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Capacitance help

  1. Feb 6, 2008 #1
    1. The problem statement, all variables and given/known data

    In lightning storms, the potential difference between the Earth and the bottom of the thunderclouds can be as high as 35,000,000 V. The bottoms of the thunderclouds are typically 1500 m above the Earth, and can have an area of 90 km^2. For the purposes of this problem, model the Earth-cloud system as a huge capacitor. Calculate the capacitance of the Earth-cloud system in µC.

    2. Relevant equations

    C = q/V and C = K eo A/d

    k= 9e9
    eo = 8.85e-12

    3. The attempt at a solution

    9e9 * 8.85e-12 8 (90 000 000 m^2 / 1500) = 4779 C = 4779000000 µC
    but this is not right
  2. jcsd
  3. Feb 6, 2008 #2
    Why'd you include Coulomb's constant? The equation should be:

    [tex]C=\epsilon_0 \frac{A}{d}[/tex]
  4. Feb 6, 2008 #3
    thanks. i dont know why put my teacher has it like that on his equation sheet

    Calculate the charge stored in the "capacitor" = 18.585

    Calculate the energy stored in the "capacitor" =

    should be 35000000 * 18.585 right?
    Last edited: Feb 6, 2008
  5. Feb 6, 2008 #4
    Point that out to him. It's probably a typo. If it weren't, the equation would simplify to

    [tex]\frac{A}{4\pi d}[/tex]


    [tex]k = \frac{1}{4\pi\epsilon_0}[/tex]
  6. Feb 6, 2008 #5
    i will.

    to find energy you do w =Vq right?
  7. Feb 6, 2008 #6
    Yup. I find it really helpful to check equations by making sure the units on both sides agree.
  8. Feb 6, 2008 #7
    so the energy of the capacitor should be (35000000 * 18.585) 650475000 J??
  9. Jan 18, 2012 #8
    For those that are still interested in this post, there seems to be some confusion about the k in the C=κε0A/d equation. This isn't the "Coulomb constant" k = 1/(4π ε0), but rather kappa. Here κ is the dielectric coefficient of the material between the conducting plates. In our cloud example, we're generally talking about the dielectric constant of air.
  10. Feb 24, 2012 #9
  11. Feb 25, 2012 #10
    EDIT: Will make a new thread. Realized the question OP posted was different
    Last edited: Feb 25, 2012
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