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A capacitor filled with dielectric of pe...

A capacitor filled with dielectric of permittivity `epsilon = 2.1` losses half the charge acquired during a time interval `tau = 3.0 min`. Assuming the charge to leak only thorugh the dielectric filler, calculate its resistivity.

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The chagre decays according to the formula
`q = q_(0) e^(-1//RC)`
Here, RC = mean life = Half-life/In 2
So, half life `= T = R C In 2`
But, `C = (epsilon epsilon_(0) A)/(d), R = (rho d)/(A)`
Hence, `rho = (T)/(epsilon epsilon_(0) In 2) = 1.4xx10^(13) Omega m`
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