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Pair of parallel wires capacitance - QuickField simulation example

Engineering question

How to calculate the mutual capacitance per unit length of two unequal parallel wires and verify?

Answer
Treat the wires as floating conductors, calculate capacitance from applied charge over potential difference, and compare it with the analytical expression.

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Typical applications
two-wire transmission lines, overhead conductors, parallel wire systems

Parallel wires capacitance

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Simulation problem

Problem Type
Plane-parallel problem of electrostatics.

Geometry
Pair of parallel wires capacitance Finding the mutual capacitance between two infinitely long parallel wires +q -q Air a1 a2 b

a1 = 4 mm, a2 = 1 mm, b = 10 mm.

Given
Relative permittivity of air ε = 1,
Charge value q = 1 nC.

Task
Find the mutual capacitance between two parallel wires and compare its value with analytical solution*:
C/Lz = 4π·ε·ε0 / 2·LN [ (b²-a1²-a2² + 2bc) / 2·a1·a2 ] [F/m],
where 2bc = √[b² - (a1+a2)²] · [b² - (a1-a2)²] [m²]

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 q is applied. The charge is then redistributed along the conductor surface automatically.
QuickField calculates electric potential U distribution. The capacitance is calculated as C = q / ΔU.
In problem properties we set model depth to be Lz = 1 m.

Results
Analytical solution 2bc = √[0.01² - (0.004+0.001)²] · [0.01² - (0.004-0.001)²] = 8.261e-5 m².
C/Lz = 4·3.142·1·8.854e-12 / 2·LN [ (0.01²-0.004²-0.001² + 8.261e-5) / (2·0.004·0.001) ] = 18.36 pF/m.

The measured potential difference in QuickField is ΔU = 54.41 V.
The capacitance is C = 1 nC / 54.41 V = 18.38 pF per model depth Lz = 1 m.
Electric potential distribution around parallel wires calculated in QuickField:
parallel wires capacitance

*Reference: Chester Snow Formulas for Computing Capacitance and Inductance, U.S. Government Printing Office, 1954

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