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 v_y = 500 000 m/s; v_z = v_φ = 0 m/s.
Emitter position (0; 0; 0).

Task
Calculate the charged particle trajectory neglecting relativistic effects.

Solution
The analytical solution is a parabola
x(t) = 0,
y(t) = v_y * t,
z(t) = 0.5 * F_z / m * t²,
where F_z - 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 E_z = -20 V/m, E_r = 0 V/m

Lorentz force F_z = q * E_z = -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.

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