Effect of earth on line capacitance

The earth is a conductor at zero potential. Its presence changes the electric field of an overhead line and so increases the line capacitance slightly.

Method of images

The earth's surface is an equipotential plane. Its effect can be modelled by removing the earth and placing an image conductor as far below the ground surface as the real conductor is above it, carrying an equal and opposite charge. For a conductor at height h with charge +q, the image is at depth h with charge −q. The field above ground is then calculated from the real and image charges together.

     a (+qa)  <-- D -->  b (+qb)
     |                   |
     h                   h
 ====|======= earth =====|======
     h                   h
     |                   |
     a' (-qa)            b' (-qb)
  distance a to b' = H = sqrt(D^2 + 4h^2)

Single-phase line

Without earth:

$$C_{ab} = \frac{\pi\varepsilon_0}{\ln(D/r)}\ \text{F/m}$$

With earth effect (images included):

$$C_{ab} = \frac{\pi\varepsilon_0}{\ln\left(\dfrac{D}{r}\cdot\dfrac{2h}{\sqrt{D^2 + 4h^2}}\right)}$$

Because $\frac{2h}{\sqrt{D^2+4h^2}} < 1$, the denominator is smaller and the capacitance is larger.

Three-phase transposed line

$$C_n = \frac{2\pi\varepsilon_0}{\ln\dfrac{D_{eq}}{r} - \ln\dfrac{\sqrt[3]{H_{12}H_{23}H_{31}}}{\sqrt[3]{H_1H_2H_3}}}\ \text{F/m to neutral}$$

Here $H_{ij}$ is the distance from conductor i to the image of conductor j, and $H_i = 2h$ is the distance to its own image.

Points to note

• The earth brings charge of the opposite sign closer to the conductor, which raises the capacitance.
• The effect is small for normal overhead lines, because $h \gg D$, so it is usually neglected. It becomes important for low-height lines, large spacings, and in unbalanced-fault (zero-sequence) calculations.
• Earth has a negligible effect on line inductance under balanced conditions.
• Higher capacitance means slightly more charging current and charging kVAR.