This page focuses on the detailed Electric Charges and Field question answers for Class 12 Physics Electric Charges and Field, addressing the 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 × 10−22 C/m2. 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?'. The solution provides a thorough breakdown of the question, highlighting key concepts and approaches to arrive at the correct answer. This easy-to-understand explanation will help students develop better problem-solving skills, reinforcing their understanding of the chapter and aiding in exam preparation.

Question 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?

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 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, σ = 17.0 × 10^{−22} C/m^{2}

Charge density of plate B, σ = −17.0 × 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,

Where,

∈0 = Permittivity of free space = 8.854 × 10^{−12} N^{−1}C^{2} m^{−2}

∴

= 1.92 × 10^{−10} N/C

Therefore, electric field between the plates is 1.92 × 10^{−10 }N/C.

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adeel
2019-05-24 12:16:18

marine is wrong as the original formula is sigma by 2 epsilon not but as he is using 2 plates he used sigma by epsilon not

Anu
2019-04-03 23:18:25

The formula should be sigma/2absolutenot

Nidhi
2018-01-04 20:03:27

Electric field between the plates will be 1.92 Ã 10â10 N/C. This is because between the plates the electric field will be towards negative plate that will results what shown above

Vishal yadav
2017-09-21 08:16:19

Yes marine is right

marine
2016-02-04 21:30:43

field due to both plates will be added answer will be 3.84*10^-10

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