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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

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  • $\begingroup$ It's the line integral of the electric field, not the potential. $\endgroup$ – Rob Jeffries Oct 31 '16 at 21:36
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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!

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  • $\begingroup$ 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 Aug 9 '16 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 Aug 9 '16 at 23:16
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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 $V(b)=0$.

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  • $\begingroup$ Yup that is a much easier way of doing it. $\endgroup$ – Josh Pilipovsky Aug 9 '16 at 17:41

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