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# Wire ring capacitance

QuickField simulation example

Problem Type

Axisymmetric problem of electrostatics.

Geometry

R = 100 mm, a = 8.74 mm.

Given

Relative permittivity of vacuum ε = 1,
The charge q = 10-9 C

Find the wire ring capacitance and compare it with analytical solution:
C = 4πεε0 · πR / ln(8R/a), [F]. *

Solution

The capacitance can be calculated as C = q /U, where U is a body potential and q is a body charge. Conductor's surface is marked as 'floating conductor', i.e. isolated conductor with unknown potential. At some point on conductor's surface the charge is applied. The charge is then redistributed along the conductor surface automatically.
It is supposed that field fades to zero far away from the ring. The calculation area is limited by the outer boundary, where zero potential is applied.

Results

Potential distribution. As the potential of the body is calculated with respect to the zero potential of the 'infinity', the result depends on the distance to the outer boundary. The set of simulations where made using automation tool LabelMover.

Distance to the outer boundary Voltage, V Capacitance, pF
10*R 119.2 8.39
20*R 123.7 8.08
40*R 125.9 7.94
80*R 127.0 7.87
160*R 127.6 7.84
320*R 127.8 7.82
Theory 7.74

*"The Capacitance of an Anchor Ring" by Thomas, T. S. E, Australian Journal of Physics, vol. 7, p.347. 