# What are the units when solving Transmission Line iterativelly?

I would like to know the conductancee,$$\C\$$, and capacitance,$$\G\$$, of my lossy transmission line which is configured as an open circuit.

In such a case, I know that the expression that defines the input impedance of the line, $$\ Z_{in}\$$ is given by the following expression

$$\ Z_L = \infty \Rightarrow Z_{in} = Z_{o}*coth \left( \gamma*l \right) =$$ $$\ = Z_{o}*coth \left( \ \left( \alpha+j\beta \right)*l \right) \tag 1$$

where

$$\ \gamma = \sqrt{ \left( R+j\omega L \right)* \left( G+j\omega C \right) } =$$ $$\ = \sqrt{ \left( RG-\omega^2LC \right)+ j\omega* \left(RC+LG \right) } = \alpha + j\beta$$ $$\ Z_o = \sqrt{ \frac{R+j\omega L}{G+j\omega C} } = \frac{\sqrt{ \gamma}}{G+j\omega C}$$

I can use a non-linear method to solve $$\(1)\$$ since I know $$\ Z_{in}, L, R\$$ and $$\ \omega\$$

My question is related to the units of $$\C\$$ and $$\G\$$ obtained when solving, are these units per length, $$\\left[C\right]= \frac{F}{m}\$$ and $$\\left[G \right]=\frac{S}{m}\$$, or the total of the line $$\\left[C\right]= F\$$ and $$\\left[G\right]= S \$$?