QuickField

A new approach to field modelling
 Language: Global English Deutch Espanol Francais Italiano Danmark Ceske Chinese Pycckuü

>> >>

# Pair of parallel wires capacitance

Problem Type:
Plane problem of electrostatics.

Geometry:

a1 = 4.72 mm, a2 = 1.13 mm, d = 100 mm.

Given:
Relative permittivity of vacuum ε = 1,
The charge q = 10-9 C
Model depth l = 1 m.

Find the mutual capacitance between two parallel wires and compare its value with analytical solution:
C = 2π·ε·ε0 l / ln( d2/a1·a2 ) [F]. *

Solution:
Wire's surfaces are marked as 'floating conductor', i.e. isolated conductors with unknown potential. At some point on each of wire's surface the charge is applied. The charge is then redistributed along the conductor surface automatically.

Results:

Potential distribution. The capacitance can be calculated as C = q / (U2-U1). The measured potential difference is U2-U1 = 135 V.
The capacitance is C = 10-9 / 135 = 7.41·10-12 F

 QuickField Theoretical result C, pF(model depth l = 1 m) 7.41 7.37
• Video: