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Capacitance between two plates

  1. Sep 29, 2009 #1
    1. The problem statement, all variables and given/known data
    An air filled capacitor consists of two parallel plates, each with an area of 8.39cm^2, seperated by a distance of .81mm. A 12.3V potential difference is applied to the plates. Find the capacitance. Answer in units of pF (pico Farad)

    2. Relevant equations
    1. E = kV/d
    2. C=q/v
    3. q=E0EA E0 being permittivity constant, 8.85e-12

    3. The attempt at a solution
    using 1. E = 12.3e-3 / 8.1e-4 = 15.185
    using 3. q = 8.85e-12 * 15.185 * 8.39e-2 = 1.12807e-11
    using 2. C=1.12807e-11 / 12.3 = 9.17136e-13
    so wouldn't it be 91.7136 pF?
    not sure where i went wrong?

  2. jcsd
  3. Sep 29, 2009 #2
    why do you have 12.3e-3 for V here instead of just 12.3
  4. Sep 29, 2009 #3


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    Homework Helper

    Capacitance of a capacitor does not depends on the voltage across the parallel plates.
  5. Sep 29, 2009 #4
    Willem: i had used 12.3e-3 because part A of the question asked for the electric field between the plates in kV/m, which 15.185 was the answer. i tried doing this part with 12.3v for the whole thing and it still came out wrong.

    rl.bhat: i also tried C=E0A/d E0 being permittivity constant, 8.85e-12 and it still came out with the same values which is being said is wrong. with the work for that being...
    (8.85e-12 * 8.39e-2) / (.00081) = 9.16685e-10 to pF 916.685

    i have 3 more chances to get it right, with each chance decreasing the points i can get... :/
  6. Sep 29, 2009 #5


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    Homework Helper

    Area should be 8.39*10^-4 m^2. Try this.
    For electric field
    E = V/d = 12.3/0.81*10^-3 =...?
  7. Sep 29, 2009 #6
    well that is just aggravating. i had the decimal in the wrong place on 3 of those 4 times i got it wrong. it is correct now. thank you!

    where i went wrong was with the area. it is 8.39e-4 instead of 8.39e-2 (going from cm to m) because it's two dimensions, right?

    once again, thank you.
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