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Distance from potential difference

  1. Apr 2, 2016 #1
    1. The problem statement, all variables and given/known data
    A metal sphere of radius 15 cm has a net charge of 4.2x10^-8C. At what distance from the sphere's surface has the electric potential decreased by 500V?

    2. Relevant equations
    V = kQ/r

    3. The attempt at a solution
    [itex]\Delta V\quad =\quad \frac { kQ }{ r } \\ \\ r\quad =\quad \frac { kQ }{ \Delta V } \\ r\quad =\quad \frac { 9e9*4.2e-8 }{ 500 } \\ =\quad 0.756\quad m[/itex]

    What is wrong?
     
  2. jcsd
  3. Apr 2, 2016 #2

    gneill

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    Staff: Mentor

    You are looking for a potential difference, not a particular potential. Find the potential difference between a point located adjacent to the sphere (ro ≅ 15 cm) and one located at some larger radius r1. Once you know r1 you should be able to find the required distance.
     
  4. Apr 4, 2016 #3
    In a metal sphere the charge is distributed to the surface right?
    Can I use V = kq/r for a sphere? It seems a bit weird to me, can you explain? Maybe I just don't understand the voltage equation..
     
  5. Apr 4, 2016 #4

    gneill

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    Staff: Mentor

    For a charged conducting sphere, all the charge will be located at its surface. For any spherically symmetric charge distribution the electric field external to the sphere behaves as though the total charge were a point charge located at the center of the sphere.

    So yes, you can use V = kq/r for the sphere.
     
  6. Apr 4, 2016 #5
    If it's not too much trouble, why does it behave like a point charge? Is it through some kind of mathematical proof?
     
  7. Apr 4, 2016 #6

    gneill

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    Staff: Mentor

    Yes. It's the same method as for Newton's gravitational shell theorem.
     
  8. Apr 8, 2016 #7
    Sorry, for the late response. The answer that i'm getting is 0.0371 m?
     
  9. Apr 8, 2016 #8

    gneill

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    Staff: Mentor

    Looks good. You should probably express it in cm since the sphere radius was given in those units.
     
  10. Apr 8, 2016 #9
    Good advice, the tricky part was just one of the steps involved putting d+r in the voltage equation.
    Thanks for the help!
     
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