A parallel-plate capacitor of plate area `40 cm^2` and separation between the plates `0.10 mm` is connected to a battery of `emf 2.0 V` through a `16 Omega` resistor. Find the electric field in the capacitor `10 ns` after the connections are made.

Explanation

To find the electric field in a parallel-plate capacitor 10 ns after it is connected to a battery, we can follow these steps:

### Step 1: Convert Units
Convert the given area and separation into standard SI units.
- Area \( A = 40 \, \text{cm}^2 = 40 \times 10^{-4} \, \text{m}^2 = 4.0 \times 10^{-2} \, \text{m}^2 \)
- Separation \( d = 0.10 \, \text{mm} = 0.1 \times 10^{-3} \, \text{m} = 1.0 \times 10^{-4} \, \text{m} \)

### Step 2: Calculate Capacitance
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