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I have a very basic question on the units for the equation of the potential at a distance: $z$ from a single line charge.

The electric potential due to a very long line charge at some distance $z$ is given by:

$$V(z)=-\frac{\lambda}{2\pi\epsilon_0}\ln(z)+C$$

If I look at the units of the terms, $\lambda$ is a line charge; so the units are $\frac{C}{m}$.

The denominator: $2\pi\epsilon_0$ is expressed in units of $\frac{F}{m}$.

This makes the units of $\frac{\lambda}{2\pi\epsilon_0}$: $\frac{C}{F}$ = $V$.

This implies that the term $\ln(z)$ is unitless and that our constant $C$ is in $V$.

For this to be true; $z$ must be a ration of distances; so that the meters cancel.

Would a verbose version of this equation be:

$$V(z)=-\frac{\lambda}{2\pi\epsilon_0}\ln(\frac{z}{1})+C$$

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  • $\begingroup$ The second log is hidden in the constant C, which will be proportional to $\log{(z_0)}$ which is your reference point of some known potential, the log will combine and be the logs of a ratio of distances. $\endgroup$ – Triatticus Nov 9 '18 at 16:45
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Yes, and in fact this isn't just true for this specific case, but it should be true in general that any argument of $\ln$ should be dimensionless.

One intuitive reason for this was explained to me by my professor, when he said to consider what might happen if you Taylor expand $\ln(1+x)$ (using this for simplicity):

$$\ln(1+x)=x-\frac{1}{2}x^2+\frac{1}{3}x^3-\frac{1}{4}x^4+...$$

If $x$ is not dimensionless, then (a) you shouldn't be adding it to $1$ in the first place, but more importantly (b) the dimensions of the series expansion explode to $\infty$, which is not good!

Many people (myself included sometimes) are guilty of writing things like $\ln(z)$, and I think it's fine in most cases, as long as you're aware that you should actually have a reference length (e.g. of $z=1\text{ m}$) implicitly included in the argument.

Note that this is also true for arguments of $\sin$, $\cos$, $\exp$ etc - the arguments should always be dimensionless.

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