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Electromagnetism: Gauss's Law

  1. Sep 27, 2011 #1
    1. The problem statement, all variables and given/known data
    Consider an infinitely long charged cylinder of radius R, carrying a charge whose density varies with radius as ρ(r) = ρ[itex]_{o}[/itex] r. Derive expressions for the electric field (a) inside the cylinder (i.e. r<R), and (b) outside the cylinder (i.e. r>R).

    2. Relevant equations
    Gauss's Law
    q=[itex]\rho[/itex] [itex]\delta[/itex][itex]\tau[/itex]

    3. The attempt at a solution
    (a) E inside cylinder
    I sketched a Gaussian surface inside of the cylinder.
    I believe that E is parallel to ds ( [itex]\vec{E}[/itex]||d[itex]\vec{s}[/itex] )
    So, gauss's law becomes E[itex]\oint[/itex]ds = q/[itex]\epsilon[/itex] for the side

    I believe the integral of ds is 2[itex]\pi[/itex] r L (L being the length of the cylinder even though it is infinite.
    And q = ρ[itex]_{o}[/itex] r [itex]\pi[/itex] r[itex]^{2}[/itex] L
    derived from q=[itex]\rho[/itex] [itex]\delta[/itex][itex]\tau[/itex]

    So we have E (2[itex]\pi[/itex] r L) = ρ[itex]_{o}[/itex] r [itex]\pi[/itex] r[itex]^{2}[/itex] L /[itex]\epsilon[/itex]
    Simplifying to E = ρ[itex]_{o}[/itex] r[itex]^{2}[/itex]/ 2[itex]\epsilon[/itex]

    Is this correct for (a)?
    And for (b) would it be the same idea but with a gaussian surface outside of R?

    Thanks!
     
  2. jcsd
  3. Sep 30, 2011 #2

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    I responded to this in your other thread, but I'll repeat here for completeness.
    So close! :cry: Try the final simplification once more. I think you forgot to cancel something out.
    Yes. The trick is to just be careful about determining q. When outside the cylinder, is the total charge q within the Gaussian surface a function of r or a function of R?

    =====================
    Edit:

    I think I see the problem now. Take another look at what I've highlighted in red:
     
    Last edited: Sep 30, 2011
  4. Sep 30, 2011 #3
    Thanks for the post!
    I don't know what is wrong with the 'r'. Are you saying it should not be there? But, ρ(r) = ρ[itex]_{o}[/itex] r. So I put in the 'r'. Hmm are you saying that [itex]\delta[/itex][itex]\tau[/itex] is equal to: [itex]\pi[/itex] r L?
     
  5. Sep 30, 2011 #4

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    Actually, let me give you a better hint than my previous one (admittedly, it probably wasn't a very useful hint).

    Evaluate q again. I think you pulled an r out from under the integral sign when you shouldn't have.

    dq = ρ dL dr rdθ

    Where

    ρ = ρ0r

    Remember, ρ is a function of r so you can't pull it out from under the integral. You can pull ρ0 out, but not ρ. (i.e., make sure you substitute ρ0r in for ρ *before* you integrate. :wink:
     
    Last edited: Sep 30, 2011
  6. Sep 30, 2011 #5
    Where did you get delta from?

    For (a)

    [tex]Q_{en} = 2 \pi \rho_0 \ell \int_{0}^{R} r^2 dr[/tex]

    Do you see why it is from 0 to R?
     
  7. Sep 30, 2011 #6
    I will give it a shot tomorrow (its 12am here...)

    flyingpig, who is the question directed to?
     
  8. Sep 30, 2011 #7
    Who else but you...?
     
  9. Sep 30, 2011 #8

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    Actually, that sounds great for part (b). But for part (a), I would write,

    [tex]Q_{en} = 2 \pi \rho_0 \ell \int_{0}^{r} r'^2 dr'[/tex]
    where [itex] r' [/itex] is a dummy variable.
     
  10. Sep 30, 2011 #9
    Oh wait you are right, it's near my bedtime too...
     
  11. Oct 3, 2011 #10
    So then for a) E = (ρr^2)/ε ?
     
  12. Oct 4, 2011 #11
    I don't understand where the 2 is coming from in the integral... is it part of the integration of a cylinder's volume?
     
  13. Oct 4, 2011 #12
    I have E = ρ[itex]_{o}[/itex] r[itex]^{2}[/itex]/ε

    Where r < R.

    Is that right?
     
  14. Oct 4, 2011 #13

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    Not quite. (See below)

    You need to integrate the charge over the volume, meaning there's three dimensions involve. We'll use cylindrical coordinates, since that's the easiest for this particular problem.

    One dimension is along the length of the cylinder, along the length of L. Let's call the differential length [itex] d \ell [/itex]

    Another dimension is along the radius r. Let's call the differential length of the radius [itex] dr' [/itex] (where [itex] r' [/itex] is a dummy variable, so as not to confuse it with [itex] r [/itex], one of the integration limits [the radius of the Gaussian surface]).

    One remaining dimension is the direction perpendicular to [itex] r' [/itex], along [itex] \theta [/itex]. But this is a special differential length, because the length of this one not only varies with [itex] d \theta [/itex], but also with [itex] r' [/itex]. So this differential length is [itex] r' d \theta [/itex].

    [tex] dq = \rho \ d \ell \ dr' \ r'd \theta [/tex]

    Here, [itex] \rho [/itex] is the charge density, a function of r', and should not be confused with [itex] \rho_0 [/itex]. We haven't made the substitution yet.

    Integrating both sides we have,

    [tex] q = \int_{r' = 0}^r \int_{\theta = 0}^{2 \pi} \int_{\ell = 0}^L \rho \ d \ell \ dr' \ r' d \theta = 2 \pi L \int_0^r \rho r' dr'[/tex]

    Now make your [itex] \rho = \rho_0 r' [/itex] substitution and evaluate the last integral. :wink:
     
  15. Oct 4, 2011 #14
    so it is E = (ρ°r^2)/(3ε°) ?
     
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