
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.
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) = 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:
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 
See the ACElec2.pbm problem in the Examples folder.