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Question
A parallel-plate capacitor of plate area 40 cm2 and separation between the plates 0.10 mm, is connected to a battery of emf 2.0 V through a 16 Ω resistor. Find the electric field in the capacitor 10 ns after the connections are made.
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Solution
Given:-
Area of plates, A = 40 cm2 = 40 × 10−4 m2
Separation between the plates, d = 0.1 mm = 1 × 10−4 m
Resistance, R = 16 Ω
Emf of the battery,
V0 = 2V
The capacitance C of a parallel plate capacitor,
\[C = \frac{\in_0 A}{d}\]
\[ = \frac{8 . 85 \times {10}^{- 12} \times 40 \times {10}^{- 4}}{1 \times {10}^{- 4}}\]
\[ = 35 . 4 \times {10}^{- 11} F\]
So, the electric field,
\[E = \frac{V}{d} = \frac{Q}{Cd} = \frac{Q_0}{A \in_0} \left( 1 - e^{- \frac{t}{RC}} \right)\]
\[ = \frac{C V_0}{A \in_0} \left( 1 - e^{- \frac{t}{RC}} \right)\]
\[ = \frac{35 . 4 \times {10}^{- 11} \times 2}{8 . 85 \times {10}^{- 12} \times 40 \times {10}^{- 4}} \left( 1 - e^{- 1 . 76} \right)\]
\[ = 1 . 655 \times {10}^4 \]
\[ = 1 . 7 \times {10}^4 V/m\]
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