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Method of images for line capacitance

Short answer

Method of images replaces the earth’s conducting plane with imaginary conductors carrying equal and opposite charges, placed symmetrically below ground. Capacitance is calculated from potentials due to real and image charges. This accounts for the ground effect, which slightly increases line capacitance compared to neglecting earth.

What is the method of images? How can it be used to take into account the presence of ground in calculating the capacitance of a line?

The method of images is a technique for finding the electric field (and hence capacitance) of charged conductors above a conducting plane. The earth is treated as a perfectly conducting, zero-potential plane. Its effect is replaced by imaginary "image" conductors placed as far below the ground as the real conductors are above it, carrying equal and opposite charges.

Basis

  • The earth's surface is an equipotential (zero potential) surface.
  • A charge +q+q at height hh and a charge −q-q at depth hh below the surface together produce zero potential everywhere on the plane midway between them.
  • So, for the region above ground, the field of "conductor + earth" is the same as the field of "conductor + image conductor" with the earth removed.
     +q o  a           +q o  b      (real conductors)
        |  h              |
 ====================================  earth (V = 0)
        |  h              |
     -q o  a'          -q o  b'     (images)

Use in capacitance calculation

  1. Replace the earth by image conductors: for each conductor of charge qq at height hh, add an image of charge −q-q at depth hh.
  2. Write the voltage of each conductor due to all real and image charges, using V=q2πε0ln⁡D2D1V = \dfrac{q}{2\pi\varepsilon_0}\ln\dfrac{D_2}{D_1} for each charge.
  3. For a single-phase line (conductors a, b, spacing DD, height hh), this gives
Vab=qπε0ln⁡2hDr4h2+D2V_{ab} = \frac{q}{\pi\varepsilon_0}\ln\frac{2hD}{r\sqrt{4h^2+D^2}} Cab=πε0ln⁡(Dr1+D2/4h2) F/mC_{ab} = \frac{\pi\varepsilon_0}{\ln\left(\dfrac{D}{r\sqrt{1+D^2/4h^2}}\right)}\ \text{F/m}
  1. For a transposed three-phase line, average over the transposition cycle:
Cn=2πε0ln⁡Deqr−ln⁡H12H23H313H1H2H33 F/mC_n = \frac{2\pi\varepsilon_0}{\ln\dfrac{D_{eq}}{r} - \ln\dfrac{\sqrt[3]{H_{12}H_{23}H_{31}}}{\sqrt[3]{H_1H_2H_3}}}\ \text{F/m}

where H1,H2,H3H_1, H_2, H_3 are the distances from each conductor to its own image and H12H_{12} etc. are the distances to the images of the other conductors.

Effect

The extra term subtracted in the denominator is positive, so the denominator becomes smaller and the ground increases the line capacitance slightly. The effect is small when the conductors are high above ground compared with their spacing, so it is usually neglected except for low lines or accurate studies.

✓ Answer checked by Claude Opus 5.5 (high effort)

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