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Gauss' Law on a sphere

  1. Jul 18, 2008 #1
    Consider an insulating sphere with radius of 7 cm. A charge of 13.8561 µC is uniformly distributed throughout this sphere. It is surrounded by a conducting shell. Denote the charge on the inner surface of the shell to be q2 and the charge on the outer surface of the shell to be q3. The total charge q on the shell is of 64.0681 µC.

    [​IMG]

    Find q2 and q3? Answer in units of µC

    For this problem I am not sure where to start because I only have one radius.
    View attachment 14744
     
  2. jcsd
  3. Jul 18, 2008 #2

    Doc Al

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    You don't even need that radius. See what you can deduce by putting Gaussian surfaces in various places and applying Gauss's law. Hint: Take advantage of the fact that it's a conducting shell.
     
  4. Jul 18, 2008 #3
    I am not sure what you mean by conducting shell. Because isn't E = -q/(4*π*ε0*r2).
     
    Last edited: Jul 18, 2008
  5. Jul 18, 2008 #4

    Doc Al

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    An electrical conductor--like a metal. Hint: What's the field within a conductor?
     
  6. Jul 18, 2008 #5
    I know that the field within a conductor is zero. But I need the charge of the inner an outer surface of the shell.
     
  7. Jul 18, 2008 #6

    Doc Al

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    Use Gauss's law!
     
  8. Jul 18, 2008 #7
    Yes if I apply Gauss' Law E = q/(4*π*ε0*r2) but I don't know both E and r. and if I use E = 0 wouldn't that make q2 = 0
     
  9. Jul 18, 2008 #8

    Doc Al

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    What does Gauss's law tell you? (Use a Gaussian surface inside the conducting shell, where E = 0.)
     
  10. Jul 18, 2008 #9
    I used E = 0 but then q2 = 0.
     
  11. Jul 18, 2008 #10

    Doc Al

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    The q in Gauss's law stands for the total charge. (q2 is just the charge on the inside of the shell. You'll use the total charge to determine q2.)
     
  12. Jul 18, 2008 #11
    So how do I apply Gauss Law to find q2 and q3 with out knowing either E nor r?

    P.S.
    I know I am asking very basic stuff but I just learned the concept and I haven't fully understand on how to use Gauss Law properly yet.
     
  13. Jul 18, 2008 #12

    Doc Al

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    But you do know E inside the conducting shell. So what does that tell you about the total charge contained within a Gaussian surface inside that shell? (Hint: q2 is just one of the charges inside that surface. What's the other?)
     
  14. Jul 18, 2008 #13
    So the charge contained within the Gaussian surface = q1
     
  15. Jul 18, 2008 #14

    Doc Al

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    Nope. (But that is one of the charges within the Gaussian surface.) Draw a diagram showing where the Gaussian surface goes. The problem only defines three charges: q1, q2, and q3. Which are inside the Gaussian surface? Which are outside?
     
  16. Jul 18, 2008 #15
    Inside are q1 and q2. Outside are q3
     
  17. Jul 18, 2008 #16

    Doc Al

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    Yes! So what's the total charge inside the surface? What must it equal from Gauss's law?
     
  18. Jul 18, 2008 #17
    So [(q1+q2) /(4*π*ε0*r2)] = [q3/(4*π*ε0*r2)]

    Does q2 + q3 = 64.0681 µC
     
  19. Jul 18, 2008 #18

    Doc Al

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    :confused:

    Answer these questions:
    (1) What's the total charge within the Gaussian surface in terms of q1, q2, and q3?
    (2) What does E equal at all points on the Gaussian surface?
    (3) What does Gauss's law tell you about the total charge?
    (4) Write a simple equation expressing the above.
     
  20. Jul 18, 2008 #19
    q1 = -q2
    q3 = 2*q1
     
  21. Jul 18, 2008 #20

    Doc Al

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    I should have said:
    :confused:
    Good. This doesn't follow from Gauss's law, but from the fact that you were given the total charge of the shell.
    Yes! This does follow from Gauss's law, since Qtotal = 0 = q1 + q2.
    Where does this come from?
     
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