A leaky parallel plate capacitor is filled completely with a material having dielectric constant K - 5 and electric conductivity - - 7-4 - 10-12-1m-1- If the charge on the plate at the instant t - 0 is q - 8-85 -C- then the leakage current at the instant t - 12 sec is approximately x- 10-1-A where x is round off to nearest integer

As in case of discharging of a capacitor through a resistance q=q_0e^ t/CR ⇒ i= dqdt=q_0CRe^ t/CR Here, CR= ( ϵ _0KAd ) ( ρdA )=ϵ _0Kσ [as ρ=1/σ] i.e. CR=8.85× ...

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Standard XI

Physics

Charging of Capacitors

A leaky paral...

Question

A leaky parallel plate capacitor is filled completely with a material having dielectric constant $K = 5$ and electric conductivity $σ = 7.4 \times 10^{- 12} Ω^{- 1} m^{- 1}$ . If the charge on the plate at the instant $t = 0$ is $q = 8.85 μ C$ , then the leakage current at the instant $t = 12 sec$ is approximately $x \times 10^{- 1} μ A$ where $x$ is
(round off to nearest integer)

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Solution

As in case of discharging of a capacitor through a resistance
$q = q_{0} e^{- t / C R}$
$\Rightarrow i = - \frac{d q}{d t} = \frac{q_{0}}{C R} e^{- t / C R}$

Here, $C R = (\frac{ϵ_{0} K A}{d}) (ρ \frac{d}{A}) = \frac{ϵ_{0} K}{σ}$
[as $ρ = 1 / σ$ ]
i.e. $C R = \frac{8.85 \times 10^{- 12} \times 5}{7.4 \times 10^{- 12}} \approx 6$

So, $i = \frac{8.85 \times 10^{- 6}}{6} e^{- 12 / 6}$
$\Rightarrow i = \frac{8.85 \times 10^{- 6}}{6 \times 7.39} \approx 0.20 μ A$
[As $e = 2.718, e^{2} = 7.39$ ]

Hence, $x = 2$ .

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As in case of discharging of a capacitor through a resistance q=q0e−t/CR ⇒i=−dqdt=q0CRe−t/CR Here, CR=(ϵ0KAd)(ρdA)=ϵ0Kσ [as ρ=1/σ] i.e. CR=8.85×10−12×57.4×10−12≈6 So, i=8.85×10−66e−12/6 ⇒i=8.85×10−66×7.39≈0.20 μA [As e=2.718, e2=7.39] Hence, x=2.