Inkjet droplet deflection - QuickField simulation example

Simulation problem

Geometry

Given
Electric potential difference ΔU = 1000 V.
Charged ink droplet charge q = 1e-13 C, mass m ≈ 6.5e-11 kg (a water-like-density droplet with a radius of about 25 µm).
Droplet initial position and speed: (x₀, y₀, z₀) = (0, 0, 0) mm, v₀ = (20, 0, 0) m/s.

Task
Calculate the droplet trajectory through the 20 mm deflection region and compare the numerical result with the analytical parabolic estimate.

Solution
QuickField computes the essentially uniform plane-parallel electrostatic field between the two plates.
The QuickField Particle Tracer integrates the equation of motion m·d²r/dt² = q·E(r) for 200 steps of 5e-6 s.

Analytical estimate (neglecting drag):
Electric field stress E_y = ΔU / 0.002 = +5.0e5 V/m
The transverse acceleration is
a_y = qE/m = (1e-13 C · 5.0e5 V/m) / 6.5e-11 kg ≈ 769.231 m/s².
The continuous parabolic solution is y(x) = 0.5·a_y·(x/v₀)².
At x = 20 mm and v₀ = 20 m/s the transit time is t = x/v₀ = 0.001 s
Transverse displacement y(20 mm) = 0.5 · 769.231 · (0.001)² ≈ 3.846e-4 m ≈ 0.385 mm.

Results
The droplet moves at an essentially constant velocity of 20 m/s in the +X direction while accelerating toward the negative electrode.
The trajectory is a smooth upward-opening parabola. It reaches a deflection of y = +0.387 mm after traveling 20 mm. This is well inside the 2 mm electrode gap.

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