A parallel-plate capacitor of plate area 40cm² and separation between the plates 0.10 mm is connected to a battery of emf 2.0 V through a 16Ω resistor. Find the electric field in the capacitor 10 ns after the connections are made.

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51 µV / m

17 µV / m

$1.7 \times 10^{4} V / m$

34 µV / m

answer is C.

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Detailed Solution

$\begin{array}{l} Q_{0} = CV \\ \Rightarrow Q_{0} = \frac{\in_{0} A}{d} v \\ \Rightarrow \frac{Q_{0}}{A \in_{0}} = \frac{v}{d} = \frac{2}{10^{- 4}} = 2 \times 10^{4} \\ \Rightarrow C = \frac{\in_{0} A}{d} = \frac{8 .85 \times 10^{- 12} \times 40 \times 10^{- 4}}{10^{- 4}} \\ \Rightarrow E = \frac{Q}{A \in_{0}} \Rightarrow C = \frac{\in_{0} A}{d} \\ E = \frac{Q_{0}}{A \in_{0}} (1 - e^{- t / CR}) \end{array}$
$\begin{array}{l} = 2 \times 10^{4} (1 - e^{\frac{- 10 \times 10^{- 9}}{8 .85 \times 10^{- 12} \times 40 \times 16}}) \\ \Rightarrow E = 1 .7 \times 10^{4} v / m \end{array}$