A parallelplate capacitor has plates of area 10 cm2 separate

A parallel-plate capacitor has plates of area 10 cm^2 separated by a 0.10-mm layer of glass insulation with resistivity rho = 1.2 times 10^13 ohm. M and dielectric constant kappa = 5.6. Because of the finite resistivity, charge leaks through the insulation. How can such a leaky capacitor be represented in a circuit diagram? Find the time constant for this capacitor to discharge through its insulation.

Solution

diagram 3 is best representation for the leaky capacitor.

the time constant is, t = CR = C[p*d/A]

Capcitance of the capacitor: C = ke₀A/d

So, the time t is calculated as follows:

    t = C[p*d/A]

      = [ke₀A/d][p*d/A]

      = ke₀p

      = 5.6*8.85x10^-12*1.2x10¹³

      = 594.72 s