Voltage across the slabs is Vt

  1. Dielectric

    If two electrodes are sandwitching two dielectric materials with very
    different dielectric constants (but the same thickness), say, water
    and glass. Would the new dielectric constant lies in between the
    original two?

    What would be the electric field in between the dielectric materials?
    I suppose not half of the total electric field imposed by the
    electrodes. Would the larger dielectric constant material take up more
    of it?

    If the water contains ions, would that change its dielectric constant
    from that of its pure form (about 80)?

    Is there any relation between dielectric constant and dielectric
    strength?
     
  2. jcsd
  3. Re: Dielectric

    rambotrout wrote:
    >
    > If two electrodes are sandwitching two dielectric materials with very
    > different dielectric constants (but the same thickness), say, water
    > and glass. Would the new dielectric constant lies in between the
    > original two?
    >
    > What would be the electric field in between the dielectric materials?
    > I suppose not half of the total electric field imposed by the
    > electrodes. Would the larger dielectric constant material take up more
    > of it?
    >
    > If the water contains ions, would that change its dielectric constant
    > from that of its pure form (about 80)?


    Water with ions is electrically conductive. A better example would be
    two solid slabs, perhaps contrasting polyethylene foam (about 1.3,
    coax cable) and poly(vinylidene fluoride) at 12.2 or potassium
    tantalate niobate at 6000.

    What if you insulated your DC electrodes with a couple of microns
    thickness of Parylene-C film then dipped them in electrolyte solution
    or placed a copper slab in-between?

    > Is there any relation between dielectric constant


    electric field attenuation

    > and dielectric
    > strength?


    breakthrough voltage/thickness

    --
    Uncle Al
    http://www.mazepath.com/uncleal/
    (Toxic URL! Unsafe for children and most mammals)
    http://www.mazepath.com/uncleal/lajos.htm#a2
     
  4. Re: Dielectric

    Please correct me if I am wrong.

    Say the voltage across the slabs is Vt, then Vt = V1 + V2, where V1,
    V2 are voltages across the two different dielectric materials.
    Therefore,

    Vt = Q*d1/(E1 * A) + Q*d2/(E2 * A),

    where Q = charge in Coulomb, d = thickness of material, E = dielectric
    constant, and A = area.
     
  5. On Sun, 15 Jun 2008 19:01:13 +0000, rambotrout wrote:

    > Please correct me if I am wrong.
    >
    > Say the voltage across the slabs is Vt, then Vt = V1 + V2, where V1, V2
    > are voltages across the two different dielectric materials. Therefore,
    >
    > Vt = Q*d1/(E1 * A) + Q*d2/(E2 * A),
    >
    > where Q = charge in Coulomb, d = thickness of material, E = dielectric
    > constant, and A = area.


    You are on the right track. In order to calculate the equivalent
    dielectric constant for the whole assembly, you can treat the system as
    two capacitors in series, each with a single kind of dielectric. I
    usually imagine an infinitesimally think conductor between them.

    Remember that capacitors in series add in reciprocals (1/C_eq = 1/C_1
    + 1/ C_2), and you can work out what the effective dielectric constant
    is (which would be relatively simple if d1 == d2, by the way).
     
Know someone interested in this topic? Share a link to this question via email, Google+, Twitter, or Facebook

Have something to add?