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What is the electric field on a thin nonconducting shell

  1. Feb 10, 2016 #1
    1. The problem statement, all variables and given/known data
    A nonconducting thin spherical shell of radius 6.80 cm has a uniform surface charge density of 9.28 nC/m2.

    i) What is the total charge on the shell?
    ii) Find the electric field at r = 2.43 cm.
    iii) Find the electric field at r = 6.61 cm.
    iv) Find the electric field at r = 7.17 cm.
    v) Find the electric field at r = 11.3 cm.

    2. Relevant equations
    p = Q/V (where p = charge density, Q = total charge and V = volume.

    and Guass's law

    3. The attempt at a solution

    Part i) What is the total charge on the shell?

    To determine the total charge, we have surface charge density of 9.28 nC/m2 and radius of .068m
    Convert 9.28nC to C, 9.28x10^-9 and convert radius of 6.8cm to m, .068 m

    We will use the equation p = q/V. Rearrange: Q = QV
    The surface area of the spherical shell is 4 pi r^2

    Q = pV
    Q = (9.28 x 10^-9) (4 pi (.068)^2)
    Q = 5.39 x 10^-10 C

    Part ii) Find the electric field at r = 2.43 cm.
    Inside the this spherical shell, the electric field is zero everywhere.

    Part iii) Find the electric field at r = 6.61 cm.
    Inside the this spherical shell, the electric field is zero everywhere.

    iv) Find the electric field at r = 7.17 cm.
    The flux through the Gaussian surface again is:

    E dA = E (4 pi r ^2)

    Applying Guass's law, we can derive this formula.

    E = Q / (4 pi Eo r^2) (Where Eo is the permittivity of free space)
    E = (5.39 x 10^-10)/ (4 pi Eo (.0717)^2)
    E = 942 N/C

    iv) Find the electric field at r = 11.3 cm.

    E = Q / (4 pi Eo r^2) (Where Eo is the permittivity of free space)
    E = (5.39 x 10^-10)/ (4 pi Eo (.113)^2)
    E = 379 N/C

    As expected a weaker electric field as you go further from the surface of the sphere.

    Question : The actual answer is 379.387 etc. However I am reducing the answer to 3 significant figures. Is that correct?
     
  2. jcsd
  3. Feb 10, 2016 #2
    You can double check by assuming all the charge is located at a point at the center and computing the electric field of a point charge at the locations of interest.
     
  4. Feb 11, 2016 #3
    I'm now sure part i) is correct where the total charge on the thin sphere is Q = 5.39 x 10^-10 C.

    But could I please get guidance on if my logic is correct with part ii) with my earlier post.
     
  5. Feb 11, 2016 #4
    Solved this in the end. My calculations were all correct.
     
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