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Conducting Sphere Problem

  1. Sep 13, 2007 #1
    1. The problem statement, all variables and given/known data
    A 32 cm diameter conducting sphere is charged to 500 V relative to V=0 and r = infinity.
    (a) What is the surface charge density sigma?
    (b) At what distance will the potential due to the sphere be only 10 V?


    2. Relevant equations



    3. The attempt at a solution

    I know that to find the surface charge density I have to find charge/area. (Sigma = q/A) Finding the area just means finding the surface area of the sphere right? 4(pi)((0.16m)^2)
    Is the charge 500 V? It seems like charge should be expressed in coulombs. I'm not really sure how to get the charge of the sphere.


    Then to find the distance at which the potential is only 10 V
    Do I use V= (1/4(pi)(eo)* sigma/r? where r is that distance and V = 10V?
     
  2. jcsd
  3. Sep 13, 2007 #2

    learningphysics

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    Hint: The field outside the sphere is the same as the field due to a point charge where the center of the sphere is... where the point charge is the total charge on the sphere...

    Using this, you can find the total charge on the sphere... then the charge density.
     
  4. Sep 13, 2007 #3

    Dick

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    To get the charge, remember Gauss' law?? The rest of what you're doing seems about right. Remember you are given the sphere diameter in cm not m.
     
  5. Sep 13, 2007 #4
    The field outside the sphere is given by the equation E=k(Q/r^2)
    I solved for Q and got units in C*m. Was it wrong for me to set E=500V?
     
  6. Sep 13, 2007 #5

    learningphysics

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    Cool. That's right.

    Yeah, that's wrong... what's the voltage due to a point charge... taking the voltage at r = infinity to be 0.
     
  7. Sep 13, 2007 #6

    Dick

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    Once you've straightened all of this out, you may want to think about the Gauss law approach. Charge contained in a sphere at the surface is (500V)*A*e0 where A is the area. To convert that to surface charge just divide out the A. Using the point charge will (of course) give you the same thing. It's just a little bit longer way around.
     
  8. Sep 13, 2007 #7

    learningphysics

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    But the charge is E*A*e0 according to Gauss law... I'm probably missing something...
     
  9. Sep 13, 2007 #8

    Dick

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    Sorry. I meant surface charge density, which is what the OP is asked for.
     
  10. Sep 13, 2007 #9

    learningphysics

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    But the 500V is voltage not electric field...
     
  11. Sep 13, 2007 #10

    Dick

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    Right, J/C not N/C. Please ignore me, I'm apparently having a fuzzy day. Thanks.
     
  12. Sep 13, 2007 #11

    learningphysics

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    No prob. Your fuzzy days are quite rare from what I've seen of your posts. :smile:
     
  13. Sep 13, 2007 #12
    I'm a little confused about what to set E equal to. The voltage due to a point charge is when r is infinity is 0. But how does that relate to the electric field?
     
  14. Sep 13, 2007 #13

    learningphysics

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    Voltage due to a point charge = kQ/r.

    You have voltage... you have r, you can solve for Q.
     
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