Electric field of infinite line VS very thin infinite cylinder

Question

The electric field outside a charged full infinite cylinder is (by Guass's law's integral form):
$$E(r)=frac{rho R^2}{epsilon r}$$
And the field induced by an infinite charged line:
$$E(r)=frac{sigma}{2pi epsilon r}$$
Setting $rho=sigma=1$ and $Rleq frac{1}{2pi}$ we find that the field from a 3-d charge is equal or weaker than the field of a 1-d charge. To me that seems like a contradiction, because "less charge" creates a stronger field. Maybe I can't claim that $rho = sigma$ since they have different units, maybe this while thing is not a  problem, either way I'd like to understand this better. Thanks.

d_b · Accepted Answer

I will stick with your notation and write the linear charge density as $sigma$, instead of the more common $lambda$. The choice of $rho = sigma = 1$ is not wrong exactly, since you can choose the units of charge and length to be whatever you want. However, this probably does not capture what you actually want to look at.
To see why, let's write the volume charge density $rho$ of the cylinder in terms of its linear charge density, which I'll call $sigma_{c}$. The charge of a length $L$ of the cylinder is
begin{align}
q = rho pi R^2 L,
end{align}
so the charge per unit length is
$sigma_c = q/L = rho pi R^2$
and
begin{align}
rho = frac{sigma_c}{pi R^2}
end{align}
By choosing units such that $rho = sigma$, you are choosing
begin{align}
sigma = frac{sigma_c}{pi R^2}.
end{align}
If you then impose $R leq 1 /(2 pi)$, you are guaranteeing that the line charge has a larger linear charge density than the cylinder, so it is not surprising that the field of the line is stronger.
What you probably want to do instead is compare the fields of both configurations written in terms of their linear charge density. If you do this, you don't need to mess around with the units. You will find
begin{align}
E_{text{line}}(r) = frac{sigma}{2pi epsilon r}
end{align}
and
begin{align}
E_{text{cyl}}(r) = frac{sigma_c}{2pi epsilon r}.
end{align}
That is, the dependence of the field strength on the linear charge density is the same for both charge configurations (assuming we observe the cylinder from outside). Whichever one has a larger linear charge density will have a larger electric field.
The only sense in which the cylinder can have "more charge" than the line is if the linear charge density is larger. The dimensionality of the charge configurations is irrelevant, and the cylinder doesn't have "more charge" simply because it's extended over more dimensions.

march · Answer

There's some confusing bits about the question, which we'll sort out first.

Conventionally, we use $rho$ for a volumetric charge density, $sigma$ for a surface charge density, and $lambda$ for a linear charge density.  So I prefer to write the expression for the field of the wire as
$$
E_{textrm{wire}}=frac{lambda}{2pi epsilon_0 r}.
$$

Your expression for the other electric field is wrong, whether $rho$ is interpreted as a volumetric or surface charge density.  The correct expressions are
$$
E_{textrm{surface of cylinder}}=frac{sigma R}{epsilon_0 r},
$$
and
$$
E_{textrm{volume of cylinder}}=frac{rho R^2}{2epsilon_0 r},
$$

With those things out of the way, we can proceed with the question.  The point of taking some limit is to see that these all coincide with each other.  To that end, let's consider a cylindrical Gaussian surface of length $l$ that contains all the charge and enforce the restriction that we want the total amount of charge in these equal-length segments of the objects to be equal.
For the cylinder with the volume charge, the total charge is the product of the volumetric charge density and the volume of the cylinder, i.e.
$$
q=rho pi R^2 l.
$$
For the cylinder with the surface charge, the total charge is the product of the surface charge density and the surface of the cylinder, i.e.
$$
q=sigma 2pi R l.
$$
For the line charge, the total charge is the product of the linear charge density and the length of the cylinder, i.e.
$$
q=lambda l.
$$
If we re-write the electric field expressions in terms of this same charge $q$, we always get the same thing:
$$
E_{textrm{vol}}=frac{rho R^2}{2epsilon_0 r}
=frac{q}{pi R^2 l}frac{R^2}{2epsilon_0 r}
=frac{q}{2piepsilon_0 l r},
$$
and
$$
E_{textrm{sur}}=frac{rho R}{epsilon_0 r}
=frac{q}{2pi R l}frac{R}{epsilon_0 r}
=frac{q}{2piepsilon_0 l r},
$$
and
$$
E_{textrm{lin}}=frac{lambda}{2piepsilon_0 r}
=frac{q}{l}frac{1}{2piepsilon_0 r}.
$$

Also see here.

Electric field of infinite line VS very thin infinite cylinder

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