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Potential of an infinite line of charge

  1. Sep 15, 2009 #1
    1. The problem statement, all variables and given/known data
    Find a potential a distance r from an infinitely long straight wire that carries a uniform line charge [tex]\lambda[/tex]. Compute the gradient of your potential and check that it yields the correct field.

    2. Relevant equations
    [tex]V=[/tex][tex]\frac{KQ}{R}[/tex]
    [tex]\oint E \bullet dS=\frac{Q}{episolon}[/tex]

    3. The attempt at a solution
    I tried doing it the V=kq/r way, but then I realized it doesnt work, since for that it assumes that potential is zero infinitely far away. But it isn't. The problem doesn't want me to find the e-field first, which would be a lot easier, any tips?

    Do i need to go to the most basic definition of potential?
     
  2. jcsd
  3. Sep 15, 2009 #2

    gabbagabbahey

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    V=kq/r gives the potential of a point charge.

    For a continuous linear charge distribution,

    [tex]V(\textbf{r})=\frac{1}{4\pi\epsilon_0}\int \frac{\lambda(\textbf{r}')dl'}{|\textbf{r}-\textbf{r}'|}[/tex]
     
  4. Sep 15, 2009 #3
    Hm, yeah I got that.
    This is what I have right now.

    [tex]V=\frac{1}{4\pi\epsilon_o}\int\frac{\lambda dx}{\sqrt{r^{2}+x^{2}}}[/tex]

    But the integral blows up when I integrate from negative to positive infinity.
     
  5. Sep 15, 2009 #4

    gabbagabbahey

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    Yes, I suppose it does.

    So, you can either compute the field first using Gauss' law and then find the potential using

    [tex]V(\textbf{r})=-\int_{\mathcal{O}}^{\textbf{r}}\textbf{E}\cdot d\textbf{l}[/tex]

    Or, if you aren't allowed to do that, use separation of variables to solve Laplace's equation in cylindrical coordinates.
     
  6. Sep 15, 2009 #5
    I can do that? Please advise on that second part.

    This is using [tex]\nabla^{2} V = \frac{\rho}{\epsilon_o}[/tex]??
     
  7. Sep 15, 2009 #6

    gabbagabbahey

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    Yes (sorry , I meant Poisson's equation, not Laplace's), based on symmetry, what variables would you expect [itex]V[/itex] to depend on? The distance from the axis? The axial coordinate? The azimuthal coordinate? You can use that to express [itex]\nabla^2V[/itex] in terms of ordinary derivatives, giving you a second order ODE to solve.
     
  8. Sep 15, 2009 #7
    That's out of the league of the class. I could do it though...

    I know the potential field is suppose to be...

    [tex]V = -\frac{\sigma}{2\pi \epsilon_o} ln(r)[/tex]

    Is there any other way to solve this? I've exhausted all possibilities.

    QUESTION:
    For the
    [tex]V=\frac{kq}{r}[/tex]
    does that assume that potential is 0 [tex]\infty[/tex] far away?
     
    Last edited: Sep 16, 2009
  9. Sep 15, 2009 #8

    gabbagabbahey

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    If you haven't been taught this method yet, then I would guess you are expected to first calculate the electric field and then integrate....the question doesn't explicitly tell you not to do that does it?

    Yes,

    [tex]V(\textbf{r})=\frac{1}{4\pi\epsilon_0}\int \frac{\lambda(\textbf{r}')dl'}{|\textbf{r}-\textbf{r}'|}[/tex]

    is a solution (in integral form) of Poisson's equation, subject to the boundary condition [itex]V\to 0[/itex] at [itex]|\textbf{r}|\to \infty[/itex].
     
  10. Sep 15, 2009 #9
    Well, I'm just going to go to the professor's office hour, thanks though!
     
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