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Parallel wires capacitance
Pair of parallel wires capacitance
Problem Type:
Plane problem of electrostatics.
Geometry:
a_{1} = 4.72 mm, a_{2} = 1.13 mm, d = 100 mm.
Given:
Relative permittivity of vacuum ε = 1,
The charge q = 10^{9} C
Model depth l = 1 m.
Task:
Find the mutual capacitance between two parallel wires and compare its value with analytical solution:
C = 2π·ε·ε_{0} l / ln( d^{2}/a_{1}·a_{2} ) [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 / (U2U1). The measured potential difference is U2U1 = 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 
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*Power System Engineering, D.P.Kothari, I.J.Nargath.