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Homework Help: Charged spheres and shells

  1. Aug 8, 2010 #1
    1. The problem statement, all variables and given/known data
    A conducting sphere of radius, R = 5.5 cm with an excess charge of Q = -35.5 nC is surrounded by a concentric, conducting, spherical shell of inner radius, Rin = 9.5 cm and outer radius, Rout = 11.5 cm that carries an excess charge of q = -13.0 nC.

    [PLAIN]http://img571.imageshack.us/img571/7821/imagex.gif [Broken]

    Determine the electric field at the following radii for the aforementioned arrangement:

    (a) r = 42.5 cm.

    (b) r = 8.5 cm.

    3. The attempt at a solution

    (a) In the indicated region r>Rout. Therefore I model the charge distribution as a sphere with charge -Q and the expression for the field in this region would be

    [tex]E=-k_e \frac{Q}{r^2}[/tex]

    [tex]-(9 \times 10^9) \frac{35.5}{42.5^2}=176885813.1[/tex]

    even if I convert r to meters I still get the wrong answer (correct answer: 2420)

    (b) Again, I can apply Gauss's law to find the electric field. Since R<r<Rin I think I should use:

    [tex]k_e \frac{Q}{r^2}=(9 \times 10^9) \frac{35.5}{8.5^2} = 4422145329[/tex]

    The correct answer is 44200 N/C, did I forget to convert something?
    Thanks.
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Aug 8, 2010 #2

    kuruman

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    What is the charge enclosed by a Gaussian surface at 42.5 cm? Don't forget that both the sphere and the shell have charge on them. Also, how many coulombs to a nano-coulomb are there?
     
  4. Aug 10, 2010 #3
    Yes, I got all the unit conversions correct but I'm still getting the wrong answer:

    [tex]-(9 \times 10^{9})\frac{35.5 \times 10^{-9}}{(0.425)^2}=-1768.8[/tex]

    Why is that?
     
    Last edited: Aug 10, 2010
  5. Aug 10, 2010 #4

    kuruman

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    It is because the charge enclosed by a spherical Gaussian surface at 42.5 cm is not -35.5 nC. What is it?
     
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