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ACElec2: Cylindrical capacitor

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:

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