Uniform fields and parallel plates

Start from one large flat sheet carrying charge per unit area σ. Its field is σ/2ε₀, perpendicular to the sheet, pointing away from it when σ is positive — and the same size at 0.1 mm as at 10 mm, because moving away shrinks the nearest charge's contribution while bringing more distant charge into view. Now face two such sheets, one at +σ and one at −σ. Between them both contributions run from the positive sheet toward the negative one, so they add: E = σ/2ε₀ + σ/2ε₀ = σ/ε₀. Outside the pair the two contributions point opposite ways and cancel, leaving E ≈ 0. Take plates of area 5.0 × 10⁻³ m² carrying ±0.40 nC: σ = 4.0 × 10⁻¹⁰ C ÷ 5.0 × 10⁻³ m² = 8.0 × 10⁻⁸ C m⁻², so E = 8.0 × 10⁻⁸ ÷ 8.85 × 10⁻¹² = 9.0 × 10³ N C⁻¹ throughout the gap. Nothing in that calculation asked how far apart the plates were.

Calling a field uniform asserts two things at once: the same magnitude everywhere and the same direction everywhere. On a field-line diagram both show up as one picture — straight, parallel lines, evenly spaced — because spacing encodes magnitude and the line's direction encodes direction. Compare a point charge, whose lines fan out and thin, giving a field that falls as 1/r² and swings around as you move. In the gap you may carry a test charge from one plate to the other, or slide it sideways, and the force vector does not change at all. That is what makes the arrangement worth building, and what makes it a measuring instrument: fix σ and you know E, with no position to keep track of.

A charge q in the gap feels F = qE, constant in size and direction, so Newton's second law gives a constant acceleration a = qE/m. Constant acceleration is the case mechanics has already solved, so every result carries over: velocity grows linearly, displacement quadratically, and a particle launched across the field traces a parabola exactly as a projectile does under gravity. Only the scale is different. An electron in E = 2.4 × 10⁴ N C⁻¹ has a = (1.60 × 10⁻¹⁹ C × 2.4 × 10⁴ N C⁻¹) ÷ (9.11 × 10⁻³¹ kg) = 4.2 × 10¹⁵ m s⁻², and the ratio of its weight to the electric force is mg/qE = 2.3 × 10⁻¹⁵, so gravity is not worth writing down. Scale up to an oil drop of radius 0.50 μm and density 900 kg m⁻³ — mass 4.7 × 10⁻¹⁶ kg — carrying three electrons: its weight of 4.6 × 10⁻¹⁵ N is balanced by a field of only 9.6 × 10³ N C⁻¹. Whether gravity matters is a question about q/m, not about the field.

Send a particle in at speed v₀ parallel to the plates, perpendicular to E, along a plate of length L. Nothing acts along the beam, so x = v₀t and the time inside is pure geometry: t = L/v₀. Across the beam the acceleration is constant, so y = ½at² = ½(qE/m)(L/v₀)² = qEL²/(2mv₀²). Two features deserve names. The deflection grows as L², so lengthening the plates pays twice over. And it falls as v₀², so a beam twice as fast is deflected four times less — not because the force is weaker, but because halving the time in the field cuts a quadratic effect by four. With v₀ = 2.4 × 10⁷ m s⁻¹, L = 40 mm and E = 1.2 × 10⁴ N C⁻¹: t = 1.67 ns, a = 2.11 × 10¹⁵ m s⁻², y = 2.9 mm, and the beam leaves at tan θ = v_y/v₀ = 0.146, that is 8.3°.

Real plates end. Within roughly one gap width of an edge the lines no longer run straight across: they bow out into the space beyond the plates, spread apart and weaken. That is the fringing field. Everything more than about d in from the edge is still uniform to within a percent or so, so the usable area of a square plate of width w is about (w − 2d)². With w = 120 mm and d = 3.0 mm that is 114² = 1.30 × 10⁴ mm² out of 1.44 × 10⁴ mm², or 90% of the gap. Open the gap to d = 30 mm with the same plates and the estimate gives (60 mm)², a bare 25%: the parallel-plate model is a small-gap model. Two consequences follow. Keep a deflected beam well clear of the ends, and read the phrase zero field outside as an approximation — the leak is small, and from far away a finite charged pair still looks like a dipole.

First, air breaks down near 3 × 10⁶ N C⁻¹. Push the field past that and a spark shorts the plates, which caps what any air gap can hold. Second, the plates are conductors, so their charge is free to move: it is not quite evenly spread, piling up slightly at the edges, and a charged particle brought close induces a redistribution that pulls it toward the surface — an effect no uniform-field formula contains. Third, the geometry has to be honest. Tilt one plate by a degree or let it bow and the gap varies along the plate, so σ/ε₀ still describes the field locally but the region is no longer uniform. Each failure is checkable before you trust a number: compare E with the breakdown strength, keep the particle several gap widths from any surface, and confirm the plates are flat and parallel.