Two large,thin metal plates are parallel and close to each other. On their inner faces,the plates have surface charge densities of opposite signs and of magnitude $17.0 \times 10^{-22} \; C/m^2$. What is $E$:
$(a)$ in the outer region of the first plate,
$(b)$ in the outer region of the second plate,and
$(c)$ between the plates?

(N/A) The situation is represented in the figure. $A$ and $B$ are two parallel plates close to each other. The outer region of plate $A$ is labeled as $I$,the outer region of plate $B$ is labeled as $III$,and the region between the plates $A$ and $B$ is labeled as $II$.
Charge density of plate $A$,$\sigma = 17.0 \times 10^{-22} \; C/m^2$
Charge density of plate $B$,$\sigma = -17.0 \times 10^{-22} \; C/m^2$
In the regions $I$ and $III$,the electric field $E$ is zero because the net charge enclosed by the Gaussian surface in these regions is zero.
In region $II$ (between the plates),the electric field $E$ is given by the relation:
$E = \frac{\sigma}{\varepsilon_0}$
Where $\varepsilon_0$ (permittivity of free space) $= 8.854 \times 10^{-12} \; C^2 N^{-1} m^{-2}$.
$E = \frac{17.0 \times 10^{-22}}{8.854 \times 10^{-12}} \approx 1.92 \times 10^{-10} \; N/C$
Thus,the electric field is $0$ in the outer regions and $1.92 \times 10^{-10} \; N/C$ between the plates.