Calculation of electric potential in Geiger-Müller tube We are given a cylindrical rod with linear charge density $λ$, and a coaxial cylinder with linear charge density $-λ$, as shown in the picture. 

We are to show that the potential difference between $r_a$ and $r_b$ is 
$$ΔV=2 k_e λ \ln\big( \frac{r_a}{r_b} \big).$$
We can easily calculate from Gauss' Law that the magnitude of the electric field at a distance $r$ from the axis of symmetry is $E=\frac{2 k_e λ}{r}.$ From the definition of potential difference we have
$$ΔV=- \int_{\vec{r_a}}^{\vec{r_b}} \vec{E} \cdot d\vec{s}.$$
Now, the field points radially outward, and $d\vec{s}=-dr \hat{r}$, so $\vec{E} \cdot d\vec{s}=-Edr$. Therefore we should get
$$ΔV=- \int_{\vec{r_a}}^{\vec{r_b}} \vec{E} \cdot d\vec{s}= \int_{r_a}^{r_b} \frac{2 k_e λ}{r} dr=2k_eλ\ln \big( \frac{r_b}{r_a}\big).$$ Why do I get the sign wrong?
 A: The misconception arose from the fact that $d\vec{s}$ was taken at a specific path (that is, the radial path from the outer cylinder to the rod). However, one need not take this specific path. $dr=ds \cos{θ}$, does not carry a sign, as the cosine takes care of that. What I meant by $\hat{r} \cdot d\vec{s}=-dr$ was $d\vec{s}\cdot\hat{r}=-ds$, and that is why I got it wrong.
A: $E=\frac{2 k_e λ}{r}$ is an electric field pointing radially outwards.
Now there are two equivalent ways to define the potential difference between two points.

The first definition is that the potential difference between two points is minus the work done by the electric field in taking unit positive charge from the first point to the second point.
Using this definition.
The work done by the electric field in moving unit positive a distance $dr$ radially outwards is $E\; dr =\dfrac{2 k_e λ}{r}\;dr$ 
So the total work done by the electric field in moving unit positive charge from $r_b$ to $r_a$ is   
$\displaystyle \int_{r_b}^{r_a} \dfrac{2 k_e λ}{r} dr=2k_eλ\ln \big( \dfrac{r_a}{r_b}\big)$
and this is $- (V_a - V_b) = (V_b - V_a) = \Delta V$

The second definition is that the potential difference between two points is  the work done by an external force in taking unit positive charge from the first point to the second point.  
Using this definition.
The work done by the external force in moving unit positive a distance $dr$ radially outwards is $-E\; dr =-\frac{2 k_e λ}{r}\;dr$ 
So the total work done by the external force in moving unit positive charge from $r_b$ to $r_a$ is   
$\displaystyle \int_{r_b}^{r_a} -\dfrac{2 k_e λ}{r} dr=-2k_eλ\ln \big( \frac{r_a}{r_b}\big) = (V_a - V_b)$ 
$\Rightarrow 2k_eλ\ln \big( \frac{r_a}{r_b}\big) = (V_b - V_a) = \Delta V$
