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Three layers of a capacitor - Voltage divider + field strength.

  1. Mar 2, 2014 #1
    1. The problem statement, all variables and given/known data

    In a plate capacitor, the plates has a area of ##100cm^2## and a distance of 3mm. The insulation between the plates is a 1mm glass plate (εr = 10), a 0.5mm thick mica plate (εr = 5) and the rest is air. The insulation layers is parallel with the capacitor plates. The capacitor is connected to a 100V voltage supplier.

    a) find the voltage across the layers.

    b) How big is the field strength in each layer?

    2. Relevant equations



    3. The attempt at a solution

    For a.

    I'm using the current divider rule.

    Vglass = [itex]\frac{\frac{10*0.01m^2}{0.004mm}}{\frac{10*0.01m^2}{0.004mm}+\frac{5*0.01m^2}{0.0035mm}+\frac{0.01m^2}{0.003mm}}[/itex] * 100V = 56.88V.

    This answer seems about 10x larger than it should be - which is why I'm pretty sure that I'm on the right track here. Well, that's only based on the fact that the random chance that I've come up with a completely wrong formula and get an answer that is almost exactly 10 times larger than what it should be, is fairly low.

    What am I missing here?

    As always, and feedback is very much appreciated.

    - Mutaja.
     
  2. jcsd
  3. Mar 2, 2014 #2

    gneill

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    The layers are stacked between the plates and have thicknesses of 1 mm, 0.5 mm, and 1.5 mm. So that models as three capacitors in series. Being in series they must all carry the same current when any current flows. So current division isn't the way to go.

    You have the information required to calculate the individual capacitor values. Why not start there?
     
  4. Mar 3, 2014 #3

    rude man

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    You don't need to compute capacitances and the area of the plates is immaterial.

    You can write an equation for the voltage drops across each region with E, the field in the air gap, as the unknown. You know what the relative E fields are in each region and you also know that the sum of the voltage drops has to equal 100V.


    You're right, your computed value for Vglass is close to 10x what it should be.
     
  5. Mar 3, 2014 #4
    I realize that, but voltage divider makes sense, which is what I thought I attempted at least.
    Ok. This is my new attempt:

    Cglass=##8.85*10^-12## * εr * [itex]\frac{A}{d}[/itex] = ##8.85*10^-12## * 10 * [itex]\frac{0.01m^2}{0.001m}[/itex] = 8,85*10-10.

    Assuming this is correct, I now know c. If I want to use this to find the voltage, I need Q. Knowing myself, though, I've probably overlooked a useful formula.

    Or are we now back at voltage dividing since I know more of the size of these capacitors related to each other (assuming, of course, I compute C for mica and air as well).

    Hmm. This has always been a challenge for me. In my initial attempt, I've written an equation for the voltage across the "glass capacitor" - wrongly. How would your method compare to that? "the field in the air gap" doesn't make much sense to me. Is it the electrical field strength (is that what the E is for?).

    Sorry for the confusion, and thanks a lot for your help.
     
  6. Mar 3, 2014 #5

    gneill

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    rude man is correct that you can write an equation for the voltage drop across each region if you know how the relative permittivity affects the net field strength (relative to vacuum) within the region. It's a good approach and quicker to achieve than what I was suggesting.

    What I was going for is a variation of the voltage divider method that takes advantage of the way charge is distributed in serially connected capacitors. The usual voltage divider formulas get messier when there are more than two capacitors in series.

    By computing each of the capacitor values and then the total equivalent capacitance, a known total voltage (100 V) applied gives you the net charge on the equivalent capacitor. For series capacitors the charges are all the same so that it will be the same charge on each of the capacitor "layers", and hence you know the potential across each layer via V = Q/C. With potential difference and thickness you then have your field strengths for each layer.
     
  7. Mar 3, 2014 #6
    Computing each of the capacitor values:

    Cglass=##8.85*10^-12## * εr * [itex]\frac{A}{d}[/itex] = ##8.85*10^-12## * 10 * [itex]\frac{0.01m^2}{0.001m}[/itex] = 8.85*10-10.

    Cmica=##8.85*10^-12## * εr * [itex]\frac{A}{d}[/itex] = ##8.85*10^-12## * 5 * [itex]\frac{0.01m^2}{0.0005m}[/itex] = 8.85*10-10.

    Cair=##8.85*10^-12## * εr * [itex]\frac{A}{d}[/itex] = ##8.85*10^-12## * 1 * [itex]\frac{0.01m^2}{0.0015m}[/itex] = 5.9*10-10.

    Ctotal=2.36*10-9

    Net charge on the equivalent total (Q?):

    Q = V * C = 100V * 2.36*10-9 = 2.36*10-7

    Using this Q across all capacitors is wrong - obviously. I get voltages of 266 and 400 using V = Q/C. Being able to see that I've done a silly mistake somewhere and not being able to spot it is unfortunate, at best.

    Hopefully it's not all wrong and I've made at least *some* progress.

    Thanks once again for helping me out.
     
  8. Mar 3, 2014 #7

    gneill

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    How do capacitors in series add? (as a check: The total should end up smaller than any of the individual capacitors).

    You're making progress. Fix up your capacitor values and net capacitance.
     
  9. Mar 3, 2014 #8
    Thank you! I figured out the voltage across each layer. My power of 10 mistake should be - 11, and capacitors in series "equals" resistors in parallel. Well, the method of computing the values at least. My C total is ##5.2*10^-11##. V=Q/C is then correct.

    I will give part b a shot in a few hours. I'm currently on the road so my apologies for that. But I guess the forum isn't going anywhere by the time I'm by my computer again heh.

    Thank you so much for all your help so far.
     
  10. Mar 3, 2014 #9

    rude man

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    Across each layer 1, 2 or 3, voltage V = distance d times parallel component of electric field E.

    So: E1 d1 + E2 d2 + E3 d3 = V.
    And then use the fact that E = Evacuor.

    When you get your answer I'll let you know if it's right.

    EDIT: NM, I see you went the other way already.
     
  11. Mar 3, 2014 #10
    Yes, I understand enough to attempt it, but seeing as I already have my answers, I'll leave this behind for now. When I get ahead of my schedule, I will look into it though. Any practice won't harm me, to say the least.

    For my 2nd problem, it was a lot easier, obviously, when I now have the voltage across each capacitor. I simply used E = V/d (which I learned about in a previous thread on this forum), and I got answers that made very much sense.

    Thanks a lot for your help, both of you.
     
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