This is an example of the cylindrical capacitor simulation, performed with QuickField software.
Problem Type: Axisymmetric problem of AC conduction.
Geometry:
Capacitor consists of ceramic tube with silver electrodes mounted at the surface.
Given:
Relative permittivity of air ε = 1.
Relative permittivity of ceramic ε = 6.
Conductivity of ceramic g = 10-8 S/m;
Voltage U = 10 V.
Frequency f = 1000 Hz.
Problem:
Determine capacitance and dissipation factor of the capacitor.
Solution:
The value of dissipation factor can be calculated as
The value of dissipation factor can be calculated as tg(δ) = PA/PR. Capacitance can be calculated as C = q/U, where U is potential difference between electrodes and q - is a charge on electrodes.
Results: Field distribution in cylindrical capacitor:
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