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Energy stored in a device (Charged Capacitor)

  1. Sep 30, 2012 #1
    1. The problem statement, all variables and given/known data
    The metallic sphere on top of a large van de Graaf generator has a radius of 2.0 m. Suppose that the sphere carries a charge of 3x10-5 C. How much energy is stored in this device?

    2. Relevant equations
    So the equation I must use is this: U = 1/2C(ΔV)^2
    Q is given; 3 x 10 ^ -5C
    To find ΔV, I'd first the formula :V = Q/(4∏ε0r)



    3. The attempt at a solution
    Then it's simply a matter of plugging the numbers. The formula I'm not sure is how to find ΔV from the information, I'm not sure I'd be using the right one.
     
  2. jcsd
  3. Sep 30, 2012 #2

    CWatters

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    What's the magic word beginning with d ?
     
  4. Sep 30, 2012 #3
    Oh, I have to find the charge density don't I? Bear with me, I'm still quite confused with all of that...

    So, charge density of a sphere: p = Q/V, V = (4∏r3)/3

    Would I use this as my Q in the above formula?
     
  5. Sep 30, 2012 #4

    CWatters

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    I meant d for differentiate .

    Somewhat informally..

    Q=VC
    V=Q/C
    dV = dQ/C

    dQ = the amount of charge when fully charged - the amount of charge when discharged

    = 3x10-5 - 0
     
  6. Sep 30, 2012 #5
    Ok I'm sorry but I'm not following you on this one, this isn't something I've seen in class
     
  7. Oct 2, 2012 #6

    CWatters

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    It's probably simpler than you think.

    The max energy case is with it charged to some voltage V, The minimium energy state is with it discharged (zero volts) so ΔV is just V. Likewise for ΔQ.

    U = 1/2C(V)2

    now
    Q=VC so
    V = Q/C

    then substitute for V giving

    U = 1/2*C*(Q/C)2
    = Q2/2C
     
  8. Oct 2, 2012 #7
    Thanks! It is simpler than I thought... So in that case I don't need the radius of the generator to solve it, it's basically an unnecessary information given?
     
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