Q:

Parameterize curve and find differential line element

This is a physics problem, but I need help with the math.
I want to calculate the potential $V=\int_\gamma \vec E \cdot d\vec l$, where $\vec E(\vec r)=\frac{Q}{2\pi r L\epsilon_0} (\hat x \cos \phi +\hat y \sin \phi)$.
I don't understand how to find the curve $\gamma$ and $d\vec l$ from this description:
A cylinder of radius $a$ and length $L$. At the point $r=b$ the potential is zero and $b>a$. Calculate the potential at $r$, when $r>b$.
According to the solution the problem can be solved without parameterization of $\gamma$, however I want to learn the math.

 From the solution: $d\vec l=\hat r \, dr$ so $ V=-\frac{Q}{2\pi L \epsilon_0} \int_b^r \frac{1}{r'} \, dr'$

Thanks!

A:

There is a small problem with your equation for the electric field, it's not $\vec{x}$ and $\vec{y}$. It should be $\hat{x}$ and $\hat{y}$, so your last terms can be rewritten as $\hat{x}\cos\phi+\hat{y}\sin\phi=\hat{r}$.
Now going back to your problem. The electric field is isotropic, in the radial direction $\vec{E}(\vec{r})=\frac{Q}{2\pi r L \epsilon_0}\hat{r}$, with $\hat{r}=\frac{\vec{r}}{|\vec{r}|}$. You can now work in either polar coordinates, or cartesian coordinates. The solution is easier to write in polar coordinates, but the answer is the same. The potential difference $V(\vec{r})-V(\vec{r_0})=\int_\gamma \vec{E}d\vec{l}$. $\gamma$ is any curve that goes from $\vec{r_0}$ to $\vec{r}$. It does not matter which path you choose, the integral has to be the same. We know that the electric field magnitude is constant on a cylinder concentric to the charge distribution in this problem. So for a given point $\vec{r}$, you should choose $\vec{r_0}=b \hat{r}$. In this case the shortest path is a simple straight line, with $d\vec{l}=\hat{r}dr$. If you plug in $V(b \hat{r})=V(b)=0$ you get the solution.
One more thing: let's suppose that we chose $\vec{r_0}$ not colinear with $\vec{r}$. As I mentioned before, you can choose any path between the two. One can go from $\vec{r_0}$ radially outward up to a distance $r$, then go along a circular arc of radius $r$, with the center at the center of the cylinder. For the first part of the path, you get the same integral as before. For the second part of the path $d\vec{l}=\hat{\phi}rd\phi$, where $\hat{\phi}$ is the tangential direction (go along the circle). We know that the dot product $\hat{r}\cdot\hat{\phi}=0$, so the integral along this path is 0. In general, for this problem, you can choose any path and decompose it into segments along the radial direction, and arcs concentric with the cylinder

