gauss law - Electrostatic in 2D: dimensional analysis

After reading this very interesting post about the electric field and the electric potential of a point charge in 2D and 1D, I've understood that, for the $2D-$case, the following formulas hold: $$ \Phi_{\operatorname{2-d}}(r) = -\frac{\lambda}{2\pi \epsilon_0} \ln(r) $$

$$ \vec{E}_{\operatorname{2-d}}(r) = \frac{\lambda}{\epsilon_0} \left(\frac{\hat{r}}{2\pi r}\right) $$

Nevertheless, I haven't fully understood how the dimensional analysis and the numerical values of the quantities that come into play (namely, the electric charge $\lambda$ and the vacuum permittivity $\epsilon_0$) change due to the reduced dimensionality of the system.

In other words:

1. Is the value of the vacuum permittivity still $8.85 \, 10^{-12}\, F/m$, as in the usual 3D world?


2. Is the value of a single electric charge still $1.60 \, 10^{-19} C$, as in the 3D world?

Answer

How you shuffle the units in different dimensions is a matter of opinion. If you keep the units of $\epsilon_0$ fixed, then the elementary charge in each dimensionality has units $\operatorname{C}\operatorname{m}^{d-3}$. Interestingly, the units of force and energy are similarly not independent of dimension. If I place two infinite line charges parallel to each other I cannot talk intelligently about the total force acting between them; it, like the total charges, is infinite. I, therefore, have to work in terms of force per unit length (similarly for energy).

The reason it works this way is because mass has a similar dimension dependence to its considerations as charges do, $\operatorname{kg} m^{d-3}$. Thus, you also have to modify the definition of $\vec{F}=m\vec{a}$. Now, $\vec{a}$ is clearly dimension independent (it's not extensive).

That's why we usually keep the units of $\epsilon_0$ dimension independent.