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Parallel Component for E over charged disk

  1. Mar 25, 2006 #1
    (NOTE: I am doing this to possibly re-enter a graduate program and earn a PhD. This is not being done for class.)

    DJ Griffiths, Prob. 2.6, solved.

    My question is about the parallel component of the [tex]\vec{E}[/tex]-field. If you integrate over cylindrical coordinates, then by symmetry this component cancels as you integrate around [tex]\phi[/tex]. So my question is about that integral - will it integrate to zero, or are we to ad hoc cancel the [tex]\phi = 0[/tex] term for [tex]dq[/tex] with the [tex]\phi = 180[/tex] term for [tex]dq[/tex]?

    Here is the integral as I set it up:

    [tex]2\pik\sigma\int{\frac{s^2}{(z^2 + s^2)^{\frac{3}{2}}}}ds[/tex],

    where [tex]s, \phi, z[/tex] are the cylindrical coordinates.

    Oh, and yes, I did try [tex]s = z \cdot tan(\theta)[/tex]

    and get [tex]\int{\frac{1}{cos(\theta)}}d\theta[/tex] plus another term which is handleable.

    Now that I think of it, the integral should integrate to zero since it is for the full disk, namely: [tex]0 < \phi < 2 \pi[/tex]. So it should naturally integrate to zero. ...no? :redface:

    Any help would be appreciated.
    Last edited: Mar 25, 2006
  2. jcsd
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