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effective capacitance

Discussion in 'Electronic Design' started by [email protected], Feb 28, 2006.

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  1. Guest

    What is the effective dielectric constant for a parallel-plate
    constructed as follows:

    | | |
    | | |
    -----| E1 | E2 |-----
    | | |
    | | |


    where both section have the same thickness for simplicity. I don't
    think you can averga E1 and E2, correct?

    The reason I ask is because I am curious about what the effective
    capacitance of a non-conducting pipe-wall would be if the measuring
    electrode was "stood-off" from the pipe using a low-dielectric material
    with the same thickness as the pipe (again, the same thickness assuming
    here is for simplicity).

    Thanks.
     
  2. Tom Bruhns

    Tom Bruhns Guest

    It's two capacitors in series. Imagine a thin conducting surface
    between E1 and E2. For equal size plates on both ends, and assuming no
    fringing effects around the edges, C1 is proportional to E1/t1 and C2
    is proportional to E2/t2, where E1 and E2 are the permittivities (or
    relative dielectric constants) and t1 and t2 are the plate spacings or
    thincknesses. Since the net capacitance is C1*C2/(C1+C2), the net
    capacitance is proportional to E1E2/(E1*t2+E2*t1). That can be
    rewritten as 1/(t2/E2 + t1/E1). In other words, you can average the
    1/E values, weighted by the thicknesses. (This also applies to the
    case where the dielectrics are in a homogenous mixture, such as the
    foam dielectrics commonly used in coaxial cables.)

    Sanity check: if you have equal thicknesses and E1=999*E2, averaging
    the permittivities yields 500*E2. But clearly the capacitance cannot
    be more than that of just the E2 section alone.

    Cheers,
    Tom
     
  3. Tim Wescott

    Tim Wescott Guest

    If you do the ascii-art thing please use fixed-point font, and no tabs.

    Your three-plate structure, with the middle plate floating, should
    pretty much have the same capacitance as the left-to-middle capacitor in
    series with the middle-to-right capacitor. You'll have some direct
    coupling from left to right through fringe effects, but by and large you
    should be able to estimate the capacitance.

    I assume you don't really have a conductor in the middle, but are just
    trying to break the problem down in approved engineering fashion. I
    expect the answer will be substantially the same with or without that
    conductor.

    --

    Tim Wescott
    Wescott Design Services
    http://www.wescottdesign.com

    Posting from Google? See http://cfaj.freeshell.org/google/
     
  4. Guest

    Thanks. My worry was just that: the legitimacy of breaking down the
    problem by assuming that conductor in the center. From the responses
    I've seen here looks like it Ok. Thanks guys.
     
  5. Guest

    Thanks. My worry was just that: the legitimacy of breaking down the
    problem by assuming that conductor in the center. From the responses
    I've seen here looks like it Ok. Thanks guys.
     
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