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Electric potential of cylinder

  1. Feb 20, 2007 #1
    1. The problem statement, all variables and given/known data

    A solid cylinder with radius R and length L has uniform charge density ro. Its base is in the x-y plane and it's axis is coincident with the z-axis (symmetrical about the z-axis)

    2. Relevant equations

    Find the electric potential at point P outside the cylinder at a distance z from the origin (P is on the z-axis)

    3. The attempt at a solution

    So I compared this to finding the potential at a distance z above a uniformly charged disk. In the case of a disk, I divided the disk into concentric rings, found the potential contribution due to one ring, and then integrated over all rings with the limits 0 to R.
    I tried to use a similar argument and divide the cylinder into equal sized disks with radius R stacked on top of one another (so the only difference is the distance from the point P). I then want to integrate this over all disks with the limits 0 to z. So can I just use my answer for the potential above a charged disk, and then integrate it with respect to z to cover all the distances?
    thanks!
     
  2. jcsd
  3. Feb 20, 2007 #2
    You idea sounds correct.
     
  4. Aug 28, 2009 #3
    hi i have the same proplem but for hight 2h and radius (a) and charge density ro inside and outside the cylinder and i tryed to use gauss's law and find fy the electric potential but i'm not sure can some one help.
    i use
    V(r)=- integrate E.da=E.2pi a (2h) =1/ebslon .Qenc
     
  5. Aug 28, 2009 #4
    I would use Gauss's Law to find the E-vectors around and inside the cylinder, when remember that [tex]U=\int_{\infty}^r \vec{E}\bullet d\vec{l}[/tex] and [tex]V=\frac{U}{q}[/tex].

    A hint: Use a cylinder for your gaussian surface, with the same axis as your charged cylinder.
     
  6. Aug 28, 2009 #5
    Thanks, but what you mean by U, is it the electric potential ?
    I'm convusing between the notations !!
     
  7. Aug 28, 2009 #6

    gabbagabbahey

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    Hi eman2009, I think espen180 is confusing electrostatic potential energy and electrostatic potential.

    As for your original question, if the cylinder only has a finite length [itex]2h[/itex], does it still have the requisite cylindrical symmetry to use Gauss' Law?

    If not, you'll have to find the potential through another means....can you think of any formulas that directly relate potential to charge density?
     
  8. Aug 28, 2009 #7
    how about

    close INT E . da=1/ebsolon Qenc

    as in Griffiths book(Introduction to Electrodynamic) p.68, equation 2.13
    and
    Q=INT RO dt t(tao) is the infinitesimal displacement

    dt for cylinder

    dt=4 pi R^2 dr
     
  9. Aug 28, 2009 #8

    gabbagabbahey

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    That's just Gauss' Law, and it is only useful in cases where symmetry allows you to pull E outside of the integral....is a cylinder of finite length one of those cases?

    Instead, try equation 2.29, it directly relates rho to V.
     
  10. Aug 29, 2009 #9
    Sorry, I mistook your cylinder as one of infinate length. As for my notation, U is the electrostatic potential energy and V is the electric potential.
     
  11. Aug 29, 2009 #10

    gabbagabbahey

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    Sure, but wouldn't you also say

    [tex]U(r)=-\int_{\infty}^r \vec{F}\bullet d\vec{l}\neq\int_{\infty}^r \vec{E}\bullet d\vec{l}[/tex]

    ?:wink:
     
  12. Aug 29, 2009 #11
    Gah! You're right! Sorry about that. It should be an F, not an E.
     
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