1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Gauss Law:Wired in a pipe!

  1. Jan 16, 2008 #1
    1. The problem statement, all variables and given/known data

    Consider an infinitely long cylindrical metallic shell with a line of charge within and
    coincides with the axis of the cylindrical shell as shown in Figure . How does E field vary
    with r?


    2. Relevant equations

    E.A=q/[tex]\epsilon[/tex]

    E=E field, A = surface area

    3. The attempt at a solution

    The Electric field within the inner circle of cylinder +Q pointing toward outer space.assuming charge at the wire is +Q. however the the area inside the thickness of the pipes has ZERO Electric field because the inner surface of the pipe is induced with -Q, while the outer surface is having total charge of +Q.

    Area surrounding the pipe surface having E field pointing outward.

    [​IMG]
    is this explanation make sense, or how should i write in a more presentable manner? thanks
     
  2. jcsd
  3. Jan 16, 2008 #2

    Doc Al

    User Avatar

    Staff: Mentor

    What you say is correct (except that you use Q instead of the charge per unit length), but you didn't do what was asked: Express the field as a function of r.

    Hint: Call the charge per unit length [itex]\lambda[/itex]. Use Gauss's law to find an expression for E(r) in each of those three regions.
     
  4. Jan 16, 2008 #3
    Well your explanation is quite correct, but you would also want to give the exact expression for the electric field at varying distances from the line of charge. Just use Gauss' law.

    Edit: I am gettin old for this lol
     
  5. Jan 16, 2008 #4
    hi Dr. Al. ok, here is the derivation of the E(r) and r relationship.

    [tex]\Phi[/tex]=E A = Q / [tex]\epsilon[/tex].

    therefore, E = [tex]\frac{\lambda}{2\pi\epsilon*r}[/tex].

    [just couldnt get the latex working]
    The E field is = lamba / (2*pi*r*epsilon) ]


    From this, we predicted the E field is inversely proportional to the distance from the wire to the pipe, r. however the only exception is when E field is inside the wall of the pipe. it is ZERO as mention above.
     
    Last edited: Jan 16, 2008
  6. Jan 16, 2008 #5


    LOL, my explanation is at post #4. thanks for the reply.
     
  7. Jan 16, 2008 #6

    Doc Al

    User Avatar

    Staff: Mentor

    Looks good to me! (I'd mention that you used a cylindrical Gaussian surface. If its length is L, the charge within that surface is [itex]\lambda[/itex] L.)
     
  8. Jan 16, 2008 #7
    yeah, q= [itex]\lambda[/itex]* length
    while derive it, length was cancel out with the cylinder length.

    btw, my final doubt is, since we are ask to consider a INFINATELY long cylindrical shape, we still consider it as length, L. which only this that my explanation can establish.
     
  9. Jan 16, 2008 #8

    Doc Al

    User Avatar

    Staff: Mentor

    Cosidering an infinitely long cylinder allows you to assume that the field is uniform and radial (by symmetry) in your Gaussian section, regardless of where along the axis you place it. If the pipe or line charge were finite, and you were near the end, you'd expect some non-radial components (edge effects).
     
  10. Jan 16, 2008 #9
    Hi Doc, i finally got the whole big picture of this. just a minor clarification, [regardless of where along the axis you place it.] are you refering to the position of the charged wires is not necessary at the centre of the pipe? because i feel this is not real intituitive, as the formula suggest, the E field do depend on the radius.correct me if i m wrong and misinterpret the meaning.


    {If the pipe or line charge were finite, and you were near the end, you'd expect some non-radial components (edge effects)} this part i can understand. thanks again.
     
  11. Jan 17, 2008 #10
    hi Doc, i finally got your meaning of your post #08. please marked this thread as solved!.thanks .
     
  12. Jan 17, 2008 #11

    Doc Al

    User Avatar

    Staff: Mentor

    Excellent. (Looks like I forgot to respond to your last post. Oops! But you have it now.)

    As far as marking the thread solved, you should be able to do it: Click on thread tools and you should see an option. :cool:
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?