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Cylindrical capacitor

This is an example of the cylindrical capacitor simulation, performed with QuickField software.

Problem Type:
Axisymmetric problem of AC conduction.

Geometry:

Cylindrical capacitor

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) = 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:
Field distribution in cylindrical capacitor:

Field distribution in cylindrical capacitor

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

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