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Potential of Concentric Cylindrical Insulator and Conducting Shell Please Help

  1. Sep 13, 2010 #1
    Potential of Concentric Cylindrical Insulator and Conducting Shell......Please Help

    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 = 40 μC/m3. Concentric with the cylinder is a cylindrical conducting shell of inner radius b = 18.9 cm, and outer radius c = 23.9 cm. The conducting shell has a linear charge density λ = -0.4μC/m. 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 = 40 μC/m3. Concentric with the cylinder is a cylindrical conducting shell of inner radius b = 18.9 cm, and outer radius c = 23.9 cm. The conducting shell has a linear charge density λ = -0.4μC/m.


    2. Relevant equations
    What is Ey(R), the y-component of the electric field at point R, located a distance d = 58 cm from the origin along the y-axis as shown?

    What is V(P) – V(R), the potential difference between points P and R? Point P is located at (x,y) = (58 cm, 58 cm).

    What is V(c) - V(a), the potentital difference between the outer surface of the conductor and the outer surface of the insulator?



    3. The attempt at a solution

    I'm having trouble converting the charge density ρ = 40 μC/m3 and λ = -0.4μC/m to Q in order to find the electric field at point R due to the insulating and conducting cylinders. Please let me know what you think. There is a diagram attached.
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     

    Attached Files:

  2. jcsd
  3. Sep 13, 2010 #2
    Re: Potential of Concentric Cylindrical Insulator and Conducting Shell......Please He

    I am also stuck on this problem, have you figured it out yet?
     
  4. Sep 14, 2010 #3
    Re: Potential of Concentric Cylindrical Insulator and Conducting Shell......Please He

    To get the E field, you need to manipulate the volume density. Say p = volume density.

    We need to get two lambdas so we can use the equation E = 2k(lambda cylinder + lambda shell)/r

    Where r is equal to the distance of the point. To get the linear density of the cylinder, whose p is 40, we need to multiply p by the area of the cylinder (just the circle, ignore length).

    Then you plug the new lambda value into the E field equation and you have the E field. The rest should be pretty easy if you understand potential.

    I'm also in this class (U of I, PHYS 212), if you guys want to get in contact about future homework, feel free.
     
  5. Sep 14, 2010 #4
    Re: Potential of Concentric Cylindrical Insulator and Conducting Shell......Please He

    Thanks I understand what you did there, but im still confused as to how to get the potential at points R and P. Could you please explain how to get those values.
     
  6. Sep 14, 2010 #5
    Re: Potential of Concentric Cylindrical Insulator and Conducting Shell......Please He

    Ok, so your equation E = 2k(lambda)/r must be integrated over the right distance to find potential.

    So the potential at point R is 2klambda(ln(r)), where r is equal to .58 meters.

    Potential at point P is the same thing, but with a different radius since it is at (.58, .58) (hint: use Pythagorean theorem).

    After you find those two, subtract.
     
  7. Sep 14, 2010 #6
    Re: Potential of Concentric Cylindrical Insulator and Conducting Shell......Please He

    Hey I was just wondering how you did part 3 of V(c)-V(a) because I tried to do it like in part 2 but it didn't work and I'm really stuck on this.
     
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