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Question

A leaky parallel plate capacitor is filled completely with a material having dielectric constant K=5 and electric conductivity σ=7.4×1012Ω1m1. If the charge on the plate at the instant t=0 is q=8.85μC, then the leakage current at the instant t=12 sec is approximately x×101μA where x is
(round off to nearest integer)

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Solution

As in case of discharging of a capacitor through a resistance
q=q0et/CR
i=dqdt=q0CRet/CR

Here, CR=(ϵ0KAd)(ρdA)=ϵ0Kσ
[as ρ=1/σ]
i.e. CR=8.85×1012×57.4×10126

So, i=8.85×1066e12/6
i=8.85×1066×7.390.20 μA
[As e=2.718, e2=7.39]

Hence, x=2.

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