Surface charge density is defined as the charge per unit surface area of surface charge
distribution. i.e.,
dq
dS
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 having magnitude of 1.70 × 10–22 Cm–2
as
shown below.
The intensity of electric field at a point is
0
E
, where, 0 = permittivity of free space.
(i) E in the outer region of the first plate is
(a) 17 × 10–22 N/C (b) 1.5 × 10–15 N/C
(c) 1.9 × 10–10 N/C (d) Zero
(ii) E in the outer region of the second plate is
(a) 17 × 10–22 N/C (b) 1.5 × 10–15 N/C
(c) 1.9 × 10–10 N/C (d) zero
(iii) E between the plates is
(a) 17 × 10–22 N/C (b) 1.5 × 10–15 N/C
(c) 1.9 × 10–10 N/C (d) zero
(iv) The ratio of E from right side of B at distance 2 cm and 4 cm, respectively is
(a) 1 : 2 (b) 2 : 1
(c) 1 : 1 (d) 1 :
2
(v) In order to estimate the electric field due to a thin finite plane metal plate, the
Gaussian surface considered is
(a) spherical (b) cylindrical
(c) straight line (d) none of these
Answers
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
Electric field in regions can be found with the help of Gauss Law. In the regions, I and III, electric field E is zero. This is because charge is not enclosed by the Gaussian surfaces of the plates.
Electric field E in region II is given by the relation,
E=
∈
0
σ
Where,
∈
0
= Permittivity of free space =8.854×10
−12
N
−1
C
2
m
−2
∴E=
8.854×10
−12
17.0×10
−22
=1.92×10
−10
N/C
Therefore, electric field between the plates is 1.92×10
−10
N/C.
Answer:
Gaussian surface
Explanation:
Given that:
Surface charge density is charge per unit surface area of surface charge distribution.
To find:
The value of each surface =?
Solution:
E in the outer region of the first plate and second plate is
Answer:
Let us consider, A and B as two parallel plates which are close to each other. The outer plate A region is I, outer Plate B region is III, and let the region between plates A and B, is II.
Let us assume that,
Charge of density of plate A is, σ = 17.0 × 10−22 C/m2
Charge density of plate B is, σ = -17.0 × 10−22 C/m2
Let the regions, I & III of an electric field E is equal to zero. It is due the charge not be being enclosed with the required respective plates.
Now, let the Electric field E in II region be the relation,
E = σ/ E0
Here, E0 is the permittivity of the free space is 8.854x10-12 N-1C2m-2
Therefore, we know that,
E = 17.-x10-22 / 8.854x10-12
E = 1.92 x 10-10 N/C
Hence, the electric field between the plates is 1.92x10 -10 N/C
We can assume that ,In order to estimate the electric field due to a thin finite plane metal plate, the Gaussian surface considered is as an:
Answer:
Cylindrical
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