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) = P_{A}/P_{R}. 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: