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Main >> Applications >> Sample problems >> Cylindrical capacitor simulation

# 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) = 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: 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

• Video: 