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In electrostatics, we have that the electric field of a charged wire is given by,

$$ E(r)=-\frac{\lambda}{r}, $$ where $\lambda$ is the wire charge density.

My question is; what about the constants in the potential? I know that the potential which gives this field can be written as $$ V(r)=\lambda\log(r), $$

however, the logarithm function needs to be dimensionless in the argument. I know that I can just put a constant with the same dimensions of $r$, like $V(r)=\lambda\log(r/a)$, and this does not affect the physics. Still, I want to know if there is a physical meaning for this constant? (in the electrostatic context).

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Still, I want to know if there is a physical meaning for this constant?

First of all the potential turn out to be $$V(r)=\lambda \ln(r)+c$$ the constant depends on the reference point, We are choosing. Not that you can not choose infinite as a point of zero potential because the wire is infinite (see here for more). Suppose we choose this constant to look like $$V(r)=\lambda \ln(r)+\lambda \ln(\alpha)=\lambda\ln(\frac{r}{\alpha})$$ There only interpretation for the constant is that it determine your reference point means the point where you have assigned the potential arbitary and all other point are relative to that point. For instance $$V(1)=\lambda\ln(1)+\lambda \ln(\alpha)\Rightarrow \lambda\ln(\alpha)=V(1)=0$$ So that $$V(r)=\lambda\ln(r)$$

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