ACElec2: Cylindrical capacitor

AC current in the cylindrical capacitor

Problem Type:
An axisymmetric problem of AC conduction.

Geometry:
Capacitor consists of ceramic tube with silver electrodes mounted at the surface.
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 air e = 1.

Relative permittivity of ceramic e = 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:

QuickField
q, C 2.77·10-11
C, F 2.77·10-12
PA, W 2.45·10-8
PR, W 8.17·10-7
tg(δ) 0.03

See the ACElec2.pbm problem in the Examples folder.