Question

Question 1.24: 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 −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?

Class 12 - Physics - Electric Charges And Fields Electric Charges And Fields Page-48

Answer 1

see the diagram , here two metal plates are parallel and close to each other . Let A and B are two metal plates and $E_A$ and $E_B$ denote electric field produce due to metal plate A and B .
as both the plates have same surface charge density e.g., $\sigma$

please see I and II areas , you observed $E_A=E_B=\frac{\sigma}{2\epsilon_0}$

(a)$\bf{in\:region-I,E_I=E_B-E_A=\frac{\sigma}{2\epsilon_0}-\frac{\sigma}{2\epsilon_0}\:or,E_I=0}$
(b) $\bf{in\:region-III,E_{III}=E_A-E_B=\frac{\sigma}{2\epsilon_0}-\frac{\sigma}{2\epsilon_0}\:or,E_{III}=0}$
(c) $\bf{in\:region-II,E_{II}=E_A+E_B=\frac{\sigma}{2\epsilon_0}+\frac{\sigma}{2\epsilon_0}=\frac{\sigma}{\epsilon_0}}$
so, electric field between plates = 17 × 10^-22/8.85 × 10^-12 = 1.92 × 10^-10 N/C

Answer 2

$\bigstar$ 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 $\sf 17.0 \times 10^{-22} \: C/m^{2}$

$\bigstar$ Solution:

The charge density of plate A can be calculated as,

$\sf \sigma = 17.0 \times 10^{-22} \: C/m^{2}$

Similarly, the charge density of plate B can be calculated as,

$\sf \sigma =-17.0 \times 10^{-22} \: C/m^{2}$

In regions, A and B, electric field E is zero because the charge is not enclosed by the respective plates.

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

$\sf E=\dfrac{\sigma}{\epsilon_0}$

Now, here

$\sf \epsilon =Permitivity \: of \: free \: space = 8.854 \times 10 - 12\: N^{ - 1} C^{ 2} \: m^{-2}$

$\sf E=\dfrac{17.0 \times 10^{-22}}{8.854 \times 10^{-12}} = 1.92 \times 10^{-10} \: N/C$

Therefore, electric field between the plates is $\sf{\bf{1.92 \times 10^{-10} \: N/C}}$