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Calculate current flowing from one cylinder to another

  1. Jul 23, 2017 #1
    1. The problem statement, all variables and given/known data
    Two long concentric cylinders of radii .04 m and .08 m are separated by aluminum. The inner cylinder has a charge per unit length of [itex] \Lambda [/itex] at any time. When the two cylinders are maintained at a constant potential difference of 2 V via an external source, calculate the current from one cylinder to the other if the cylinders are 1 m long

    2. Relevant equations
    potential difference = IR
    R = pl / A
    I = dq/dt
    J = I/A
    J = neV
    E = pJ

    3. The attempt at a solution

    So my first attempt I used:

    potential difference = IR

    (2 V ) / (R) = I

    (R) = pl / A

    length is 1 meter
    I took area by pi(r)^2 outside - pi(r)^2 inside, pi(.08)^2 - pi(.04)^2 , .020 - .005 = .015 m^2
    p is given in the book, p for aluminum is 2.655x10^-8

    plugging in, R = 1.77 x 10^-6

    so (2 V) / R = I

    (2 V ) / (1.77 x 10^-6) = 1.1 x 10^6

    But my answer is way off from my books, with the answer being: 6.8 x 10^8 A

    What did I do wrong? I think the lambda at the beginning is a hint and that I might have to use Guass law??


    My next thought was:

    E according to gauss law = (lambda)/(2pi\epsilon0r)

    E/p = J

    J * A = I

    but how am I suppose to get a value for lambda?? sorry that part gets me mixed up :/

    Can anyone comment on which method they think is leading me to the right direction? Also something I might be doing wrong??
     
    Last edited: Jul 23, 2017
  2. jcsd
  3. Jul 23, 2017 #2

    gneill

    User Avatar

    Staff: Mentor

    Your method for the calculation of the resistance of the aluminum tube is not correct. The current is flowing laterally, from the inside surface of the tube to its outside surface, not lengthwise down the tube. You'll want to find that resistance using integration -- think of the tube as consisting of nested cylinders of differential thickness.
     
  4. Jul 23, 2017 #3
    Ok this is starting to make sense to me... am I right when I say:

    I am using (dR) = d(pl / A)

    I am going to be integrating with respect to dr? my integral is going to go from inside radius to big radius?

    Since im taking the Area of the tube part, the inside tubes area is circumfrence * length right?

    so A = 2pi(r_inside) * length of the cylinders (L)

    but I'm confused on the top part.

    Since l is not the length of the cylinders, but instead its coming from the surface of the inside cylinder to the surface of the outside one.

    So if I replaced this with r, considering that I have A = 2pi(r_inside) * length of the cylinders (L) on the bottom, the r's would cancel and I would be left with dr?

    so [itex] R = (p/2piL) \int_{r_in}^{r_out} dr [/itex]

    is that my integral?
     
    Last edited: Jul 23, 2017
  5. Jul 23, 2017 #4
    Or maybe because I know that .08 m - .04 m = .04 m, I can substitute that for l, pull p and l out, a = 2 pi r l, so I pull out pl/2pi(l)


    that way I'm just integrating dr/r, which is ln(x) = ln(outside/inside radius)
     
  6. Jul 23, 2017 #5
    Ok, ignore that top post.

    I have R = pL / A

    so I must integrate

    p dr / 2pirL

    p/2piL integral dr/r from .04 to .08

    p/2piL ln(.08/.04) = R

    wow guys.

    After subbing that value of R into I = RV, i finally got the same answer as my text book.

    I can't believe I made the integration right :DD thanks as usual Gniel your a gangsta bro
     
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