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Potential of Concentric Cylindrical Insulator

  1. Jan 26, 2013 #1
    1. The problem statement, all variables and given/known data

    An infinitely long solid insulating cylinder of radius a = 3.6 cm is positioned with its symmetry axis along the z-axis as shown. The cylinder is uniformly charged with a charge density ρ = 25.0 μC/m3. Concentric with the cylinder is a cylindrical conducting shell of inner radius b = 19.5 cm, and outer radius c = 23.5 cm. The conducting shell has a linear charge density λ = -0.41μC/m.

    What is V(P) – V(R), the potential difference between points P and R? Point P is located at (x,y) = (41.0 cm, 41.0 cm).

    https://www.smartphysics.com/Content/Media/Images/EM/06/h6_cylinder.png [Broken]

    2. Relevant equations

    V(r) = ∫Edr
    [itex]\Phi[/itex] = [itex]\oint[/itex]EdA
    [itex]\Phi[/itex] = qenclosed/[itex]\epsilon[/itex]o*A


    3. The attempt at a solution
    I understand that I am to solve for E, and then take the integral in terms of r. But I'm still confused. I solved for E previously to get the following:

    E[itex]\oint[/itex]dA = qenclosed/[itex]\epsilon[/itex]o
    E*A = qenclosed/[itex]\epsilon[/itex]o

    A = 2*pi*r*h
    qenclosed = λ(h) + ρ(volume of cylinder)

    Therefore:

    E(2*pi*r*h) = [λ(h) + ρ(volume of cylinder)]/[itex]\epsilon[/itex]o
    E = [λ(h) + ρ(volume of cylinder)]/([itex]\epsilon[/itex]o)*(2*pi*r*h)

    I'm confident about taking the integral of that equation above, but I'm just not sure what interval I'm supposed to take it over. Any help is appreciated. :)
     
    Last edited by a moderator: May 6, 2017
  2. jcsd
  3. Jan 26, 2013 #2

    haruspex

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    You know what direction that field is in at any given point. At a point x along the the line from R to P, what is the strength and direction of the field? What is its component in the direction RP? (This can also be done vectorially of course.)
     
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