Electrostatics Ques 146
- A leaky parallel plate capacitor is filled completely with a material having dielectric constant $K=5$ and electrical conductivity $\sigma=7.4 \times 10^{-12} \Omega^{-1} \mathrm{~m}^{-1}$. If the charge on the capacitor at instant $t=0$ is $q=8.85 \mu \mathrm{C}$, then calculate the leakage current at the instant $t=12 \mathrm{~s}$.
(1997 C, 5M)
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Answer:
Correct Answer: 146.0.198 $\mu A$
Solution:
Formula:
Charging and Discharging of a Capacitor:
- The problem is basically of discharging of $C R$ circuit, because between the plates of the capacitor, there is capacitor as well as resistance.
$ \begin{array}{rlrl} R & =\frac{d}{\sigma A} & R=\frac{l}{\sigma A} \\ \text { and } & C =\frac{K \varepsilon_{0} A}{d} & \end{array} $
$\therefore$ Time constant, $\tau_{c}=C R=\frac{K \varepsilon_{0}}{\sigma}$
Substituting the values, we have
$ \tau_{c}=\frac{5 \times 8.86 \times 10^{-12}}{7.4 \times 10^{-12}}=5.98 \mathrm{~s} $
Charge at any time decreases exponentially as
$ q=q_{0} e^{-t / \tau_{c}} $
Here, $q_{0}=8.85 \times 10^{-6} \mathrm{C}($ Charge at time $t=0)$
Therefore, discharging (leakage) current at time $t$ will be given by
$ i=-\frac{d q}{d t}=\frac{q_{0}}{\tau_{c}} e^{-t / \tau_{c}} $
or current at $t=12 \mathrm{~s}$ is
$ \begin{aligned} i & =\frac{\left(8.85 \times 10^{-6}\right)}{5.98} e^{-12 / 5.98} \\ & =0.198 \times 10^{-6} \mathrm{~A}=0.198 \mu \mathrm{A} \\ i & =0.198 \mu \mathrm{A} \end{aligned} $
BETA
