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Gauss' Law - Cylindrical Shell

  1. Oct 25, 2004 #1
    I'm stuck on two problems. I hope someone can help me. Here they are...

    1) For 1a I thought Q would be [tex]Q=\rho \pi L (b^2-a^2)[/tex] but since [tex]\rho=\frac{k}{r}[/tex] so [tex]Q=\frac {k \pi L (b^2-a^2)}{r}[/tex]. After being stumped on 1a I'm not sure how to go about 1b.

    2) I've derived about 4 equations for this problem (all wrong of course) and I get numbers like 4.7 N/C or so but never 7.2 N/C. I think the wire inside the cylinder is really screwing me up.

    I'd really appreciate some help. Even a little nudge in the right direction would be great. Thank you thank you thank you
  2. jcsd
  3. Oct 25, 2004 #2

    Claude Bile

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    I get

    [tex] Q = 2 \pi L k \ln{(b-a)} [/tex]

    for question 1 a), which disagrees with the answer you quoted. Perhaps somebody could verify this.

  4. Oct 26, 2004 #3
    Hey thanks for the reply; could you give me a quick explanation on how you got that answer? Just understanding the process you went through would really help me a lot. Thanks.
  5. Oct 26, 2004 #4


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    I find that the quoted answer is correct.

    [tex]Q=\int\rho dV=L\int\rho dA=L\int\rho (2\pi r)dr=2\pi Lk\Delta r[/tex]
  6. Oct 26, 2004 #5

    Claude Bile

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    Yes, I see my error now. Thanks krab.

    You need to start with definitions, in a question such as this one;

    [tex] Q = \int_{V} \rho d\tau [/tex]

    is a good place to start. Instead of performing a triple integral, utilize the symmetry of the cylinder to express dV in terms of dr, thus reducing it to a single integral.

  7. Oct 28, 2004 #6
    Ok I'm pretty sure I understand the first problem now.

    Like Krab said for 1a) [tex]Q=\int^b_a \frac{k}{r} 2 \pi r L dr = k 2 \pi L \int^b_a dr = k2\pi L (b-a)[/tex]
    Since the total charge is asked for you must integrate from a to b.

    To find the charge for 1b) you have to integrate from a to r so [tex] q=k2 \pi L(r-a)[/tex] and [tex]E=\frac{k2 \pi L(r-a)}{\varepsilon_0 2 \pi L r}[/tex] which is simply [tex]E=\frac{k(r-a)}{\varepsilon_0 r}[/tex]

    Thanks for the help. :biggrin: I'm still stuck on deriving an expression for #2 though. I'd really appreciate even just a hint to get me started. Thanks again!
  8. Nov 3, 2004 #7
    Ok, I've been working on this problem for an insane amount of time but I never get 7.2 N/C. Can someone check the expression I came up with and tell me if it's right? This is for the second question...

    [tex] \lambda L + \rho \pi L(b^2-a^2) = \varepsilon_0 E 2 \pi r L[/tex]


    [tex] E= \frac{\lambda + \rho \pi (b^2-a^2)}{\varepsilon_0 2 \pi r} [/tex]

    where lambda is the charge density of the wire, rho is the charge density of the cylinder, b is the outside radius, a is the inside radius, and r is the distance from the axis (r is greater than b).

    If I plug in the numbers from the problem I do not get the answer quoted. What am I doing wrong here?
  9. Nov 3, 2004 #8

    Doc Al

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    Your expression is no good. It also looks much more complicated than it needs to be. Hint: Since you only care about the field outside the cylinder, the net linear charge density is just the sum of that for the wire and the cylinder.
  10. Nov 7, 2004 #9
    Is this correct?

    [tex] \lambda L + \rho L = \varepsilon_0 E 2 \pi r L[/tex]


    [tex] E= \frac{\lambda + \rho }{\varepsilon_0 2 \pi r} [/tex]

    I just noticed that the problem gives the charge per unit LENGTH. Oops. Thanks for the help.
  11. Nov 7, 2004 #10

    Doc Al

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    Looks good to me. (Assuming that [itex]\lambda + \rho[/itex] represents the two linear charge densities.)
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