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Electric field of a uniform finite cylinder

  1. Feb 2, 2012 #1
    1. The problem statement, all variables and given/known data

    I have a solid cylinder of uniform charge density whose axis is centered along the z-axis. I am trying to calculate the electric field at a point on the z-axis.

    What I'm trying to do is to start by first calculating the field of a disk centered on the z-axis at a point on the z-axis, then sum up a bunch of disks to find the field of the cylinder.

    2. Relevant equations

    I find that the electric field of a uniform disk at a point on the z-axis is given by:

    [itex]2\pi\rho[1-\frac{z}{\sqrt{z^{2}+R^{2}}}][/itex]

    Where R is the radius of the disk and rho is the charge density.

    Now I want to write this as:

    [itex]2\pi\rho[1-\frac{z-z'}{\sqrt{(z-z')^{2}+R^{2}}}][/itex]

    Where z' is the position on the z-axis of the nth disk, and z-z' is the distance between the disk and the point of interest. Integrating the above expression over the length of the cylinder, however, leads to an electric field which increases with z and approaches a constant value, rather than decreasing and dropping off to zero at infinity.

    I'm not sure what I'm doing wrong, but I know that I'm evaluating the above integral correctly, and I know the expression for the field of uniform disk is correct, so I must not be summing the disks properly... any help?
     
  2. jcsd
  3. Feb 2, 2012 #2

    rude man

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    Inside and outside the cylinder?
     
  4. Feb 4, 2012 #3
    I think it'll be easier if you start by finding the potential at z and checking it for consistency. As for summing the disks, you can choose easy points as the limits for your integral. For example, you can set the point z where you want to know the E-field at the origin and and say that the left end of the cylinder is at z. Where would the right end then be? ;)
     
  5. Feb 4, 2012 #4
    I remember doing this problem at one point or another, and I remember that the integrals had the potential to be pretty nasty. What answer are you getting? Explicitly, what integral are you doing?

    You may be getting the right answer, just in a different form.
     
    Last edited: Feb 4, 2012
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