ only e^-2≈13.5% of initial homogeneous bulk excess charge remains.

Engineering context: Transient charge decay

Take an initially charged bulk dielectric with known ε and small finite σ. Under a simplistic homogeneous relaxation model its charge density falls exponentially. Plot rho(t)/rho(0) over five relaxation times and verify 1/e at t=τ. Then discuss the missing mechanisms: surface conduction, traps, injection, humidity, electrode polarization. These omissions explain why a real dielectric can display much slower tails than an ideal single-exponential model.

Sensitivity and design interpretation

Doubling σ halves the homogeneous relaxation time ε/σ, when ε is held fixed.

Why the model can fail

Relaxation time depends strongly on humidity and impurity. A material called an insulator may still have measurable leakage, especially on contaminated surfaces. Do not equate div J=0 to zero current.

Two resolved conceptual checkpoints

Question A. After one relaxation time, what fraction of excess bulk charge remains in the homogeneous model?

Reasoned answer. The exponential factor is e^-1≈0.368, not zero. This is a model prediction for a specified ε/σ.

Question B. How can steady current exist with div J=0?

Reasoned answer. The divergence measures local net source or sink, not the current magnitude; equal inflow and outflow are allowed.

Next steps

[Unit 5 interactive model](../simulations/model-05.html) · [Assessment](../assessments/unit-05.html) · [Project](../projects/unit-05.html) · [Shared study guidance](../STUDY-GUIDE.html).

Unit 5, lesson 3 — Resistance fields and the conduction–electrostatic analogy

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

Many steady conductive structures can be analyzed with the same mathematics as electrostatic potential problems. If σ is uniform and there are no internal sources, V satisfies Laplace’s equation because J=-σgrad V and div J=0. Current flow lines are parallel to E in an isotropic medium and orthogonal to equipotentials. The analogy is powerful for field plotting, but boundary conditions distinguish conducting electrodes, insulated surfaces and material interfaces.

Governing relationships and derivation pathway

div(σgrad V)=0 in steady source-free conduction. For constant σ this becomes ∇²V=0. Across an interface, potential is continuous under ideal contact and the normal component of J is continuous if no charge accumulates. An insulated wall satisfies J·n=0.

Worked numerical illustration

A uniform slab of thickness 2 mm and electrode area 10 cm² has σ=1e-3 S/m. Its resistance is d/(σA)=0.002/(1e-3×0.001)=2000 Ω. With 10 V the current is 5 mA and the uniform current density is 5 A/m².

Independent numerical variant

A uniform 1 mm thick slab with area 20 cm² and σ=1e-4 S/m has R=5 kΩ. At 10 V it carries 2 mA.

Engineering context: Distributed current solver

A resistor slab clamped between large planar electrodes can be solved analytically, while tiny surface contacts create spreading-current patterns. Treat the planar case as the solver benchmark and then replace one electrode with a small circular pad. Check that integrated current across every cut plane is the same at steady state. Document how the mesh resolves electrode edges; otherwise computed peak current density may be dominated by a geometric singularity.

Sensitivity and design interpretation

Doubling conductivity at the same fixed field doubles J and local loss σE²; at fixed terminal current it reduces loss instead.

Why the model can fail

Potential-continuity at a contact is not guaranteed if appreciable contact resistance exists. A finite-element conductivity model should include electrode spreading resistance when electrode area is much smaller than the sample cross section.

Two resolved conceptual checkpoints

Question A. When does the conduction potential obey Laplace’s equation?

Reasoned answer. For steady, source-free current in a region of constant isotropic conductivity. Variable conductivity requires div(σgrad V)=0.

Question B. At a dielectric-insulated wall in a steady conduction solver, what condition applies?

Reasoned answer. The normal conduction current J·n is zero. Prescribing arbitrary fixed voltage there may create an unintended current sink.

Next steps

[Unit 5 interactive model](../simulations/model-05.html) · [Assessment](../assessments/unit-05.html) · [Project](../projects/unit-05.html) · [Shared study guidance](../STUDY-GUIDE.html).

Unit 5, lesson 4 — Field-based thermal loss and measurement

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

Power dissipation is a spatial integral, not merely a circuit-level I²R value. A field solver can predict where heat generation is concentrated using p=J·E; coupled thermal simulation is needed to translate those losses into temperatures. Conductivity measurements at low voltages should verify contact condition and instrument loading. Four-terminal sensing separates voltage leads from current-carrying contact drops when high precision is needed.

Governing relationships and derivation pathway

P=∫V J·E dV. For a uniform resistive bar, this reduces to VI=I²R=V²/R. The integral is nonnegative for passive ohmic materials σ>0. The appropriate convection and conduction heat paths determine temperature rise.

Worked numerical illustration

For a 10 Ω resistive film at 12 V, P=14.4 W. If that heat is deposited uniformly in volume 2 cm³, the average source density is 7.2e6 W/m³. The result gives a thermal source, not a finished junction-temperature estimate.

Independent numerical variant

A 25 Ω film powered from 10 V dissipates 4 W. If its volume is 1 cm³, average p is 4e6 W/m³, with no temperature predicted.

Engineering context: Heat-source consistency

Compare three independent routes to resistive power in a uniform conductor: VI, I²R and ∫σE² dV. Agreement indicates consistency of fields, boundary voltages and cross-sectional geometry. If one route differs, first check unit conversion, contact resistance or whether heat is generated outside the modeled volume. In a real heating study, couple field-derived volumetric power to thermal conduction, convection and radiation equations before predicting steady temperature.

Sensitivity and design interpretation

At fixed resistance, doubling terminal voltage multiplies Joule power by four.

Why the model can fail

A thermal camera measures surface radiance and requires emissivity interpretation. Resistive loss plots are often most sensitive to assumed conductivity; always report assumed temperature and resistance values with model results.

Two resolved conceptual checkpoints

Question A. Which is the correct local heat source: J·E or

Reasoned answer. J

Question B. Would a 14.4 W heating value determine the film temperature?

Reasoned answer. Not without geometry, thermal conductivities, ambient temperature and boundary heat-transfer mechanisms.

Next steps

[Unit 5 interactive model](../simulations/model-05.html) · [Assessment](../assessments/unit-05.html) · [Project](../projects/unit-05.html) · [Shared study guidance](../STUDY-GUIDE.html).

Unit 5 engineering project: Nonuniform current and loss review

Problem statement and intended decision

A resistive sensing film has uniform thickness but narrowing width. Build a one-dimensional width-dependent approximation that conserves total current and evaluate local J, E and Joule heating along its length. Compare integrated power with terminal VI. Then explain where the approximation fails near abrupt width transitions, requiring a two-dimensional conductivity solve. Use at least three widths and document source voltage, conductivity and thermal boundary assumptions. Finish with a measurement plan using four-wire voltage sensing to separate contact drop.

Detailed investigation plan

Treat a uniform conductive bar as a resistor at first. Calculate R, I, J, E and dissipated power using terminal V and known σ. Verify the local power integral matches VI to rounding precision. Now divide the same bar into two serial segments of equal length but different cross-sectional areas. Because steady current must be equal in both, the narrower segment has greater J and E. Integrate voltage drop along the length, rather than attempting to assign one uniform gradient to the full device. This introduces field-based resistance as a physical consequence of boundary and conservation laws.

Extend the discussion to a rounded constriction in a two-dimensional conductive sheet. There is no simple exact one-dimensional current-density profile near the taper; show the limitations of a series-slab estimate. Explain why contacts contribute spreading resistance and how a four-terminal measurement can isolate the sensing region. The project is complete only when it states the ambient temperature used to select σ and identifies whether electrode contact resistance is included.

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 5 assessment — Current flow and conduction fields

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 — Conduction current density and Ohm’s law

Prompt: If a conductor length doubles but area and conductivity stay fixed, how does R change?

Reveal worked reasoning

It doubles under uniform-bar approximation, and at fixed current the Joule loss doubles. At fixed voltage the loss falls by half. The matching lesson and shared study guide provide additional verification questions.

Question 2 — Conduction current density and Ohm’s law

Prompt: Why do sharp trace corners threaten a uniform-J assumption?

Reveal worked reasoning

Current trajectories crowd near some edges, raising local J and J·E above average values. The matching lesson and shared study guide provide additional verification questions.

Question 3 — Conduction current density and Ohm’s law numerical and decision challenge

Prompt: Independently derive and check this scenario: A 1 m copper-like rod with σ=5.8e7 S/m and A=2 mm² has R≈8.62 mΩ; at 3 A its ideal loss is about 0.0776 W. Then choose a parameter to change, predict its effect from doubling L doubles resistance; doubling A halves resistance, for constant σ and negligible contacts. State which quantities remain fixed and identify one important excluded physical effect.

Reveal worked reasoning and review criteria

Reference calculation: A 1 m copper-like rod with σ=5.8e7 S/m and A=2 mm² has R≈8.62 mΩ; at 3 A its ideal loss is about 0.0776 W.

Sensitivity logic: Doubling L doubles resistance; doubling A halves resistance, for constant σ and negligible contacts.

Engineering qualification: Electrical conductivity changes with temperature and microstructure. A simulation using constant σ must not claim to predict thermal runaway; current crowding can produce local hot spots even when average J appears acceptable. 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 — Charge continuity and relaxation

Prompt: After one relaxation time, what fraction of excess bulk charge remains in the homogeneous model?

Reveal worked reasoning

The exponential factor is e^-1≈0.368, not zero. This is a model prediction for a specified ε/σ. The matching lesson and shared study guide provide additional verification questions.

Question 5 — Charge continuity and relaxation

Prompt: How can steady current exist with div J=0?

Reveal worked reasoning

The divergence measures local net source or sink, not the current magnitude; equal inflow and outflow are allowed. The matching lesson and shared study guide provide additional verification questions.

Question 6 — Charge continuity and relaxation numerical and decision challenge

Prompt: Independently derive and check this scenario: If ε=4epsilon0 and σ=2e-9 S/m, τ≈17.7 ms, and after 2τ only e^-2≈13.5% of initial homogeneous bulk excess charge remains. Then choose a parameter to change, predict its effect from doubling σ halves the homogeneous relaxation time ε/σ, when ε is held fixed. State which quantities remain fixed and identify one important excluded physical effect.

Reveal worked reasoning and review criteria

Reference calculation: If ε=4epsilon0 and σ=2e-9 S/m, τ≈17.7 ms, and after 2τ only e^-2≈13.5% of initial homogeneous bulk excess charge remains.

Sensitivity logic: Doubling σ halves the homogeneous relaxation time ε/σ, when ε is held fixed.

Engineering qualification: Relaxation time depends strongly on humidity and impurity. A material called an insulator may still have measurable leakage, especially on contaminated surfaces. Do not equate div J=0 to zero current. 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 — Resistance fields and the conduction–electrostatic analogy

Prompt: When does the conduction potential obey Laplace’s equation?

Reveal worked reasoning

For steady, source-free current in a region of constant isotropic conductivity. Variable conductivity requires div(σgrad V)=0. The matching lesson and shared study guide provide additional verification questions.

Question 8 — Resistance fields and the conduction–electrostatic analogy

Prompt: At a dielectric-insulated wall in a steady conduction solver, what condition applies?

Reveal worked reasoning

The normal conduction current J·n is zero. Prescribing arbitrary fixed voltage there may create an unintended current sink. The matching lesson and shared study guide provide additional verification questions.

Question 9 — Resistance fields and the conduction–electrostatic analogy numerical and decision challenge

Prompt: Independently derive and check this scenario: A uniform 1 mm thick slab with area 20 cm² and σ=1e-4 S/m has R=5 kΩ. At 10 V it carries 2 mA. Then choose a parameter to change, predict its effect from doubling conductivity at the same fixed field doubles J and local loss σE²; at fixed terminal current it reduces loss instead. State which quantities remain fixed and identify one important excluded physical effect.

Reveal worked reasoning and review criteria

Reference calculation: A uniform 1 mm thick slab with area 20 cm² and σ=1e-4 S/m has R=5 kΩ. At 10 V it carries 2 mA.

Sensitivity logic: Doubling conductivity at the same fixed field doubles J and local loss σE²; at fixed terminal current it reduces loss instead.

Engineering qualification: Potential-continuity at a contact is not guaranteed if appreciable contact resistance exists. A finite-element conductivity model should include electrode spreading resistance when electrode area is much smaller than the sample cross section. 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 — Field-based thermal loss and measurement

Prompt: Which is the correct local heat source: J·E or

Reveal worked reasoning

J| by itself?|J·E in W/m³. Current density alone omits the material response and cannot determine dissipation. The matching lesson and shared study guide provide additional verification questions.

Question 11 — Field-based thermal loss and measurement

Prompt: Would a 14.4 W heating value determine the film temperature?

Reveal worked reasoning

Not without geometry, thermal conductivities, ambient temperature and boundary heat-transfer mechanisms. The matching lesson and shared study guide provide additional verification questions.

Question 12 — Field-based thermal loss and measurement numerical and decision challenge

Prompt: Independently derive and check this scenario: A 25 Ω film powered from 10 V dissipates 4 W. If its volume is 1 cm³, average p is 4e6 W/m³, with no temperature predicted. Then choose a parameter to change, predict its effect from at fixed resistance, doubling terminal voltage multiplies Joule power by four. State which quantities remain fixed and identify one important excluded physical effect.

Reveal worked reasoning and review criteria

Reference calculation: A 25 Ω film powered from 10 V dissipates 4 W. If its volume is 1 cm³, average p is 4e6 W/m³, with no temperature predicted.

Sensitivity logic: At fixed resistance, doubling terminal voltage multiplies Joule power by four.

Engineering qualification: A thermal camera measures surface radiance and requires emissivity interpretation. Resistive loss plots are often most sensitive to assumed conductivity; always report assumed temperature and resistance values with model results. 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.