capacitor and frequency

Discussion in 'General Electronics Chat' started by shreyas_bhat, Jul 19, 2005.

  1. shreyas_bhat

    Thread Starter Active Member

    Jul 26, 2004
    I have a 2 plates each coated with a layer of insulator (0.05 mm thick and dielectric constant 6), and the gap between these insulated plates (.5 mm) is filled with an ionic solution of conductivity 5 S/m and dielectric constant 80. Is it possible for me to conclude that I have equivalently 3 capacitors in series (insulator + solution + insulator) and compute the net capacitance.

    I tried to do it, but the computed value seems to be less than the experimental result. Wat cud be the reason.

    Also, if I wud appreciate if someone could suggest or send an equivalent circuit model for such a capacitor.

  2. JoeSykora

    New Member

    Jul 29, 2005
    A capacitor in it's simplest form is basiclly two conductors seperated by an insulator. The static field is setup in the dielectric or insulator. In your case, the insulation around your plates. Electrons are stripped from one plate and deposted on the other to setup this static field.

    The conductive matterial between the two insulated plates is esentially doing nothing except adding distance between the plates. A greater distace = less capacitance. You will inscrease your dielectric strenght with greater distance.

    C = A x εr / d

    C= Capacitance in farads
    A = Plate area in square meters
    εr = Relative permittivity
    d = Distance in meters

    εr = ε / ε o

    ε o = 8.85 x 10 -12 F/m

    F = farads
    m = meters