University Physics II · Capacitance and Dielectrics · 5.8
Dielectric Strength & Capacitor Design
Every capacitor has a field its insulator cannot survive. That one number fixes the voltage rating, caps the energy stored per unit volume, and forces the thickness trade the rest of the design is built around.
Build the model
Connect the measurement to the mechanism.
Geometry sets a capacitor's capacitance; the dielectric sets its limits. Every insulator has a dielectric strength Eₘₐₓ, the field above which it stops insulating, and in a parallel-plate design that converts straight into a voltage ceiling, Vₘₐₓ = Eₘₐₓ d. Thickness therefore cuts both ways: halve d and the capacitance doubles while the voltage rating halves.
Combine the two and the trade collapses into one statement, Uₘₐₓ = ½κε₀Eₘₐₓ² × (volume of dielectric). The most energy you can store is fixed by the material and by how much of it you use, not by how you split that volume between area and thickness. Everything else in capacitor design subtracts from that ideal: leakage that drains the charge, tolerances that spread the capacitance, defects that break down early, and the safety margin you keep to stay clear of the cliff.
- Simple definition
- Dielectric strength is the largest electric field an insulator can hold without conducting. Multiplied by the plate separation it gives the breakdown voltage, and it caps the energy density any design using that material can reach.
- Example
- Air breaks down near 3 × 10⁶ V m⁻¹, so a 1.0 mm air gap arcs at about 3 kV. A 6.0 μm polypropylene film holds roughly 6 × 10⁸ V m⁻¹, so it survives about 3.6 kV across a film one hundred and sixty-seven times thinner.
The insulator fixes the field; the thickness converts it into a voltage.
Uniform field only; Eₘₐₓ in V m⁻¹, usually quoted in kV mm⁻¹
Thinner film buys capacitance and hands back exactly that much voltage.
d cancels; the largest free surface charge the plates can carry
Total energy is uₘₐₓ × A d, so only the dielectric volume counts.
J m⁻³; running at Eₘₐₓ/n cuts the usable energy by a factor n²
Datasheets quote this product directly, in ohm-farads.
Geometry cancels; ρ is the dielectric resistivity in Ω m
Thickness control dominates, and it sets the voltage rating too.
Independent errors add in quadrature: ±3 %, ±2 %, ±1 % → ±3.7 %
A gas pocket feels κ times the field the solid around it feels.
Flat void with faces normal to E, where the normal part of D is continuous
Dielectric strength is where insulating stops
An insulator holds its electrons bound, but only up to a point. Raise the field and an electron freed by chance gains energy eEλ between collisions; once that exceeds the ionisation energy of the molecules around it, every collision liberates more carriers and the current avalanches. As an order of magnitude, 10 eV over a 1 nm mean free path needs about 10¹⁰ V m⁻¹, and measured strengths sit well below that because defects, trapped gas and local heating trigger the cascade first. Typical figures: dry air 3 × 10⁶ V m⁻¹, paper about 1.6 × 10⁷, Pyrex 1.4 × 10⁷, PTFE 6 × 10⁷, mica above 10⁸, and thin polypropylene film up to 6 × 10⁸. Treat these as design guides, not constants. Dielectric strength rises in thinner films, falls as temperature rises and as the stress is held for longer, and differs between DC, AC and impulse testing. Breakdown in a solid is also permanent: a puncture leaves a conducting channel, so the part does not recover the way an air gap does.
Thickness trades capacitance against voltage rating
For parallel plates the two design equations pull in opposite directions: C = κε₀A/d rewards a thin dielectric and Vₘₐₓ = Eₘₐₓ d punishes it. Their product removes d entirely, Qₘₐₓ = C Vₘₐₓ = κε₀ A Eₘₐₓ, so the largest charge a design can hold is set by plate area and material, never by how thin the film is. Take metallised polypropylene, κ = 2.2 and Eₘₐₓ = 6.0 × 10⁸ V m⁻¹, rolled to A = 0.20 m² at d = 6.0 μm. Then C = 2.2 × 8.854 × 10⁻¹² × 0.20 / (6.0 × 10⁻⁶) = 0.65 μF, and the film punctures at V = 6.0 × 10⁸ × 6.0 × 10⁻⁶ = 3.6 kV. Halve the thickness to 3.0 μm at the same area and C doubles to 1.30 μF while the breakdown voltage halves to 1.8 kV. Neither is the better design; they are the same physics divided differently.
Energy per unit volume is the real ceiling
Put the two limits together: Uₘₐₓ = ½ C Vₘₐₓ² = ½ (κε₀A/d)(Eₘₐₓ d)² = ½ κε₀Eₘₐₓ² A d. Thickness cancels again, leaving energy proportional to the dielectric volume with uₘₐₓ = ½κε₀Eₘₐₓ² as the constant of proportionality. The polypropylene above gives uₘₐₓ = ½ × 2.2 × 8.854 × 10⁻¹² × (6.0 × 10⁸)² = 3.5 × 10⁶ J m⁻³, and its 1.2 cm³ of film holds 4.2 J at the instant it fails. That one expression settles which material to use. A Class-2 ceramic with κ ≈ 3000 has 1400 times the permittivity, but its usable field is nearer 1.5 × 10⁷ V m⁻¹, giving uₘₐₓ = 3.0 × 10⁶ J m⁻³ — slightly worse. Permittivity is not the figure of merit; κEₘₐₓ² is, and the square is what decides it.
Leakage, and a time constant that ignores geometry
No dielectric insulates perfectly. Model the loss as a resistance Rₗₑₐₖ = ρd/A in parallel with the capacitor, ρ being the bulk resistivity, and the stored charge decays as exp(−t/τ) with τ = Rₗₑₐₖ C = (ρd/A)(κε₀A/d) = ρκε₀. Area and thickness cancel: the self-discharge time constant is a property of the material alone, which is why datasheets specify insulation resistance as an RC product in ohm-farads rather than in ohms. Polypropylene at ρ ≈ 10¹⁶ Ω m and κ = 2.2 gives τ ≈ 1.9 × 10⁵ s, about 54 hours, so a charged film capacitor is still dangerous a day later. A lossier dielectric with ρ = 10¹² Ω m and κ = 3 gives τ ≈ 27 s, useless for holding a sampled voltage. Real parts fall short of the bulk figure, because surface films, humidity and terminations shunt the dielectric, and resistivity drops steeply as temperature rises.
Capacitance averages; breakdown finds the worst point
Capacitance integrates over the whole plate area, so a few thin patches barely move it and the tolerance follows the mean. Write ΔC/C = Δκ/κ + ΔA/A − Δd/d; with independent spreads of ±2 % in κ, ±1 % in A and ±3 % in d, quadrature gives about ±3.7 %, which is why ±5 % film parts are routine while Class-2 ceramics are graded −20/+80 %. Breakdown obeys the opposite statistic. It occurs at the weakest point in the roll, so it follows the extreme of the thickness and defect distributions, not their mean — and doubling the area, which reliably doubles C, lowers the expected breakdown voltage, because there is twice as much film in which to find a flaw. Trapped gas is the usual flaw: across a flat void the normal component of D is continuous, so Evoid = κE, and a pocket inside polypropylene sees 2.2 times the field the polymer feels while air withstands far less. Micron-scale voids survive better than that suggests, since Paschen's law raises the breakdown of very short gaps, but partial discharge still erodes film capacitors over years.
What a real capacitor also does
All of this is why ratings are derated in use: run at Eₘₐₓ/2 and the usable energy falls by four, so the 0.65 μF film above drops from 4.2 J to 1.05 J at 1.8 kV. Four further effects matter. Series resistance dissipates Iᵣₘₛ²·ESR, and that heat raises leakage and lowers dielectric strength, a loop that can run away. Class-2 ceramics have a strong voltage coefficient, and an X7R part can lose more than half its capacitance at rated voltage, so the printed value is a small-signal value. Dielectric absorption returns charge trapped in slow polarisation after a short, so a discharged high-voltage capacitor can recover hundreds of volts on its own. And metallised film self-heals: a puncture vaporises the thin electrode around it and isolates the fault, so the part loses a little capacitance instead of shorting, whereas foil electrodes and electrolytics fail outright, the latter also demanding correct polarity.
Change one variable at a time
Make the relationship visible.
Drag the thickness: with the 1.2 cm³ of film fixed the slab stretches or squats over the same drawn area, and the plate area falling as 1/d takes the capacitance down as 1/d² while the rating rises only as d, so ½CV² — the energy bar — never moves. Then drag the dielectric strength: it is the only thing that lifts that bar, and it lifts it as the square.
PLATE AREA A0.200 m²
CAPACITANCE C0.65 μF
BREAKDOWN VOLTAGE3600 V
ENERGY AT BREAKDOWN4.21 J
Live interpretationPLATE AREA A: 0.200 m². CAPACITANCE C: 0.65 μF. BREAKDOWN VOLTAGE: 3600 V. ENERGY AT BREAKDOWN: 4.21 J
Catch the common trap
Explain before calculating.
Two capacitors use the same dielectric (κ = 2.2, dielectric strength 6.0 × 10⁸ V m⁻¹) and the same volume of it. X has A = 0.20 m² with d = 6.0 μm; Y has A = 0.40 m² with d = 3.0 μm. Each is charged to its own breakdown voltage. Which stores more energy?
Choose an answer to test the model.
Practice & worked examples
Reason from the model, then test the result.
EasyA capacitor is wound from 25 μm polyester film, κ = 3.3, dielectric strength 1.6 × 10⁸ V m⁻¹, with a plate area of 0.045 m². Find its capacitance, its breakdown voltage, and the energy it holds at that voltage.
- C = κε₀A/d = 3.3 × 8.854 × 10⁻¹² F m⁻¹ × 0.045 m² ÷ (25 × 10⁻⁶ m) = 5.259 × 10⁻⁸ F = 52.6 nF.
- The insulator fixes the field, the thickness converts it into a voltage: Vₘₐₓ = Eₘₐₓ d = 1.6 × 10⁸ V m⁻¹ × 25 × 10⁻⁶ m = 4.0 × 10³ V.
- U = ½CVₘₐₓ² = ½ × 5.259 × 10⁻⁸ F × (4.0 × 10³ V)² = ½ × 0.8415 J = 0.421 J.
- Cross-check the same number from the material alone: uₘₐₓ = ½κε₀Eₘₐₓ² = ½ × 3.3 × 8.854 × 10⁻¹² × (1.6 × 10⁸)² = 3.74 × 10⁵ J m⁻³, and the film volume is Ad = 0.045 m² × 25 × 10⁻⁶ m = 1.125 × 10⁻⁶ m³, so U = 3.74 × 10⁵ × 1.125 × 10⁻⁶ = 0.421 J.
AnswerC = 52.6 nF, Vₘₐₓ = 4.0 kV, U = 0.42 J — and ½κε₀Eₘₐₓ² × Ad returns the same 0.42 J without touching C or V.
MediumThe same film is rerolled to 12.5 μm at the same 0.045 m² area, and the finished part is then operated at half its rated voltage. Find the new capacitance, the new rating, the energy at rating, and the energy in service.
- C = κε₀A/d with d halved: C = 2 × 52.6 nF = 105 nF.
- Vₘₐₓ = Eₘₐₓ d = 1.6 × 10⁸ V m⁻¹ × 12.5 × 10⁻⁶ m = 2.0 × 10³ V — halved, exactly as the capacitance doubled.
- At rating: U = ½ × 1.052 × 10⁻⁷ F × (2.0 × 10³ V)² = ½ × 0.4207 J = 0.210 J. The same answer from the volume: the film is now Ad = 5.625 × 10⁻⁷ m³, half of before, and 3.74 × 10⁵ J m⁻³ × 5.625 × 10⁻⁷ m³ = 0.210 J.
- Derated to 1.0 kV: U = ½ × 1.052 × 10⁻⁷ F × (1.0 × 10³ V)² = 0.0526 J, one quarter of the rating energy because U goes as V².
AnswerC = 105 nF, Vₘₐₓ = 2.0 kV, 0.210 J at rating and 0.0526 J at 1.0 kV. Thinning the film at fixed area halved the dielectric volume, so it halved the ceiling; the 2:1 derating then removed three quarters of what was left.
HardA polypropylene film capacitor (κ = 2.2, bulk resistivity ρ = 1.0 × 10¹⁶ Ω m, dielectric strength 6.0 × 10⁸ V m⁻¹) uses 8.0 μm film, is charged to 2.40 kV and left open-circuit. (a) How long until the terminals read 1.00 kV? (b) A flat air-filled void 0.50 μm thick lies across the film with its faces perpendicular to the field. What field does the gas inside it feel at 2.40 kV?
- Self-discharge: τ = Rₗₑₐₖ C = (ρd/A)(κε₀A/d) = ρκε₀ = 1.0 × 10¹⁶ Ω m × 2.2 × 8.854 × 10⁻¹² F m⁻¹ = 1.95 × 10⁵ s. Area and thickness cancel, so neither C nor d is needed.
- V = V₀e(−t/τ) → t = τ ln(V₀/V) = 1.95 × 10⁵ s × ln(2.40) = 1.95 × 10⁵ × 0.8755 = 1.71 × 10⁵ s = 47 h.
- For the void, the normal component of D is continuous across its faces: ε₀Evoid = κε₀Esolid, so Evoid = 2.2 Esolid.
- The 2.40 kV divides between the void and the 7.50 μm of solid left: 2400 V = Esolid[2.2 × 0.50 × 10⁻⁶ m + 7.50 × 10⁻⁶ m] = Esolid × 8.60 × 10⁻⁶ m, so Esolid = 2.79 × 10⁸ V m⁻¹ and Evoid = 2.2 × 2.79 × 10⁸ = 6.14 × 10⁸ V m⁻¹.
- Compare each to its own limit: the polymer sits at 2.79 × 10⁸ ÷ 6.0 × 10⁸ = 0.47 of Eₘₐₓ and is comfortable, while the gas is at 205 times the 3.0 × 10⁶ V m⁻¹ bulk figure for air. Paschen's law lifts the breakdown of a 0.50 μm gap far above that bulk figure, so the void need not arc outright — but it, not the film, is the stressed part of the design.
Answerτ = 1.95 × 10⁵ s, so about 47 hours to fall from 2.40 kV to 1.00 kV; the trapped gas sees 6.1 × 10⁸ V m⁻¹ while the polymer around it sees only 2.8 × 10⁸ V m⁻¹.