## 4 Effect of a Non-Perfect Conducting Ground  
The finite conductivity of the soil has repercussions on the terms of Equation (5) involving improper integrals of the Sommerfeld type. As previously seen, these terms become simplest when considering the soil as a perfect conductor.  
\[\begin{array}{l}A_{x}^{\sigma}(x,y,z,\omega)=-\frac{\mu_{0}\delta I}{4\pi} \int\limits_{0}^{\infty}\frac{2\lambda}{\lambda_{0}}\cdot\frac{\lambda_{1}}{ \lambda_{0}+\lambda_{1}}e^{-\lambda_{0}(z+z_{p})}J_{0}(\lambda r)d\lambda\ \underset{\sigma\to\infty}{\rightarrow}-\frac{2\mu_{0}\delta I}{4\pi}\cdot \frac{e^{-\mu_{0}\delta I}}{R^{\prime}}\\ A_{z}^{\sigma}(x,y,z,\omega)=\frac{\mu_{0}\delta I}{4\pi}\frac{x-y}{r}\int \limits_{0}^{\infty}\frac{\gamma_{0}^{2}-\gamma_{1}^{2}}{\gamma_{0}^{2}\lambda _{1}+\gamma_{1}^{2}\lambda_{0}}\cdot\frac{2\lambda^{2}}{\lambda_{0}+\lambda_{ 1}}e^{-\lambda_{0}(z+z_{p})}J_{1}(\lambda r)d\lambda\ \underset{\sigma\to \infty}{\rightarrow}\ 0\end{array} \tag{16}\]  
The main problem with the terms associated with finite soil conductivity (16) is that they do not have a closed expression in the time domain and must be evaluated in the Fourier domain. Additionally, Sommerfeld-type integrals are difficult to evaluate and require special numerical techniques [14]. In this paper, the objective is not to numerically calculate the contribution of these terms but to estimate bounds on the values of the induced potentials in the extreme cases associated with infinite soil conductivity and zero soil conductivity. A soil with finite conductivity lies between these extreme situations, and therefore, the induced potential will fall within the bounds associated with these extreme cases.  
When the soil is considered a medium with zero conductivity, such as air, and with electric parameters similar to air, the Equation (5) along with property (4) yield the following results,  
\[\begin{array}{l}A_{x}^{\sigma}(x,y,z,\omega)=-\frac{\mu_{0}\delta I}{4\pi} \int\limits_{0}^{\infty}\frac{2\lambda}{\lambda_{0}}\cdot\frac{\lambda_{1}}{ \lambda_{0}+\lambda_{1}}e^{-\lambda_{0}(z+z_{p})}J_{0}(\lambda r)d\lambda\ \underset{\sigma\to 0}{ \rightarrow}-\frac{\mu_{0}\delta I}{4\pi}\cdot\frac{e^{-\mu_{0}\delta I}}{R^{ \prime}}\\ \varepsilon_{1}=\varepsilon_{0}\\ \mu_{1}=\mu_{0}\end{array} \tag{17}\]  
Thus, one bound on the induced potential value will be obtained by replacing the soil with air, where the only contribution to the fields comes from the vector potential of the actual horizontal lightning channel. On the other hand, the other bound will be obtained by considering the soil as a perfect conductor, and the contribution to the fields will be the difference between the contribution from the vector potential of the actual channel and the contribution from the image channel with respect to the soil surface. For a soil with finite conductivity, the potential will lie between the two mentioned bounds.  
It can be expected that the influence of finite soil conductivity will be more pronounced in points close to the surface since both the real lightning channel and the image channel are at comparable distances. However, in points near the real lightning channel, the influence of the soil is heavily limited by the distance.  
Figure 10 depicts the upper and lower bounds of the induced potential on the vertical rod shown in Figure 3. The blue line corresponds to soil with zero conductivity, equivalent to replacing the conductive medium of the soil with air. The red line, representing the minimum bound of the induced potential, corresponds to a perfectly conducting soil. The graph associated with a soil of finite and non-zero conductivity will lie between the two curves shown in the figure.