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Homework Help: Calculate capacitance with a dielectric

  1. Oct 2, 2012 #1
    1. The problem statement, all variables and given/known data
    An isolated charged parallel-plate capacitor with the plate separation d=2 mm and area of A=10 cm2 is initially empty. One slides a polystyrene dielectric slab with thickness equal to half the plate separation and with only half the area of the plates into the region between the plates. What is the final capacitance?

    2. Relevant equations
    C = κ((ε0A)/d)

    Since no specification is made, I assume there is air in-between, thus use K of 1.00059 (given in a table)

    Since the dielectric slab is only half the area, I convert it to 5 cm2, or 0.0025 m2

    3. The attempt at a solution
    (8.854e-12 X 0.0025) / 0.0001m

    = 2.2135e -10
  2. jcsd
  3. Oct 2, 2012 #2
    I can't tell 100% what you've actually done. I'd have to say this isn't correct considering it looks like you've neglected the rest of the capacitor after the dielectric was inserted.

    What it looks like to me is that you'll need to break down the capacitor into what looks like 3 parts

    The part that is the same size as the dielectric, but is the air that is above it.
    The dielectric part itself
    The rest of the plate which isn't covered by the dielectric at all

    C = εA/d2 {area is halved} + εA2/2d {area and d are halved} + kε2A/2d {area and d are halved, but has the dieletric}

    It doesn't say to only focus on the area which is occupied on the dielectric
  4. Oct 3, 2012 #3
    If I try this, it gives me:

    C = ε0A/2d + ε0A/d + κε0A/d (since the 2nd and third expression the 2's cancel out)

    But then the unit give out something weird: C2/Nm2 X m2/m = C2m/N

    I must be doing something wrong, because it's not any unit I can use. Although if the units were re-arranged, they'd give a Farad, considering I have Nm which gives J, it could theoretically give C / V, or C / J/C
  5. Oct 3, 2012 #4


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    Yes, you do need to break the capacitor into what looks like three parts.

    However, you can't simply add all three capacitances arithmetically. You must consider which parts act as if they're in parallel, and which parts act as if they're in series.
  6. Oct 3, 2012 #5
    Wouldn't the whole system be in parallel? Considering the two plates are parallel, and then the dielectric is slid in-between the two
  7. Oct 3, 2012 #6


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    In a word: No.

    The half with the dielectric is as if you have a capacitor, area, A/2, plate separation, d/2, with the dielectric, in series with a similar size capacitor filled with air.

    By the way, considering significant digits and all, we usually take the dielectric constant or air as being 1 .
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