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Charge inside metallic shell

  1. Nov 6, 2012 #1
    1. The problem statement, all variables and given/known data
    A +30 mC point charge sits at the center of a spherical thick metallic shell, of 4.0 cm inner radius and 2.0 cm thickness. The net charge of the shell is -50mC. a.) Determine the location of the 0 volt equipotential surface. b.) calculate the electric potential of the metallic shell. c.) Give the expression for the electric potential in the entire space, V(r) as a function of the distance from the center. Sketch the graph of V(r)


    2. Relevant equations
    V = kQ/r
    E = kQ/r2



    3. The attempt at a solution

    a.) I believe the location will be anywhere inside the metallic shell, because it is a conductor.

    b.) V = kQ/r = ((8.99 x 109 N(m/C2)(-50mC))/(2.0cm)
    = 2247.5V
    i dont think this is right. i think my biggest issue is converting units.

    c.) For this i think it is the sum of the electric potential of the inside and the electric potential of the outside.
    V = ((8.99 x 109 N(m/C2)(+30mC))/(2.0cm) + ((8.99 x 109 N(m/C2)(-50mC))/(2.0cm)
    = ?
     
  2. jcsd
  3. Nov 6, 2012 #2

    haruspex

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    I'm not sure whether you mean within the space enclosed by the metallic shell, within the body (annulus) of the shell, or both. But the answer is wrong in each interpretation.
    In the annulus there is no field, but there would be a potential. I would suggest starting by figuring out the potential just outside the shell, then thinking about what that implies for the potential in the annulus, and finally consider the potential within the cavity.
    As a hint for the first step, what is the potential (or what is the field if that's easier) outside a uniformly charged thin shell?
     
  4. Nov 6, 2012 #3
    If i were to find the field just outside i would do:

    Qtotal/(4πε0r2) ??

    what would i use for r? my teacher has told us in the past if we were to solve for something that is slightly larger than something else, in this case the gaussian surface, to use +1.0 cm larger than the radius. would it be safe to solve using 7.0cm as the r value?
     
    Last edited: Nov 6, 2012
  5. Nov 6, 2012 #4

    haruspex

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    Yes. By symmetry, the charges at the shell can be considered as many thin spherical shells with uniform charge. (Actually, all the charge will be on either the outer surface or the inner surface, right?). So all charges less then r from the centre can be treated as though at the centre for this calculation.
    Strange advice. I mean at an infinitesimal distance outside, so just use 6cm.
     
  6. Nov 6, 2012 #5
    would the answer for letter a.) be the center then because if you were to find the potential the r value would be 0 which would cause the answer to be infinite?

    for b.) would you use Va - Vb?

    when i go to plug in my equation in my calculator i try to do conversions on the fly. when i find the answer its an outrageous amount, over a million usually, so i think i am doing something wrong.
     
  7. Nov 6, 2012 #6

    haruspex

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    The potential must be zero because it's infinite? Weird logic.
    Let's get past (a) first.
    And if you happen to get something reasonable you presume it's right? Not a recommended approach. Monkeys on typewriters come to mind.
     
  8. Nov 6, 2012 #7
    Ok so working on a then.

    am i wrong in saying that the center will cause no electric potential? the value of r would be 0 which would cause the whole denominator to be 0, which would cause the equation to be undefined.

    from what i understand the metallic shell is a conductor. and conductors do not carry a charge on the inside. and if you were to use the equation and plug 0 in for Q it would cause the equation the equal 0.

    i dont know if any of this is correct. I have been having problems in physics recently because most of my attention was focused on one of my other classes.
     
  9. Nov 6, 2012 #8

    haruspex

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    Yes (that's wrong).
    No charge on the interior of the conductor. In this case that is the annulus between the inner surface and the outer surface. Those two surfaces will carry charges.
     
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