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# Charged particle trajectory in the uniform static electric field. Case: cylindrical.

QuickField simulation example

Problem Type
Axisymmetric problem of electrostatics.

Geometry

Given
Relative permittivity of vacuum ε = 1;
Positive potential U+ = 20 V.
Charge (electron) q = -1.602e-19 C
Mass (electron) m = 9.109e-31 kg
Initial velocity vy = 500 000 m/s; vz = vφ = 0 m/s.
Emitter position (0; 0; 0).

Calculate the charged particle trajectory neglecting relativistic effects.

Solution
The analytical solution is a parabola
x(t) = 0,
y(t) = vy * t,
z(t) = 0.5 * Fz / m * t²,
where Fz - is the z-component of the Lorentz force,
t - is time.

Particle trajectory may be calculated using the built-in function in QuickField Electrostatic postprocessor or by the free tool TrajectoryTracer.

Results
Electric field strength Ez = -20 V/m, Er = 0 V/m

Lorentz force Fz = q * Ez = -1.602e-19 * -20 = 3.204e-18 N. x(t) = 0 m,
y(t) = 500000*t m,
z(t) = 0.5*3.204e-18/9.109e-31 *t² m.

 Particle coordinates (x; y; z) m time Theory QuickField TrajectoryTracer tool 0 s (0; 0; 0) (0; 0; 0) (0; 0; 0) 1e-7 s (0; 0.050; 0.018) (0; 0.050; 0.018) (0; 0.050; 0.018) 2e-7 s (0; 0.100; 0.070) (0; 0.100; 0.070) (0; 0.100; 0.070) 3e-7 s (0; 0.150; 0.158) (0; 0.150; 0.158) (0; 0.150; 0.158) 4e-7 s (0; 0.200; 0.281) (0; 0.200; 0.281) (0; 0.200; 0.281) 5e-7 s (0; 0.250; 0.440) (0; 0.250; 0.440) (0; 0.250; 0.440)