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Exercise with a total charge null

  1. Mar 15, 2016 #1
    1. The problem statement, all variables and given/known data
    A point charge ##q = 30 mC## is located in the center of a conductor spherical shell with an internal radius of ##a = 10 cm## and an external radius of ##b = 20 cm## and total charge null.
    Find the surface density charge of both the internal surface and external surface of the conductor and draw a graph of the module of the electric field as a function of the distance from the center of the sphere.

    2. Relevant equations
    Gauss Theorem
    ##\Phi_S(\vec E_0) = \int_S \vec E_0 \cdot d\vec S = \frac{Q_{TOT}^{int}}{\epsilon_0}##

    3. The attempt at a solution
    I started this way and applied the Gauss Theorem on the internal surface and external surface, so
    ##E 4 \pi r^2 = \frac{Q_{TOT}^{int}}{\epsilon_0} = \frac{\sigma 4 \pi a^2}{\epsilon_0}##
    ##E 4 \pi r^2 = \frac{\sigma 4 \pi b^2}{\epsilon_0}##
    which become
    ##\sigma_{int} = \frac{E r^2 \epsilon_0}{a^2}##
    ##\sigma_{ext} = \frac{E r^2 \epsilon_0}{b^2}##
    and I thought I did it right, but when the exercise asks me to draw a graph of the electric field as a function of the distance form the center I noticed that the exercise says "total charge null" in the conductor. This means that ##Q_{TOT}^{int} = 0##, right? So the two surface density charges I found are wrong? And how should I draw the graph?
     
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  3. Mar 15, 2016 #2

    Doc Al

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    If you apply Gauss' law to a gaussian surface inside the conductor, what must the total charge be within that surface? What does that tell you about the charge on the inner surface of the conductor?
     
  4. Mar 15, 2016 #3
    I know that the electric field inside a conductor is ##\vec E = 0## and outside it is ##\vec E_0## and orthogonal to the conductor's surface . The total charge should be ##Q = \int_S \sigma(x,y,z) dS = \sigma 4 \pi R^2##, with ##R## a radius of a sphere between the internal and external radius(so ##a < R < b##).
    I don't know. I know there is a single point charge in the middle. Does it affect the inner surface?
     
  5. Mar 15, 2016 #4

    TSny

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    If you choose your Gaussian surface inside the conducting material (Doc Al's suggestion), then what is the value of ∫E⋅dA for that surface?
     
  6. Mar 15, 2016 #5

    mfb

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    Where do you get E from, in particular for the outer surface?
    How did you define ##Q_{TOT}^{int}##?
     
  7. Mar 15, 2016 #6

    Doc Al

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    Since I'm talking about a gaussian surface within the conducting material, say at r = 11 cm, ##\vec E = 0##. So what must be the total charge within that gaussian surface? (Hint: No calculation needed!)

    Absolutely. The charge in the middle is part of the total charge within that gaussian surface, as is the charge on the inner surface. If the charge in the middle is q, what must be the charge on the inner surface? (Hint: No calculation needed!)
     
  8. Mar 15, 2016 #7
    It should be ##0## since ##E = 0## inside the conductor.

    Oh! The complete induction! The outer surface can't block the effects of the internal point charge, so it would be ##\vec E_0 = \frac{1}{4 \pi \epsilon_0} \frac{q}{r^2}## for the outer surface, right?

    I think I took the description too literally(I got it wrong). It's not that there are no charges in the conductor, the conductor simply has a electric field equal to ##0## and a flux of the electric field equal to ##0## too. So there is a ##Q_{TOT}^{int}##. Would this be ##Q_{TOT}^{int} = \int_{\tau} \rho(x,y,z) d\tau##?

    Should it be ##0##?

    It should be ##-q##, because the conductor must stay at a null internal charge, right?
     
  9. Mar 15, 2016 #8

    Doc Al

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    Yes, since the field on that gaussian surface (inside the conductor) is zero the flux is zero and thus the total charge within it must be zero.

    Yes. You know the total charge must be zero within the gaussian surface, so the charge on the inner surface of the conductor must exactly cancel that of the charge in the middle.
     
  10. Mar 15, 2016 #9
    So, from what we just said, the internal surface charge comes from this ##-q = \sigma_{int} 4 \pi a^2## that becomes this ##\sigma_{int} = - \frac{q}{4 \pi a^2}## while the external surface charge would be the opposite with ##b## instead of ##a##? Like this ##\sigma_{ext} = \frac{q}{4 \pi b^2}##?
     
  11. Mar 15, 2016 #10

    mfb

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    Right.
     
  12. Mar 15, 2016 #11
    Thank you so much! All of you! I think I got how to do the graph too.
     
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