1
$\begingroup$

I having a problem with the line integral of electric potential.

I have a cylinder of radius $a$ and length $L$ with a uniformly surface charge. At point $b$ the potential is zero. I want to calculate the potential a $r$, i.e. $r>b$.

The integral is $\int_\gamma \vec E \,\, \cdot d\vec l$ and the electric field is $\vec E(\vec r)=\frac{Q}{2\pi r L\epsilon_0} \hat r$ (cylindrical coordinates, $\hat r=\hat x\cos\phi+\hat y\sin \phi$).

In multivariable calculus I use $\int_L \vec E \, \, \cdot d\vec r$ and parameterize the curve $L$ so $\int_L \vec E \, \, \cdot d\vec r= \int_P \vec E (\vec r(t)) \, \, \cdot \frac{d\vec r(t)}{dt}dt$.

But what is $d\vec l$?

I'm lost here, how can I find a curve to parameterize?

http://i67.tinypic.com/md2slj.jpg

Improve this question
$\endgroup$
1
  • $\begingroup$ It's the line integral of the electric field, not the potential. $\endgroup$
    – ProfRob
    Commented Oct 31, 2016 at 21:36

2 Answers 2

Reset to default
2
$\begingroup$

You want to parametrize as you said. Your starting point is $r = b$ and the endpoint is some $r = r_0$.

One way you can parametrize is by doing something like: $$ \vec{r}(t) = (1 - t) <b> + t < r_0>$$ where $t$ goes from 0 to 1. The rest I'll let you do!

Improve this answer
$\endgroup$
2
  • $\begingroup$ It was actually this method I was looking for. With cartesian coordinates, if $\vec E(\vec r)=\frac{Q}{2\pi r L\epsilon_0}( \hat x\cos\phi+\hat y\sin \phi)$ and $\vec r(t)=b+t(r_0-b)$ and $\frac{d\vec r}{dt}=r_0-b$ so $d\vec r=(r_0-b)dt$. But here, what is $\vec E(\vec r(t))$? And also, $\vec r(t)$ don't have any $\vec x$- and $\vec y$-components for the dot product? $\endgroup$
    – JDoeDoe
    Commented Aug 9, 2016 at 18:21
  • $\begingroup$ Good question! When I write $\vec{r}(t)$, I really mean $\vec{r}(t) = (1-t)(b\hat{r} + \phi_0 \hat{\phi} + z_o \hat{z}) +t(r_o \hat{r}+\phi_0 \hat{\phi} + z_0 \hat{z}) $, for some $\phi_0$ and $z_0$, they are completely arbitrary. Remember we are using cylindrical, not cartesian coordinates! Anyhow, you get your expression for $\vec{r}(t)$, which you got correctly but you did not put any direction to it, namely $\vec{r} (t) =( b + t(r_0 - b))\hat{r}+... $ Now take the derivative to get your velocity vector. $\vec{E}(\vec{r}(t))$ is just the expression you get when you plug in $\vec{r}(t)$. GL! $\endgroup$
    – Josh Pilipovsky
    Commented Aug 9, 2016 at 23:16
1
$\begingroup$

You don't need to parametrize. The electric field is radially symmetric, so you will integrate along a radius. $dl$ is the same as $dr$, so you have to integrate $E(r)dr$ between suitable limits - one of which will be $r=b$ where $V(b)=0$.

Improve this answer
$\endgroup$
1
  • $\begingroup$ Yup that is a much easier way of doing it. $\endgroup$
    – Josh Pilipovsky
    Commented Aug 9, 2016 at 17:41

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.