A leaky parallel plate capacitor is filled completely with a material having dielectric constant K = 5 and electrical conductivity σ=7.4×1012Ω−1m−1. If the charge on the capacitor at the instant t=0 is q0=8.55μC, then calculate the leakage current at the instant t=12s in μA.

τC=CR=(Kε0Ad)(ρdA)=Kε0ρ=5×8.86×10−127.4×10−12≈6sInitial current,i0=q0/CR=q0CR=q0τC=8.556=1.425μANow, current as function of time i=iie−t/τC Or i=(1.425)e−12/6=0.193μA

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A leaky parallel plate capacitor is filled completely with a material having dielectric constant K = 5 and electrical conductivity $σ = 7.4 \times 10^{12} Ω^{- 1} m^{- 1}$ . If the charge on the capacitor at the instant $t = 0$ is $q_{0} = 8.55 μ C$ , then calculate the leakage current at the instant $t = 12 s$ in $μ A$ .

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Detailed Solution

$τ_{C} = C R = (\frac{K ε_{0} A}{d}) (\frac{ρ d}{A}) = K ε_{0} ρ$

$= 5 \times 8.86 \times \frac{10^{- 12}}{7.4 \times 10^{- 12}} \approx 6 s$

Initial current,

$i_{0} = \frac{q_{0} / C}{R} = \frac{q_{0}}{C R} = \frac{q_{0}}{τ_{C}} = \frac{8.55}{6} = 1.425 μ A$

Now, current as function of time $i = i_{i} e^{- t / τ_{C}}$

Or $i = (1.425) e^{- 12 / 6} = 0.193 μ A$

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