Class 12 Physics Electric Charges and Field: NCERT Solutions for 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/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 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/m2
Charge density of plate B, σ = −17.0 × 10−22 C/m2
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−1C2 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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5 Comment(s) on this Question
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
The formula should be sigma/2absolutenot
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
Yes marine is right
field due to both plates will be added answer will be 3.84*10^-10
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