Effect of neglecting earth on capacitance

The earth is included by the method of images. For a transposed line,

$$C_n = \frac{2\pi\varepsilon_0}{\ln\dfrac{D_{eq}}{r} - \ln\dfrac{\sqrt[3]{H_{ab}H_{bc}H_{ca}}}{\sqrt[3]{H_aH_bH_c}}}$$

Without the earth, the second term is dropped.

   a o-----6 m-----o b-----6 m-----o c
     |13 m         |13 m           |13 m
 =========================================  earth
     |13 m         |13 m           |13 m
   a'o             o b'            o c'

Data: $r = 0.015$ m, $D_{ab} = D_{bc} = 6$ m, $D_{ca} = 12$ m, $h = 13$ m.

$$D_{eq} = \sqrt[3]{6 \times 6 \times 12} = 7.5595\ \text{m}, \qquad \ln\frac{7.5595}{0.015} = 6.2225$$

Neglecting earth

$$C_n = \frac{55.63\times10^{-12}}{6.2225} = 8.940\times10^{-12}\ \text{F/m} = 8.940\ \text{nF/km}$$

Considering earth

$$\begin{aligned} \sqrt[3]{H_{ab}H_{bc}H_{ca}} &= \sqrt[3]{26.683^2 \times 28.636} = 27.319\ \text{m} \\ \text{Correction} &= \ln\frac{27.319}{26} = 0.04948 \\ C_n &= \frac{55.63\times10^{-12}}{6.2225 - 0.0495} = \frac{55.63\times10^{-12}}{6.1730} \\ &= 9.012\times10^{-12}\ \text{F/m} = 9.012\ \text{nF/km} \end{aligned}$$

Effect of neglecting earth

$$\frac{9.012 - 8.940}{8.940}\times100 = 0.80\%$$

Neglecting the earth gives a capacitance about 0.8% lower than the true value. The earth slightly increases the capacitance because the opposite-sign image charges lower the potential of each conductor, so more charge is needed for the same voltage. The error is small when the height (13 m) is large compared with the spacing, so the earth effect is usually neglected.

Answer: without earth $8.940$ nF/km; with earth $9.012$ nF/km; error from neglecting earth ≈ 0.8%.