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Homework Help: Cylindrical Capacitor Problem

  1. Mar 8, 2005 #1


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    For some reason, I am having a lot of difficulty finding the electric field between two co-axial cylinders. In fact, I pretty much know that it should be:

    [tex]\frac{Q}{2\pi \epsilon _0 Lr}[/tex]

    Where Q and -Q are the charges on the two cylinders, L is the length of the cylinders, and r is the distance from the axis. But I don't know how to come to this formula. It appears to me from this that the electric field of one cylinder would be:

    [tex]\frac{Q}{4 \pi \epsilon _0 L r}[/tex]

    where, again, r is the distance from the axis, and it wouldn't matter if r were greater than the radius or less than it. Anyways, I've been trying to find the electric field of one cylinder, then I'll add the two to get the electric field between the two. If we place the cylinder so that it's axis is the z axis, and we place a test charge at some point (r, 0, 0), the only field at this point will point in the x direction.

    So I get:

    [tex]E(r,0,0) = \int \frac{\sigma da}{4\pi \epsilon _0}\frac{1}{((r - x)^2 + y^2 + z^2)^{3/2}}(r - x, y, z)[/tex]

    But since only the field in the x direction will really matter, we can say:

    [tex]E(r,0,0) = \hat{x}\frac{\sigma}{4\pi \epsilon _0}\int \frac{da (r - x)}{((r - x)^2 + y^2 + z^2)^{3/2}}[/tex]

    Now, da will not vary with the z position or the azimuthal angle [itex]\phi[/itex], so [itex]da = c\, dz\, d\phi [/itex], for some constant real c. Since the area is [itex]2\pi RL[/itex], where R is the radius of the cylinder, L is its length, then c = R. Therefore, we get:

    [tex]E(r,0,0) = \hat{x}\frac{R \sigma}{4\pi \epsilon _0}\int _0 ^{2\pi} \int _{-L/2} ^{L/2} \frac{dz\, d\phi \, (r - R\cos \phi)}{((r - R\cos \phi)^2 + (R\sin \phi)^2 + z^2)^{3/2}}[/tex]

    Am I on the right track? From here, I've tried substiting something in the form z = Atan(phi), and I get somewhere, but it eventually gets very ugly and unsolvable. I've been trying this for hours, help is greatly appreciated. Thanks.
  2. jcsd
  3. Mar 8, 2005 #2
    Try to use the Gauss's law instead.....because is simpler!
  4. Mar 9, 2005 #3


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    Thanks clive, somehow, what you said helped. I actually tried using Gauss's Law before, but wasn't using the right kind of Gaussian surface; somehow, after reading your post, I instantly knew what type of surface to use. Thanks.
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