A parallel-plate capacitor having a plate area of 40 cm^2 and a...

Area of the plates, $A = 40 \text{ cm}^2 = 4 \times 10^{-3} \text{ m}^2$

Separation between plates, $d = 0.10 \text{ mm} = 10^{-4} \text{ m}$

Capacitance, $C = \dfrac{\epsilon_0 A}{d} = \dfrac{8.85 \times 10^{-12} \times 4 \times 10^{-3}}{10^{-4}} = 3.54 \times 10^{-10} \text{ F}$

Time constant of the $RC$ circuit, $\tau = RC = 16 \times 3.54 \times 10^{-10} = 5.664 \times 10^{-9} \text{ s} = 5.664 \text{ ns}$

The voltage across the capacitor at time $t$ is given by:

$V(t) = V_0 \left(1 - e^{-t/\tau}\right)$

Substituting $t = 10 \text{ ns}$and $\tau = 5.664 \text{ ns}$:

V(10 \text{ ns}) =2.0 \left(1 - e^{-\dfrac{10}{5.664}}\right) = 2.0 \left(1 - e^{-1.765}\right)

$V(10$\text{ ns}) \approx$2.0 (1 - 0.171) = 2.0$\times$0.829 = 1.658$\text{ V}

The electric field inside the capacitor is:

$E = \dfrac{V}{d} = \dfrac{1.658}{10^{-4}} =1.658$\times$10^4$\text{ V/m} \approx$1.7$\times$10^4$\text{ V/m}