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Cylindrical capacitor dissipation factor

QuickField simulation example

This is an example of the cylindrical capacitor simulation, performed with QuickField software. Capacitor consists of ceramic tube with silver electrodes mounted at the surface.

Problem Type
Axisymmetric problem of AC conduction.

Geometry
Cylindrical ceramic capacitor dissipation factor Capacitor consists of ceramic tube with silver electrodes mounted at the surface. Ø2 mm 3 mm ceramic silver silver

Given
Relative permittivity of ceramic ε = 6.
Conductivity of ceramic g = 10 nS/m;
Voltage U = 10 V.
Frequency f = 1 kHz.

Task
Determine capacitance and dissipation factor of the capacitor.

Solution
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
Charge q = 27.7 pC.
Capacitance C = 2.77 pF.
Active power PA = 24.5 nW.
Reactive power PR = 817 nW.
Dissipation factor tg(δ) = 0.03.

Field distribution in cylindrical capacitor:
Field distribution in cylindrical capacitor