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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?
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 $A , \sigma=17.0 \times 10^{-22} \,C / m ^{2}$
Charge density of plate $B , \sigma=-17.0 \times 10^{-22} \,C / m ^{2}$
In the regions, $I$ and $III$, 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,
$E=\frac{\sigma}{\varepsilon_{0}}$
Where, $\varepsilon_{0}=$ Permittivity of free space $=8.854 \times 10^{-12}\, N ^{-1} \,C ^{2} \,m ^{-2}$
$=1.92 \times 10^{-10} \,N / C$
$E=\frac{17.0 \times 10^{-22}}{8.854 \times 10^{-12}}$
Therefore, electric field between the plates is $1.92 \times 10^{-10}\; N / C$
