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Homework Help: Insulating sphere inside a hollow, conducting sphere (electric fields, Gauss's Law)

  1. Apr 21, 2012 #1
    1. The problem statement, all variables and given/known data
    A solid, insulating sphere of radius a has a uniform charge density of ρ and a total charge of Q. Concentric with this sphere is an uncharged, conducting hollow sphere whose inner and outer radii are b and c, as shown.


    A). Find the magnitude of the electric field in the following regions:

    r < a (Use the following as necessary: ρ, ε0, and r.)

    a < r < b

    b < r < c

    r > c

    B). Determine the induced charge per unit area on the inner and outer surfaces of the hollow sphere.

    2. Relevant equations
    Gauss's law for electric field (?):

    E∫dA = (q in) / (ε0)

    E = ke q / r2

    E = 0 inside a CONDUCTOR

    q = σdA (surface area?)

    q = ρdV (volume? )

    3. The attempt at a solution
    I'm just concerned about the first part r< a and hopefully I will understand the rest

    This is just really tough for me... so many things I need to look out for and it is really confusing.

    It is confusing when thinking about dimensions as the uniform charge is concerned with volume ( q = ρdV) yet I thought Gauss's Law ( flux = E ∫ dA ) was just concerned with 2D surface area only. Or am I missing something?

    Also I am not sure how the inner sphere, being an insulator, has any affect on the electric field.


    for r < a, the electric field is not 0 and so

    E ∫ dA = q in / ε0

    E (4πr2) = q in / ε0

    E = q in / (4πr2ε0)

    which turns out wrong for r < a. It is still wrong even when I substitute q in as pV.
  2. jcsd
  3. Apr 22, 2012 #2
    Re: insulating sphere inside a hollow, conducting sphere (electric fields, Gauss's La

    Let me try

    I think you were doing fine but substituted the wrong charge:

    Q=ρV=ρ((4/3)πa3) therefor ρ=Q/((4/3)πa3)
    for the new volume inside the sphere then q=ρ/((4/3)πr3)
    if you substitute those to what you found then you get E = q in / (4πr2ε0) = ρ((4/3)πr3)/(4πr2ε0)
    this equals ρr/3ε0 and if you substitute for ρ then Q/((4/3)πa3)r/3ε0 = Qr/(4/3)πa30 which equals Qr/4πε0a3
    Last edited: Apr 22, 2012
  4. Apr 22, 2012 #3


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    Re: insulating sphere inside a hollow, conducting sphere (electric fields, Gauss's La

    If, for example, the inner sphere was a conductor, it would be impossible to have a uniform charge density (as then the electric field inside the sphere would not be zero). This is the only reason they're stressing that it is an insulator
  5. Apr 22, 2012 #4
    Re: insulating sphere inside a hollow, conducting sphere (electric fields, Gauss's La

    I got all the answers to it now, thanks for your help.

    How would I go about finding the induced charges now? How would I start?
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