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Physics Question on Electric Charges and Coulomb's Law

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 17.0×10−22Cm−217.0 × 10^{−22} Cm^{-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?

Answer

The situation is represented in the following figure.

A and B are two parallel plates close to each other. Outer region of plate A is labelled as II, outer region of plate B is labelled as IIIIII, and the region between the plates, A and B, is labelled as IIII.
Charge density of plate A, σ= 17.0×10−22Cm−2σ =\, 17.0 × 10^{−22} Cm^{-2}
Charge density of plate B, σ=−17.0×10−22Cm−2σ = −17.0 × 10^{−22} Cm^{-2}
In the regions, II and IIIIII, electric field E is zero. This is because charge is not enclosed by the respective plates.
Electric field E in region IIII is given by the relation,
E=σε0E = \frac{σ }{ ε_0}
Where,
ε0ε_0 = Permittivity of free space = 8.854×10−12N−1C2m−28.854 × 10^{-12} N^{-1}C^2m^{-2}
E=17.0×10−228.854×10−12E = \frac{17.0 × 10^{-22} }{ 8.854 × 10^{-12}}
=1.92×10−10NC−1= 1.92 × 10^{-10} NC^{-1}
Therefore, electric field between the plates is 1.92×10−10NC−11.92 × 10^{-10} NC^{-1}.