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Homework Help: Calculating V Involving a Tube

  1. Sep 20, 2010 #1
    1. The problem statement, all variables and given/known data

    You need to calculate the electric potential for a thin, long tube with a uniform surface charge density (σ). This tube has a length L and a radius R. Find the electric potential along the tube axis for points outside the tube.

    2. Relevant equations



    3. The attempt at a solution

    I am having trouble getting this problem. If I am understanding it correctly, we are supposed to calculated V either above or below the tube, as long as it is along the tube's axis. I am not sure how to set up my problem. Should I set the origin of my axes between L/2 and -L/2? I am not sure where to start. Thanks.
     
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  3. Sep 21, 2010 #2
    I would just find E using Gauss's law, usually if a problem specifies that a tube is long then you are free to ignore the ends of the tube (i.e. treat as if it were infinite). Then use V=-∫E*dl
     
  4. Sep 21, 2010 #3
    But it says that it has a length of L, so it isn't infinite (I think)? Also, I think he wants E along the axis of the tube, so wouldn't the ends be the only parts that contribute. This is where I'm confused.
     
  5. Sep 21, 2010 #4
    My teacher told us today in class to calculate V directly using the formula for it. He told me not to use the integral of E dl.
     
  6. Sep 21, 2010 #5

    gabbagabbahey

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  7. Sep 21, 2010 #6
    Ok so I agree with everything you said. I tried to use the equation v=k Integral[(sigma*Pi*R^2)/z]dr. When i convert to cylindrical coords, everything that was an 'r' term, becomes an 's' term. Does it then become a tripe integral where I integrate with respect to S, phi, and z?
     
  8. Sep 21, 2010 #7

    gabbagabbahey

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    The general form for a volume charge distribution [itex]\rho(\textbf{r})[/itex] is

    [tex]V(\textbf{r})=\frac{1}{4\pi\epsilon_0}\int\frac{\rho(\textbf{r}')d^3r'}{|\textbf{r}-\textbf{r}'|}[/tex]

    which is a triple integral (volume integral). In the case of surface charge distributions [itex]\sigma(\textbf{r})[/itex], the integral simplifies to a double integral (surface integral):

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

    and for linear charge distributions [itex]\lambda[/itex], the integral simplifies to a single integral (line or path integral):

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


    So, for the surface charge distribuion in this problem, you will want to use the double integral form....
     
  9. Sep 21, 2010 #8
    Oh yeah that's right. I keep wanting to use a volume charge, even though the problem gave us a surface charge.

    So the denominator of that integrand is just 'z' and dA becomes dtheta dr?

    If that is correct, then I'm not really accounting for the non-endcaps part of the cylinder, am I?
     
  10. Sep 21, 2010 #9

    gabbagabbahey

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    No, the field point (the point(s) at which you want to determine the potential) is at [itex]\textbf{r}=z\mathbf{\hat{z}}[/itex], but what about the source points (the points where there is charge that creates the potential)? Where are they? What does that make [itex]\textbf{r}'[/itex] ( their position(s) )? And [itex]|\textbf{r}-\textbf{r}'|[/itex] (the distance from each source point to the field point)?

    Close, but you're missing something (what are the units of area? what are the units of your expression?) :wink:
     
  11. Sep 21, 2010 #10
    Would the source points be (R, Phi, Z) where Z is -L/2<z<L/2?

    And would dA become r*dtheta*dr since the r*dr factor give the proper units?
     
  12. Sep 21, 2010 #11

    gabbagabbahey

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    Right, in unit vector notation, that means [itex]\textbf{r}'=R\mathbf{\hat{s}}+z'\mathbf{\hat{z}}[/itex] (where I use [itex]s[/itex] as the radial coordinate, so you don't confuse it with the distance from the origin [itex]r=|\textbf{r}|[/itex] )...if you are wondering where the [itex]\phi[/itex] dependence is, just draw a picture...you will see that the direction of [itex]\mathbf{\hat{s}}[/itex] depends on [itex]\phi[/itex].

    So, what does that make [itex]|\textbf{r}-\textbf{r}'|[/itex]? What does that make your integral?
     
  13. Sep 21, 2010 #12
    Ok before I start the integral, I want to make sure that value for the denominator is correct. Would that value become (z-z', -Phi, -R)?
    I tried to draw a picture to better visualize this, but I think the Phi is still throwing me off a little bit.
     
  14. Sep 21, 2010 #13

    gabbagabbahey

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    If the phi is throwing you off, Google "position vector in cylindrical coordinates", you'll probably find some useful sites.

    I'm not sure what you mean by the ordered triplet (z-z', -Phi, -R)? An ordered triplet usually represents a vector, and in cylindrical coordinates, it is usually in the order [itex](s,\phi,z)[/itex]. The value in the denominator shouldn't be a vector, it should be the magnitude of the vector [itex]\textbf{r}-\textbf{r}'[/itex] (the distance from the source point to the field point....distance is a scalar quantity, not a vector).
     
  15. Sep 21, 2010 #14
    Oh ok I think I got the phi now.

    I am trying to set it up as the correct vector, before I take the magnitude. Is the difference in the s-components [0-R]? And for the Phi compononet is it [0-Phi] (this is the most confusing one to me), and for the z-components [z-z']?
     
  16. Sep 21, 2010 #15

    gabbagabbahey

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    Well, there is no phi comonent for a position vector (some vector fields, like an electric/magnetic field will have phi components, but porsition vectors do not), but your other two are correct.

    If [itex]\textbf{r}=z\mathbf{\hat{z}}[/itex] and [itex]\textbf{r}'=R\mathbf{\hat{s}}+z'\mathbf{\hat{z}}[/itex], then you should immediately be able to say [itex]\textbf{r}-textbf{r}'=-R\mathbf{\hat{s}}+(z-z')\mathbf{\hat{z}}[/itex]... this is simple vector addition/subtraction in unit-vector notation.

    So, what is the magnitude of that vector (usually called the separation vector) then?
     
  17. Sep 21, 2010 #16
    Would it be sqrt[(R^2) + (z^2) -2zz' + (z'^2)]?

    And I have to go to campus now. I will try to respond when I am over there, but if I can't, I will definitely respond ASAP this evening.
     
  18. Sep 21, 2010 #17

    gabbagabbahey

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    Yup...now just calculate your integral
     
  19. Sep 21, 2010 #18
    Would all those values in the integral be constant except for 'r'? Wouldn't k, sigma, R, z, and z' all be constant? If that's true, then it just becomes the integral of 'r' dtheta dr where 'r' goes from 0 to R, and theta goes from 0 to 2Pi.
     
  20. Sep 22, 2010 #19

    gabbagabbahey

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    Why would [itex]z'[/itex] be constant....it's the z-coordinate of a given source point, and there are source points all along the cylinder from [itex]z'=-L/2[/itex] to [itex]z'=+L/2[/itex]...since [itex]z'[/itex] and [itex]\phi'[/itex] vary over the surface of the cylinder, you need to determine the area subtended when you vary each by a differential amount, you should find [itex]dA'=Rd\phi'dz'[/itex].
     
  21. Sep 22, 2010 #20
    Ok I understand now why z' isn't constant. I should have recognized that.

    I am a bit confused on your value of dA'. If I use that value, does it incorporate the surface charge on the endcaps? Charge is located all along the endcaps, not just at R distance from the axis.

    I guess I am still a bit confused on how you got dA'.
     
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