Polarisation, dielectric response and breakdown

A dielectric responds without steady metallic conduction

An electric field can displace charges or reorient polar units, creating polarisation. Different mechanisms respond on different timescales. Permittivity describes the electrical response associated with polarisation; dielectric loss describes energy dissipated during a changing field. Both depend on frequency, temperature and material state. Real dielectrics can also carry leakage current, so polarisation does not establish perfect insulation.

Place a neutral slab between two conducting plates and charge the upper plate positively and the lower negatively. The electric field points downwards between them. In a simple induced dipole, the negative electronic charge shifts slightly upwards relative to positive charge. The two charge centres separate while the unit can remain neutral. Its dipole moment points from negative towards positive, downwards here. This is not an electron travelling continuously from one electrode to the other.

Other structures have permanent molecular dipoles whose orientations can change in a field; thermal motion and surrounding constraints oppose perfect alignment. Relative positive/negative ion displacement in an ionic structure supplies another mechanism. Electronic distortion, ionic displacement and orientational rearrangement need not have the same response time or strength. A picture of rigid identical units is a representative cartoon, not a claim that every dielectric contains freely rotating molecules.

Polarisation P is net electric dipole moment per unit volume, in C/m². In the uniform, plate-normal example, it produces opposite bound surface charges: negative next to the upper positive electrode, positive next to the lower negative electrode. Each unit remains neutral overall; interior opposite contributions cancel in this ideal uniform account. Nonuniform polarisation can also give bound charge within the material.

Free electrode charge and bound polarisation charge are different accounts. The external circuit moves free charge onto the metal plates while the dielectric responds internally. A changing polarisation can therefore contribute to external charging current without supplying steady electronic conduction through the slab. Real mobile charges can additionally leak through it. Charge carriers and electronic energy bands prepares that carrier distinction; Conductors, insulators and semiconductors develops steady transport and leakage paths.

Specify the plate geometry and the controlled quantity

Use a homogeneous dielectric that completely fills wide, parallel, ideal conducting plates. Their common facing area is A and separation is d. Ignore fringing at the edges, free charge inside the dielectric, contact drops and structural changes. A positive voltage magnitude V across the gap then gives a uniform field magnitude E = V/d. A volt is a joule per coulomb; E has units V/m. A single voltage is not a field until its geometry is specified.

For the stipulated reversible linear response, write P = ε₀(ε_r − 1)E. Here ε₀ is vacuum permittivity, approximately 8.85 × 10⁻¹² F/m in our rounded calculations, and ε_r is dimensionless relative permittivity under these conditions. The material permittivity is ε = ε₀ε_r. Direction, temperature, frequency and field range belong with that response; ε_r is not an unrestricted number valid at every condition.

The free electrode charge per area must support both the vacuum-field and polarisation terms: Q/A = ε₀E + P = εE in this ideal geometry. Substituting E = V/d gives Q = (εA/d)V. Capacitance C = Q/V is therefore εA/d, in farads; one farad is one coulomb per volt. Q names the magnitude on either plate, not their algebraic sum, which is zero for the equal and opposite charges.

At fixed V and fixed d, inserting a linear dielectric changes C and the required free Q, while E = V/d stays the same. If an isolated capacitor instead keeps Q fixed, E = Q/(εA) decreases when ε increases, and its voltage changes. “The dielectric reduces the field” therefore needs the charge or voltage boundary condition. These are different experiments, not contradictory results.

Calculate charge and energy before calling it useful storage

All numbers in this plate example are invented teaching inputs, not specifications for a named dielectric. Let A = 100 mm² = 1.00 × 10⁻⁴ m², d = 100 μm = 1.00 × 10⁻⁴ m, ε_r = 4.00 and V = 100 V. The uniform, reversible linear assumptions are independently stipulated for this case; no breakdown or loss limit is established by them.

E = 100/(1.00 × 10⁻⁴) = 1.00 × 10⁶ V/m, or 1.00 MV/m. C = (8.85 × 10⁻¹² × 4.00)(1.00 × 10⁻⁴)/(1.00 × 10⁻⁴) = 35.4 pF, where pF means 10⁻¹² F. Thus Q = CV = 3.54 nC. The polarisation is P = 26.55 μC/m², giving bound charge magnitude AP = 2.655 nC on each dielectric face. The free-charge account is Aε₀E + AP = 0.885 + 2.655 = 3.54 nC.

During slow reversible charging of this linear capacitor from zero, its instantaneous voltage is q/C. An added charge dq requires work V dq. Summing those increments gives U = ∫₀^Q (q/C)dq = Q²/(2C) = ½CV² = ½QV. The factor one-half arises because voltage grows from zero; using Q times the final V would double this stored energy. Under the ideal reversible assumptions the same energy can be returned on discharge. Energy dissipated in a real charging circuit or dielectric is a separate contribution.

The calculated U is 1.77 × 10⁻⁷ J = 0.177 μJ. Divide by the dielectric volume Ad = 1.00 × 10⁻⁸ m³ to obtain energy density u = 17.7 J/m³. Algebraically u = ½εE² for this linear loss-free case. The total U and density u are different quantities. Do not apply this formula to a nonlinear, hysteretic or frequency-dispersive material as a complete recoverable-energy calculation.

Halve d while keeping A, ε_r and V unchanged. C doubles to 70.8 pF, E doubles to 2.00 MV/m and U doubles to 0.354 μJ. The volume halves, so u becomes four times larger, 70.8 J/m³. This does not establish that the thinner dielectric tolerates the doubled field. If field rather than voltage is kept at 1.00 MV/m, the new V must be 50.0 V; then U is 0.0885 μJ and u stays 17.7 J/m³. State which comparison is intended.

A finite response time produces lag

A process can contribute strongly in a slow measurement and weakly when the field reverses before it can respond. Relaxation means an approach toward a changed equilibrium response, rather than a resonant oscillation driven by inertia. Real materials can have several or broadly distributed times, and temperature or structure can change them. A single time is a limited illustrative model, not a consequence of the material name.

For a separate invented small-field specimen, model its electrode charge as q(t) = C_f V(t) + q_d(t). C_f is a fast response on the displayed time range, including the vacuum contribution; q_d is the charge associated with one delayed polarisation. Assume τ dq_d/dt + q_d = C_d V(t), with positive C_d and response time τ. The derivative dq_d/dt means charge change per time. If voltage is held fixed, the equation says that delayed charge approaches C_d V rather than appearing instantly.

For a voltage step from zero, q_d(t) = C_d V[1 − exp(−t/τ)]. At t = τ it reaches about 63.2% of its final delayed value. At long times a perfectly insulating specimen has no continuing charging current, although its separated plate and bound charges remain. This relaxation time is not automatically a mobile-charge leakage time, a breakdown lifetime or a switching specification.

Now drive V(t) = V₀ cos(ωt) after the initial transient has died away. V₀ is the peak voltage, not its root-mean-square value. Frequency f counts cycles per second, in hertz; angular frequency is ω = 2πf. One cycle lasts 1/f. Define x = ωτ, the dimensionless comparison between the driving and response clocks.

Try q_d = X cos(ωt) + Y sin(ωt). Substitution into the rate equation gives X + xY = C_dV₀ and Y − xX = 0, so X = C_dV₀/(1 + x²) and Y = xX. The sin term is a quarter-cycle-shifted part; together the terms give a delayed maximum. This coefficient matching uses ordinary real functions and does not require complex numbers.

Total charge becomes q = V₀[C′ cos(ωt) + C″ sin(ωt)], where C′ = C_f + C_d/(1 + x²) and C″ = C_d x/(1 + x²). The primes label the two response coefficients in farads; they are not time derivatives. C′ is the charge component following the voltage waveform, while C″ gives the lagging component. Neither is a claim that there are two separate loss-free physical capacitors. At x ≪ 1 the delayed process nearly follows; at x ≫ 1 its amplitude becomes small and C′ approaches C_f within this model’s range.

Choose C_f = 20.0 pF, C_d = 80.0 pF and τ = 1.00 ms. These are supplied model parameters for this separate specimen, not fitted data for the earlier ε_r = 4 plate. At x = 1, f ≈ 159.155 Hz, C′ = 60.0 pF and C″ = 40.0 pF. With V₀ = 10.0 V, q = 600 cos(ωt) + 400 sin(ωt) pC. At zero voltage, the two passages through the cycle have q = +400 and −400 pC. Charge is not a unique function of the instantaneous voltage: its history matters.

Lost energy per cycle is not power or useful stored energy

With leakage temporarily excluded, external current is I = dq/dt. For the periodic charge above, I = ωV₀[C″ cos(ωt) − C′ sin(ωt)]. Power entering the electrical response is V I. The average of cos² over one cycle is one-half and the average of cos times sin is zero. Therefore average dissipated power is P_loss = ½V₀²ωC″. A passive cycle returns to the same state, so positive net work over it is dissipated rather than accumulated indefinitely as stored energy.

Energy lost per cycle is W_loss = P_loss/f = πV₀²C″. Equivalently it is the signed integral ∮V dq around the charge–voltage loop; the symbol ∮ means one complete closed cycle. The finite loop area is a work account, not a new static capacitance. An ideal instantaneous loss-free q = CV line retraces itself, enclosing zero area. The second figure shows both the loop and the response coefficients of this limited model.

The assumed rate equation also gives a direct energy check. Multiplying it by (dq_d/dt)/C_d gives V(dq_d/dt) = d[q_d²/(2C_d)]/dt + (τ/C_d)(dq_d/dt)². Adding the fast part stores C_fV²/2. Thus input power changes a stored state energy and supplies a nonnegative dissipative term. The latter remains positive over a cycle even when the stored energy returns to its starting value. This is a property of the stated model, not an independently measured microscopic heat mechanism.

At x = 1 and peak 10.0 V, I = 0.400 cos(ωt) − 0.600 sin(ωt) μA. Its component in phase with V gives P_loss = 2.00 μW. Multiplying by the period 6.28319 ms gives W_loss ≈ 12.5664 nJ. This dissipated work cannot be counted as useful stored charge energy. A static ½CV² calculation alone cannot describe the recovery and heating of this loop.

The table keeps the same peak voltage and model parameters while changing only frequency. At x = 0.100 and 10.0, C″ and energy lost per cycle are equal, but the latter repeats cycles 100 times more often and dissipates 100 times the average power. C″ peaks at x = 1 in this one-time model; average power instead rises toward a plateau as x grows. A declining loss coefficient does not alone establish declining heating power.

These are exact results of a small-signal single-time model in its stipulated range. Real dielectric responses may have multiple times, interfacial charge, resonances, nonlinear polarisation, hysteresis or temperature-dependent losses. Heat removal is absent here, so even the calculated power gives no temperature rise or thermal stability by itself.

A charge-storage measurement is not every conduction path

A real slab can also have a steady leakage path. In a deliberately simple independent parallel model, add current I_leak = GV, where G is conductance in siemens. Total lead current is then dq/dt + GV, rather than dq/dt alone. Mobile charge crossing the specimen and changing bound charge can both contribute to an alternating-current reading, while representing different mechanisms.

For a separately supplied G = 1.00 nS = 1.00 × 10⁻⁹ S, a constant 10.0 V gives continuing leakage 10.0 nA after the polarisation transient has ended. Its power is 100 nW. For a sinusoid with peak 10.0 V, its average power is ½GV₀² = 50.0 nW = 0.0500 μW. At the model’s x = 1 this raises total dissipated power from 2.00 to 2.05 μW if all other parameters are kept fixed.

The same instrument may report an effective loss coefficient that includes both delayed polarisation and conduction. Electrode response, contacts and stray paths can add more. A large low-frequency apparent permittivity or loss therefore needs geometry, frequency and path evidence before being assigned to one microscopic mechanism. No single capacitance reading identifies them all.

Breakdown is a limiting event

Strong fields can trigger irreversible damage or a conducting path. Electrodes, defects, thickness, waveform and exposure time influence breakdown evidence. A high permittivity does not by itself establish high usable energy storage: losses and attainable field also matter.

Breakdown is an observed loss of the intended insulating function under the specified electrical history. Charge injection or multiplication, local damage, defects and heating feedback can contribute in different materials. A local sharp feature or void can make the field different from V/d. A film can also fail along a surface or through an unintended path; identify the event and region before naming it bulk breakdown.

A separate fictional test record reports one event at 2000 V across a 100 μm film and another at 1000 V across a 50.0 μm film. Both nominal fields are 20.0 MV/m. The second voltage is lower, but the nominal breakdown fields in these two records are equal. Nor do equal V/d values establish the same flaw distribution, field profile or breakdown probability. One event supplies neither an operating field nor a population guarantee.

Area, thickness, electrodes, waveform, exposure time, temperature, preparation and event-detection convention belong with breakdown evidence. Some metallized polymer-film capacitors can isolate a local failed spot when electrode material is removed around it. That is a particular construction and a local damage event; it does not imply that every dielectric repairs itself or that repeated events leave capacitance unchanged.

Finally compare an explicitly stipulated reversible, loss-free screening model: material A has ε_r = 4 at evaluation field 2.00 MV/m, while B has ε_r = 8 at 1.00 MV/m. Their u = ½ε₀ε_rE² values are 70.8 and 35.4 J/m³. B has twice the relative permittivity but half the model energy density because field enters squared. The evaluation fields are supplied comparison inputs, not limits deduced from the event records. Real usable storage additionally requires recoverability, loss, temperature and durability evidence under the intended waveform.

Worked examples

Try the reasoning

1. Is dielectric loss useful stored energy?

   Show the answer and reasoning

   No.

   It represents dissipative response. A closed cycle that returns to the same state can still consume positive net work, which is dissipated rather than accumulated as useful storage.

2. Why state thickness in breakdown tests?

   Show the answer and reasoning

   Geometry and defect sampling affect the field assessment.

   A single voltage does not specify the same field. The nominal E = V/d also need not equal a local peak near a void, edge or defect; one breakdown event is not an operating guarantee.

3. A new ideal fully filled linear plate has A = 50.0 mm², d = 200 μm, ε_r = 6.00 and V = 120 V. Use the same rounded ε₀. Calculate C, Q, E and U. Which assumptions let those be one uniform field and reversible storage?

   Show the answer and reasoning

   C = 13.275 pF, Q = 1.593 nC, E = 0.600 MV/m and U = 95.58 nJ.

   A/d = (5.00 × 10⁻⁵ m²)/(2.00 × 10⁻⁴ m) = 0.250 m. Multiply by ε₀ε_r for C, then use CV, V/d and ½CV². The ideal broad-plate, homogeneous, fully filled, reversible linear and loss-free assumptions are needed; these are not inferred from a material label.

4. An isolated vacuum plate capacitor has C₀ = 8.85 pF and was charged to 100 V before disconnection. Insert an ideal dielectric with ε_r = 4 without changing geometry or free charge. Find the new C and V. Would keeping 100 V describe the same experiment?

   Show the answer and reasoning

   C = 35.4 pF and V = 25.0 V. Keeping 100 V would require a different free charge and a connected voltage source.

   The retained Q is C₀ × 100 V = 0.885 nC. Divide by 4C₀ for the new V. At fixed geometry the field also falls to one-quarter. This is distinct from the earlier fixed-voltage comparison; any mechanical insertion work and full external energy balance are not solved here.

5. Compare x = 1 with x = 10 in the model table, at the same peak voltage. C″ falls from 40.0 to 7.92079 pF. Does average dissipated power fall by the same factor?

   Show the answer and reasoning

   No. Loss per cycle becomes about 0.19802 of its old value, but frequency rises tenfold, so average power becomes about 1.98020 times as large: 3.96040 μW instead of 2.00 μW.

   Use P = fW rather than treating joules per cycle as watts. Within this single-time model the two factors change together. No temperature rise follows without thermal and boundary information.

6. In a new otherwise unchanged single-time model, τ becomes 2.00 ms. At which f does C″ reach its maximum? Can the same statement locate every real material’s loss maximum?

   Show the answer and reasoning

   The model maximum is at x = ωτ = 1, so f ≈ 79.5775 Hz, half the previous value. It is not a universal real-material rule.

   C″ = C_d x/(1 + x²) is largest at x = 1; for positive x, 1 + x² ≥ 2x. Different distributions, resonances, conduction and evolving state can change a measured loss spectrum. The loss coefficient peak is also not the model power maximum.

7. The model specimen is held at constant 10.0 V for many τ, with the separate supplied leakage G = 1.00 nS. Is zero continuing polarisation rate proof of zero lead current? Find the remaining current and power.

   Show the answer and reasoning

   No. I_leak = 10.0 nA and power is 100 nW even after the polarisation charge stops changing.

   dq/dt tends to zero in the steady polarisation account, while GV persists in the stipulated independent path. This does not identify the microscopic leakage carrier or prove that a real conductance stays constant with voltage or temperature.

8. A third fictional film has one reported event at 1500 V with thickness 75.0 μm. Compare nominal E with the 2000 V, 100 μm record. Does equal E settle which material is more reliable?

   Show the answer and reasoning

   Both nominal values are 20.0 MV/m. Those two event fields do not settle a reliability ranking or an operating field.

   Divide each voltage by its own gap in metres. Active area, defects, electrodes, local field, waveform/time and event criterion can differ. Population and durability evidence cannot be reconstructed from two isolated normalised numbers.

9. In the periodic x = 1 example, V is zero twice per cycle, but q is +400 pC once and −400 pC later. A learner divides q by V and calls it a new static capacitance. What is wrong?

   Show the answer and reasoning

   The quotient is undefined at V = 0, and the two different charge states show that instantaneous q is not one loss-free function CV of the present V.

   The delayed state and history matter. The coefficients C′ and C″ are defined from the complete periodic waveform, not by taking one singular point ratio. The q–V loop encodes cycle work; it is not two contradictory static material constants.

What to carry forward

• Report frequency, field and loss together.
• Distinguish polarisation from breakdown durability.
• Bound polarisation, external charging current and leakage are different charge processes.
• Field requires geometry; fixed voltage, fixed free charge and fixed field are different comparisons.
• The formula ½CV² is reversible linear storage under its conditions, not a universal recoverable-energy rule.
• A frequency-dependent loss coefficient, energy per cycle and average power are distinct quantities.
• A conditional breakdown observation is not an automatically transferable material voltage or operating guarantee.