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Question about E-Fields and Potential of a Coaxial Cylinder

  1. May 2, 2016 #1
    1. The problem statement, all variables and given/known data
    I have posted the given question and conditions in the attached image

    2. Relevant equations
    (Q_enclosed/ epsilon_0) = closed integral (E-field) dA
    Q_encolsed = p*A
    V(r)= -Int_(from origin to r) (E-field(r'))*dl'

    3. The attempt at a solution

    a)
    E=0------>For r<a

    (p*v)/(epsilon_0) = pi*(a^2)*L*E
    E= ((p*v)/(epsilon_0))*(1/(pi*(a^2)*L)------>For the solid inside's edge r=a

    E= 0-------> Between soild inside's edge and the outer shell's edge (Q_out = -Q_in) a<r<b

    (Q_enclosed)/(epsilon_0) = 2pi*b*L*E
    E= ((Q_enclosed)/(epsilon_0))*(1/(2pi*b*L))-------For the outer shell's edge r=b

    b)As for potential, V, I am very uncertain as to what I should do.
    c) I think this portion will be much clearer once a) and b) are clearified.


    Thanks!
     

    Attached Files:

  2. jcsd
  3. May 2, 2016 #2

    BvU

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    Hello artoribio, :welcome:

    In your work for (a), is there an area where the E field is not zero ?
    I agree with ##|\vec E | = 0## for 0 < r < a and also for b < r : you can form a gaussian cylinder that contains no charge.
    What I don't understand is what you mean with
    where is the gaussian surface ? And why do you start with E = 0 ?

    PS the image is hard to read; at PF we prefer it if you type out the problem statement...
     
  4. May 2, 2016 #3
    I thought that the area would not have a zero E-field outside of the coaxial cylinder. I did not put an |⃗E| equation for b<r so I'm am not quite sure what you are referring to. As for your quoted portion, I was thinking that directly in between a and b the fields from the solid center and the shell would cancel out because Q_out + Q_in = 0.

    I apologize for the image and will write out any future questions I might post. Thank you for the heads up. I had no idea!
     
  5. May 2, 2016 #4

    BvU

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    Ah ! So in between a positive charge and a negative charge there is no electric field :woot: ?
     
  6. May 2, 2016 #5
    I was thinking in terms of two positive nodes. Is that wrong? In the case of a positive and negative set it would just be straight lines between the center and and the shell considering, right?
     
  7. May 2, 2016 #6

    BvU

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    I am afraid so. It even seems that that is a given in your exercise !?!?

    I retype (I am tired of switching to the unsharp picture) and correct the

    ---------------------------------------
    Problem statement
    • We have a coaxial cylindrical shell wth a solid center and a uniform charge density ## \rho ## C/m3
    • Inner solid cylinder has radius a. Outer shell has inner radius b.
    • Qout = - Qin
    Find
    • ##\vec E## everywhere
    • ##V##
    • Plot both
    -----------------------------------------

    That wasn't so hard ! let's also do

    2. Relevant equations
    1. ##\oint \vec E \cdot d\vec A = {\displaystyle Q_{\rm enclosed} \over \displaystyle \varepsilon_0 } ##
    2. ## Q_{\rm enclosed} = \rho A L \ \ ## Coulomb ## \ \ \ (\rho A ## has the wrong dimension ! ##)##
    3. ## V(r)= - \displaystyle \int_0^r E(r') \, dr' ##
    -----------------------------------------

    Explanation of

    • My adding density : ##\rho## is used for volume charge density (##\sigma## for area density, ##\lambda## for line density). You can use a length L = 1 for your calculations, though (see below). But writing ##Q = \rho A## is ugly.
    • ##\bf Q_{\rm out} = -Q_{\rm in}##: the charge on the inner cylinder induces an opposite charge on the inside of the outer cylinder shell. If that shell is grounded, the potential on the outside of that cylinder remains zero. And V = 0 outside both cylinders is then a solution for the electric field. Since the field is unique it is the solution.

    Your equation #1 is Gauss theorem. It works usefully if there is a symmetry to exploit. In this case there is cylindrical symmetry, so (in cylinder coordinates) ##\ \ \vec E_z = 0 \ \ \&\ \ \vec E_\phi = 0 \ \ ## (can you explain why ?) and ##\ \ \vec E = E(r) \hat r \ \ ## (as you assumed and work out). Sensible gaussian surfaces are coaxial cylinder shells and you have a nonzero E field everywhere where such a cylinder encloses charge. Back to the drawing board for b < r therefore !
     
  8. May 2, 2016 #7
    I think that I understand your explanations. As for the symmetry of the cylinder or in general if the field lines are perpendicular to the surface then you can take the E values. So in this case E_z = 0 because it is parallel to the charge. That is what I gathered from a Khan Academy video. As for E_phi = 0 There are equal and opposite sets that cancel out in the z-direction and sum in the perpendicular direction of the surface of the cylinder. Do I need to find E for the areas 0, a, and b, or just r<a, a<r<b, and b<r?
     
  9. May 2, 2016 #8

    BvU

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    How can a vector be parallel to a charge ?
    How does that make ##E_\phi = 0 ## ?
    just r<a, a<r<b, and b<r
     
  10. May 2, 2016 #9
    Maybe it was parallel to the surface?

    I think that E_phi just ends up becoming E_perpendicular(to the surface) because the lateral portions cancel, no?
     
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