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EFields of Insulating hallow sphere around conductive hallow sphere

  1. Feb 6, 2012 #1
    1. The problem statement, all variables and given/known data

    There is a insulating hollow sphere of radius who's inner radius is 0.04 m and outside radius is 0.08 m. It's density is not uniform and is represented by [itex]\rho = 0.0211 \mu C / m^{3.5} * \sqrt{r}[/itex]. It has a net electric charge of +10.0 microC. There is a concentric conducting shell of inner radius 0.12 m and outer radius 0.16 m. It's net electric charge is -20.0 micro C

    Find the inner and outer electric fields of both spheres

    2. Relevant equations

    [itex] \Phi = \oint \vec{E} \cdot d\vec{a} = \frac{q _{enclosed}}{\epsilon _{0}}[/itex]
    [itex]E = \frac{q _{enclosed}}{\epsilon _{0}}*\frac{1}{A}[/itex]

    [itex]\epsilon _{0} = 8.85 * 10^{-12}[/itex]

    [itex]q _{enclosed} = \int \rho dV[/itex]

    3. The attempt at a solution

    [itex]A _{1} = 2\pi r = 2\pi(0.04m) = 0.08m\pi[/itex]
    [itex]A _{2} = 2\pi r = 2\pi(0.08m) = 0.16m\pi[/itex]
    [itex]A _{3} = 2\pi r = 2\pi(0.12m) = 0.24m\pi[/itex]
    [itex]A _{4} = 2\pi r = 2\pi(0.16m) = 0.32m\pi[/itex]

    [itex]E_{1} = \frac{0}{\epsilon _{0}}*\frac{1}{0.08m\pi} = 0[/itex]

    [itex]E_{2} = \frac{10 \mu C}{\epsilon _{0}}*\frac{1}{0.16m\pi} = 2.25 MN/C[/itex]

    [itex]E_{3} = \frac{10 \mu C}{\epsilon _{0}}*\frac{1}{0.24m\pi} = 1.50 MN/C[/itex]

    [itex]E_{4} = \frac{-20 \mu C}{\epsilon _{0}}*\frac{1}{0.32m\pi} = -2.25 MN/C[/itex]

    I thought the outside of the insulator should be equal to inside of conductor that somehow the A won't matter?

    Also I wasn't sure if "net charge" meant the just conductor itself has a net charge of -20 uC [thus q enclosed= 10 uC (insulator) - 20 uC (conductor)= -10 uC] or if "net charge" meant all charges added up (10 uC + -30 uC = -20 uC)


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    1. The problem statement, all variables and given/known data

    Using the same example, find the potential difference in KV for the following ranges

    0 < r < 0.04 m
    0.04 m < r < 0.08 m
    0.08 m < r < 0.12 m
    0.12 m < r < 0.16 m
    0.16 m < r < infinity


    2. Relevant equations


    potential difference = [itex]ΔV = - \int \vec{E} \cdot dr = -E \int_{r_{intial}}^{r_{final}} dr = -E Δr[/itex]

    3. The attempt at a solution
     
    Last edited: Feb 6, 2012
  2. jcsd
  3. Feb 6, 2012 #2

    rude man

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    What's the formula for area A1 etc. again???
     
  4. Feb 6, 2012 #3
    A1 is just the area of the Gaussian surface, which I believe in this case would be 2 * pi * radius in the

    [itex] \Phi = \oint \vec{E} \cdot d\vec{A}[/itex]

    Since we picked a surface that E is constant we can pull it out:

    [itex] \Phi = \vec{E_{1}} \cdot \int d\vec{A_{1}}[/itex]

    [itex] \Phi = \vec{E_{1}} \cdot \vec{A_{1}}[/itex]

    [itex] \Phi = E_{1} * A_{1} * cos (0) [/itex]

    [itex] \Phi = E_{1} * A_{1} [/itex]

    Set [itex] E_{1} * A_{1} [/itex]equal to [itex]\frac{q _{enclosed}}{\epsilon _{0}}[/itex] and get E by itself:

    [itex]E_{1} = \frac{q _{enclosed}}{\epsilon _{0}}*\frac{1}{A_{1}}[/itex]

    Mmm.. I think i was suppose to use the surface area of a sphere and not a circle. 4*pi*r^2 and not 2*pi*r
     
    Last edited: Feb 6, 2012
  5. Feb 6, 2012 #4

    rude man

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    Good bet!
     
  6. Feb 6, 2012 #5
    Is it being setup right for the most part? Ignoring the Area mistake
     
  7. Feb 6, 2012 #6

    rude man

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    The question says "radii of 0.4m and 0.8m" yet then you talk about radii of 0.04m and 0.08m. Which is it?
     
  8. Feb 6, 2012 #7
    Sorry I typed the question from memory. It was 0.04 and 0.08.
     
  9. Feb 6, 2012 #8

    vela

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    Your approach is close, but not quite right. The electric field is going to vary with r. You should be finding E(r), not constant values.
     
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