**Question:**

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 $17.0 \times 10^{-22} \mathrm{C} / \mathrm{m}^{2}$. What is $\mathrm{E}$ :

(a) in the outer region of the first plate,

(b) in the outer region of the second plate, and

(c) between the plates?

**Solution:**

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 I, outer region of plate B is labelled as III, and the region between the plates, A and B, is labelled as II.

Charge density of plate $\mathrm{A}, \sigma=17.0 \times 10^{-22} \mathrm{C} / \mathrm{m}^{2}$

Charge density of plate $B, \sigma=-17.0 \times 10^{-22} \mathrm{C} / \mathrm{m}^{2}$

In the regions, I and III, electric field *E* is zero. This is because charge is not enclosed by the respective plates.

Electric field *E* in region II is given by the relation,

$E=\frac{\sigma}{\epsilon_{0}}$

Where,

$\epsilon_{0}=$ Permittivity of free space $=8.854 \times 10^{-12} \mathrm{~N}^{-1} \mathrm{C}^{2} \mathrm{~m}^{-2}$

$\therefore E=\frac{17.0 \times 10^{-22}}{8.854 \times 10^{-12}}$

$=1.92 \times 10^{-10} \mathrm{~N} / \mathrm{C}$

Therefore, electric field between the plates is $1.92 \times 10^{-10} \mathrm{~N} / \mathrm{C}$.