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# Cylindrical capacitor dissipation factor

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:

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

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: • Video: Cylindrical capacitor dissipation factor 