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Homework Help: Gaussian cylinder problem

  1. Sep 9, 2007 #1
    Hi all - I reposted this here, as I posted in the advanced forum on accident:

    Here's the problem I am having trouble with:

    A very long, solid insulating cylinder with radius "R" has a cylindrical hole with radius "a" bored along its entire length. The axis of the hole is a distance "b" from the axis of the cylinder, where a< b< R. The solid material of the cylinder has a uniform volume charge density.

    Find the magnitude and direction of the electric field inside the hole using a,b(vector), [tex]\rho[/tex], [tex]\epsilon[/tex], and R.

    The picture attached describes the surface. I really don't know where to start with this.

    I believe that the field, "E", between the outer surface and the inner hole should = 0. Given that, we have a field inside the bored hole, and on the outer surface of the full cylinder. What I am getting hung up on is the usage of uniform charge density in the cylinder - I know that [tex]\rho[/tex]= Q/V(cylinder), but because of the infinite length, I am unsure if this is a necessary step. Because of the infinite length, should I use a gaussian surface to find the field on the outside to use somehow for the inside field?

    I am quite confused with this one, and I am really only looking for a place to start chipping away at it - If anyone has any suggestions as to the approach of this problem, I would highly appreciate it.



    Attached Files:

  2. jcsd
  3. Sep 9, 2007 #2

    Doc Al

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    Staff: Mentor

    Here's a hint for you: Treat the uncharged hole as a superposition of + and - charges.
  4. Sep 9, 2007 #3
    Thanks -

    By that, do you mean that I should replace the hole with point charges that equal out to a 0 net charge? Sorry, I'm not sure if I'm following you exactly

    Thanks again

  5. Sep 9, 2007 #4

    Doc Al

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    Staff: Mentor

    Here's what I mean: Imagine there was no hole in the cylinder. You'd have no problem finding the field at any point, right? Take that uniformly charged cylinder and add a second, negatively charged cylinder to create a hole. You should be able to find the field from that second cylinder of charge. The net field is the superposition of the fields from each cylinder. Make sense?
  6. Sep 11, 2007 #5
    Doc -

    Sorry it took a bit to follow up. I think I understand what you are saying - By using superposition, we are effectively able to substitute a -charge density for the small cylinder. Because the charge density, [tex]\rho[/tex], is 0 for the small cylinder, we can establish -[tex]\rho[/tex] to allow it to equal zero?

    I am finding this problem really hard for some reason, so I appreciate your help so far. I have attached my work that I have done, and I started by separating the cylinders to make things a little more clear for myself. Is this the right direction so far? If it's a bunch of jibberish, it's because I am still a bit confused on how to proceed



    Last edited: Sep 11, 2007
  7. Sep 12, 2007 #6

    Doc Al

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    Staff: Mentor

    Here's the idea. We are going to replace the charged cylinder with a hole in it, the field of which is not obvious, with two solid cylinders of charge, for which the field is easy to find. One cylinder is the big one with charge density [itex]\rho[/itex], the other is a small one of charge density [itex]- \rho[/itex]. (The trick is to recognize that these configurations are equivalent.) Now the field at any point can "easily" be found by just adding up the field from each solid cylinder.

    You should be able to find the field at any point within a solid cylinder of charge using Gauss's law. (Solve that problem first.) Once you are able to do that, you just have to noodle around with expressing the field as a function of the given variables (including the vectors a & b). It's not clear to me exactly how they want you to describe the position within the hole--you have several ways to go.
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