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Physics II course sequence · Interactive lesson

Electrostatic Applications & Safety

Every static hazard in a workshop comes down to two numbers: the field at which the surrounding air gives way, and the energy the spark then delivers. Once you can compute both, lightning rods, ionisers, wrist straps and ore separators stop being folklore and become arithmetic you can check against a meter.

UPII-01.08 · Lesson 8 of 8 in this unit. Position is not a completion score.
Starting model: Electrostatic Applications & SafetyA polished dome of radius 0.20 m carries a small tip of radius 2.0 mm at the same potential, 100 kV. The bars give each surface field on a logarithmic scale: 0.50 MV m⁻¹ at the dome against 50.0 MV m⁻¹ at the tip. A bar past the dashed 3 MV m⁻¹ line is a surface already breaking down the air.conductor at V = 100 kVdome R = 0.20 m, tip r = 2.0 mmtip field = 100 × dome fieldtip coronas from 6.0 kVdometip10⁴ V m⁻¹10⁶10⁸dashed line = 3 MV m⁻¹ in air
Start with a picture. Change the model in the lab below.
1

Build the model

Air fails at a field, and the spark carries an energy.

The key idea

Electrostatic safety is the practice of keeping the largest field anywhere in an arrangement below the value at which the surrounding gas breaks down, and keeping the energy any unavoidable spark releases below what the surroundings can tolerate.

Example. A polished 0.25 m dome reaches 3.0 × 10⁶ V m⁻¹ at its surface when it holds 2.1 × 10⁻⁵ C, at 7.5 × 10⁵ V. Leave a 1.0 mm burr on it and the same air fails at 3.0 × 10³ V, two hundred and fifty times sooner.

The complete physical model and assumptions

Charge on a conductor is harmless until the air around it stops insulating. Dry air near 1 atm holds off about 3 × 10⁶ V m⁻¹; above that a stray electron gains enough energy between collisions to ionise a molecule, the ionisation avalanches, and a conducting channel opens. So the working model behind every electrostatic application is a threshold in field — not in charge, and not in voltage: find the largest surface field the arrangement produces, compare it with Ebreak, and the design question is answered. Geometry does the rest of the work, because a sphere's surface field is kQ/R² while joined conductors all sit at one potential, so the smallest radius carries the largest field. That is why a lightning rod, an ioniser needle and a photocopier corona wire are deliberately sharp, and why every high-voltage dome, terminal and nut is deliberately round. What the threshold cannot tell you is the damage, which is set instead by the energy released, ½CV², and by what that energy lands on. The model also has edges: 3 × 10⁶ V m⁻¹ describes clean dry air in centimetre gaps at sea level, rises steeply for gaps below a millimetre, falls with altitude, and shifts with humidity, dust and surface films. Treat it as a design threshold carrying margin, never as a guarantee.

1Breakdown is a threshold in field, not in charge

Field means force per unit charge, F/q, measured in V m⁻¹ or equivalently N C⁻¹. Air conducts when that field is strong enough that a free electron, accelerated between two collisions, arrives with enough energy to knock an electron off the next molecule; each collision then makes two carriers, and the avalanche builds a channel in under a microsecond. For clean dry air in centimetre gaps the onset is about 3.0 × 10⁶ V m⁻¹. Put a charge Q on an isolated sphere of radius R and the field is largest right at the metal, E = kQ/R², so the sphere can hold at most Qₘₐₓ = Ebreak R²/k. For R = 0.25 m that is 2.1 × 10⁻⁵ C, reached at V = Ebreak R = 7.5 × 10⁵ V. Double the radius and the charge limit quadruples while the voltage limit only doubles — which is the whole reason a demonstration generator carries a large smooth dome rather than a small one.

2Why the sharpest feature discharges first

Conductors joined by metal share one potential. For a sphere V = kQ/R, so at a common V the charge divides as Q ∝ R, and the surface field becomes E = kQ/R² = V/R. Halve a radius and you double its field at the same potential. Take a 0.20 m dome carrying a 1.0 mm burr on a bolt head: the burr's field is R/r = 200 times the dome's, so air fails at the burr when the whole assembly reaches only V = Ebreak r = 3.0 × 10³ V, while the bare dome would have held 6.0 × 10⁵ V. The burr carries just r/R = 0.5% of the charge and destroys 99.5% of the working range. Sometimes that is exactly the point: a lightning rod, an ioniser needle and a photocopier corona wire are sharpened on purpose so that they ionise air at a modest potential. Everywhere else it is a defect, which is why high-voltage hardware is deburred, rounded and polished.

3Lightning protection is a path, not a shield

A lightning rod neither repels the strike nor quietly drains the cloud; the corona current it emits is microamps against a stroke of tens of kiloamps. What it does is offer the most probable termination point and, behind it, a low-resistance route to earth that passes through nothing flammable. The resistance of the earth electrode then matters directly: a median 30 kA return stroke into a 10 Ω electrode lifts the whole down-conductor to 30 × 10³ A × 10 Ω = 3 × 10⁵ V above remote earth for the duration of the stroke. Anything metallic nearby but separately earthed is suddenly 300 kV away from it, and that difference is bridged by a side flash. This is why a protection system bonds pipework, cable armour and structural steel together instead of isolating them, and why a car or a steel-framed building protects the people inside: the current stays on the conducting shell around them.

4Turning the same force into a sorting machine

Rub two dissimilar insulators together and they exchange charge, leaving flakes of one plastic with a positive charge-to-mass ratio and the other with a negative one. Drop them between two plates that make a uniform field E and each flake feels a sideways force F = qE while gravity pulls it down, so the horizontal acceleration is a = (q/m)E and the sorting is by charge-to-mass ratio. With q/m = 6.0 × 10⁻⁶ C kg⁻¹ and E = 3.0 × 10⁵ V m⁻¹ the acceleration is 1.8 m s⁻²; a 1.2 m fall takes t = √(2h/g) = 0.49 s, so the flake drifts x = ½at² = 0.22 m sideways and the two plastics land 0.44 m apart in different bins. Note the field chosen: a tenth of breakdown, because a spark would neutralise the very charge the machine runs on. Precipitators and powder coating live on the same two lines of algebra.

5The spark you cannot feel is the one that matters

A standing person is modelled as a 100 pF capacitor with about 1.5 kΩ of body resistance. Walk across a nylon carpet in dry air and a field meter may read 8.0 kV: that is Q = CV = 0.80 μC and W = ½CV² = 3.2 mJ, delivered at a peak current of V/R = 5.3 A over a time constant RC = 150 ns. Now compare thresholds rather than adjectives. A person first feels a discharge at roughly 2 kV, which is 0.20 mJ. A hydrogen-air mixture ignites on 0.017 mJ, a typical solvent vapour on about 0.25 mJ, and a fine organic dust cloud on tens of millijoules. A thin gate oxide can fail at 100 V, which on 100 pF is 0.50 μJ — four hundred times below the faintest shock a human can detect. The nervous system is calibrated far too high to be used as an instrument.

6Choose a control by its time constant, then measure it

Charge leaves along a resistance, and τ = RC says how fast. With the 100 pF body, a dissipative floor-and-footwear path of 1 MΩ to 1 GΩ gives τ between 1.0 × 10⁻⁴ s and 0.1 s, so a person is neutral again before reaching the bench; insulating shoes on a sealed floor can exceed 1 TΩ, where τ ≈ 100 s and the charge simply travels with them. A wrist strap is built around a 1 MΩ series resistor for both halves of that trade: fast enough to bleed static, high enough that touching 250 V would pass only 250 μA. Raising humidity above about 50% helps by making surfaces conduct, and an ioniser is the fallback where the charged object is an insulator that cannot be drained at all. Then verify rather than assume: a charged-plate monitor should fall from 1000 V to 100 V in a couple of seconds, and a field meter should read the process, not the specification sheet.

2

Key equations

Read the units, assumptions and limits before substituting.

Dielectric strength of air

Ebreak ≈ 3.0 × 10⁶ V m⁻¹

Turns any charge arrangement into a yes-or-no answer about sparking.

Clean dry air near 1 atm, gaps of order 1 cm; 1 V m⁻¹ = 1 N C⁻¹.

Surface field of a charged sphere

Esurf = k Q / R²

Geometry, not the charge alone, decides how hard a conductor pushes on the air.

Q in C, R in m, k = 8.99 × 10⁹ N m² C⁻², E in V m⁻¹.

Corona onset of a sharp feature

Vₒₙₛₑₜ = Ebreak r

Reads a whole machine's safe working potential off its sharpest edge.

r is the radius of curvature in m; r = 1.0 mm gives Vₒₙₛₑₜ = 3.0 kV.

Field sharing between joined conductors

E₁ / E₂ = R₂ / R₁

A 1.0 mm burr on a 0.20 m dome sits in 200 times the dome's own field.

Two spheres at one potential and far apart, so the charge splits as Q ∝ R.

Energy released by a discharge

W = ½ C V²

Ranks one spark against ignition energies and device damage limits.

C in F, V in V, W in J; the human body model takes C = 100 pF.

Charge decay through a path to earth

τ = R C

Chooses a floor, a shoe or a wrist strap by how fast it drains charge.

R in Ω, C in F, τ in s; 1 MΩ with 100 pF gives τ = 1.0 × 10⁻⁴ s.
3

Interactive lab

Predict first. Explore the model. Compare the result.

Explore the model

0.20 m
0.05 to 0.4 m; steps of 0.05.
2.0 mm
0.5 to 20 mm; steps of 0.5.
100 kV
10 to 300 kV; steps of 10.

Try this. Drag the tip radius down and watch its bar climb while the dome's stays put: air fails at the tip once the conductor reaches {3*rtip|1} kV, far under the dome's own limit, which is why the sharpest feature decides when a machine starts sparking.

Live results

FIELD AT DOME
0.50 MV m⁻¹
FIELD AT TIP
50.0 MV m⁻¹
CORONA ONSET AT TIP
6.0 kV
CHARGE ON DOME
2.22 μC

Results follow this lesson’s model and assumptions. Displayed values are rounded.

Interactive Physics II modelA polished dome of radius 0.20 m carries a small tip of radius 2.0 mm at the same potential, 100 kV. The bars give each surface field on a logarithmic scale: 0.50 MV m⁻¹ at the dome against 50.0 MV m⁻¹ at the tip. A bar past the dashed 3 MV m⁻¹ line is a surface already breaking down the air.conductor at V = 100 kVdome R = 0.20 m, tip r = 2.0 mmtip field = 100 × dome fieldtip coronas from 6.0 kVdometip10⁴ V m⁻¹10⁶10⁸dashed line = 3 MV m⁻¹ in air
A polished dome of radius 0.20 m carries a small tip of radius 2.0 mm at the same potential, 100 kV. The bars give each surface field on a logarithmic scale: 0.50 MV m⁻¹ at the dome against 50.0 MV m⁻¹ at the tip. A bar past the dashed 3 MV m⁻¹ line is a surface already breaking down the air.

Predict → change one variable → check

Plan a change before running it. Every other input stays at the saved baseline.

Saved baseline. Dome radius R: 0.20 m; Tip radius r: 2.0 mm; Conductor potential V: 100 kV

Baseline 0.20; choose 0.05 to 0.4, in steps of 0.05.
4. Compared with the saved baseline, this result will…

This experiment is kept only while this page is open. It does not change saved course progress.

4

Catch the common trap

“I earthed it, so the charge is gone.” Earthing drains charge only along a conducting path, and only from what that path actually touches. Clip a bonding strap to a polyethylene drum and you have earthed one square centimetre of plastic; the bulk relaxes with τ = ε₀ εᵣ ρ, and with εᵣ ≈ 2.3 and ρ ≈ 10¹⁶ Ω m that is 2.0 × 10⁵ s, more than two days, so tomorrow's operator meets the same charge. The mirror-image error is just as common: a perfectly conductive steel drum standing on an insulating pallet is not earthed at all, however shiny it looks, and holds its whole charge as one isolated conductor until somebody offers it a spark. Earthing is a claim about a measured resistance to earth, not about a wire being attached.

5

Practice & worked examples

3 authored examples. Make an attempt, then reveal the reasoning one step at a time.

EasyExample 1

A demonstration generator charges a polished metal dome of radius 0.25 m in dry air, which breaks down at 3.0 × 10⁶ V m⁻¹. Find the largest charge the dome can hold, the potential it reaches at that point, and how many elementary charges that is.

Sketch your answer or reasoning, then compare it with the authored solution. Your writing is not automatically graded or saved.

0 of 4 steps revealed.

MediumExample 2

A 0.20 m dome is left with a 1.0 mm radius burr on a bolt head, in good electrical contact with it. Treat the burr as a small sphere held at the dome's potential. Find the potential at which the burr starts to corona, the charge on the dome at that moment, and the fraction of that charge sitting on the burr.

Sketch your answer or reasoning, then compare it with the authored solution. Your writing is not automatically graded or saved.

0 of 5 steps revealed.

HardExample 3

A field meter reads an operator at 8.0 kV after they cross a nylon carpet in dry air. Use the human body model, C = 100 pF in series with 1.5 kΩ. Find the charge carried, the energy a spark to a grounded rail would release, the peak current and the time constant, then judge the workstation for decanting a solvent of minimum ignition energy 0.25 mJ and size a bleed path.

Sketch your answer or reasoning, then compare it with the authored solution. Your writing is not automatically graded or saved.

0 of 5 steps revealed.

6

Quick check

Select an answer and explain it before checking.

A worker charged to 1.5 kV sparks to a grounded pipe in a hydrogen-air atmosphere of minimum ignition energy 0.017 mJ. Take C = 100 pF. What follows?
7

Review & continue

Electrostatic safety is the practice of keeping the largest field anywhere in an arrangement below the value at which the surrounding gas breaks down, and keeping the energy any unavoidable spark releases below what the surroundings can tolerate.

  • Explain the relationship without looking at the equation.
  • Check a sign, a unit and a model assumption in your working.
  • Return to the lab or worked example if a step is still unclear.

Optional supporting lessons

These chapter resources can help with earlier ideas; they are not additional required Physics II completion tasks.

Extra challenging questions

Connect several ideas, explore supplied data, and compare your reasoning with a worked solution.