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.

Answer 1

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,$V_0$= 2 V
The capacitance C of a parallel plate capacitor,
$$\begin{gathered} C=\frac{{\mathit{\in }}_{0}A}{d} \\ =\frac{8.85\times {10}^{-12}\times 40\times {10}^{-4}}{1\times {10}^{-4}} \\ =35.4\times {10}^{-11}\mathrm{F} \end{gathered}$$
So, the electric field,
$$\begin{gathered} E=\frac{V}{d}=\frac{Q}{Cd}=\frac{Q_0}{A{\mathit{\in }}_0}\left(1-{\mathrm{e}}^{\mathit{-}\frac{t}{RC}}\right) \\ =\frac{C{V}_0}{A{\mathit{\in }}_0}\left(1-{\mathrm{e}}^{-\frac{t}{RC}}\right) \\ =\frac{35.4\times {10}^{-11}\times 2}{8.85\times {10}^{-12}\times 40\times {10}^{-4}}\left(1-{\mathrm{e}}^{-1.76}\right) \\ =1.655\times {10}^4 \\ =1.7\times {10}^4\mathrm{V}/\mathrm{m} \end{gathered}$$