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A circular disc of radius 5 m with a charge density of ρs = 12 sinϕ μC/m2 is enclosed by a surface S. The net flux crossing the S is.
- a)1 μC
- b)2 μC
- c)0
- d)5 μC
Correct answer is option 'C'. Can you explain this answer?
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A circular disc of radius 5 m with a charge density of ρs = 12 sinϕ μ...
The net flux crossing,
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A circular disc of radius 5 m with a charge density of ρs = 12 sinϕ μ...
To find the net flux crossing the surface S, we can use Gauss's Law, which states that the net electric flux through any closed surface is equal to the total charge enclosed by that surface divided by the permittivity of free space (ε0).
Given:
Radius of the circular disc (r) = 5 m
Charge density (ρs) = 12 sinϕ μC/m^2
We need to find the net flux crossing the surface S.
Step 1: Calculate the total charge enclosed by the surface S
The total charge enclosed by the surface S can be found by integrating the charge density over the surface area of the circular disc.
Let's assume the charge enclosed by an infinitesimally small area element dA on the disc is dq. The charge dq can be calculated as:
dq = ρs * dA
The charge enclosed by the surface S is given by the integral of dq over the entire surface area of the circular disc.
Q = ∫dq
To perform the integration, we need to express the infinitesimal area element dA in terms of the polar coordinates (r, θ) on the disc.
dA = r * dr * dθ
The limits of integration for r are 0 to 5 m, and for θ are 0 to 2π (since it's a full circle).
Therefore, the charge enclosed by the surface S is given by:
Q = ∫(0 to 2π)∫(0 to 5)ρs * r * dr * dθ
= ∫(0 to 2π)∫(0 to 5)(12 sinϕ) * r * dr * dθ
Step 2: Calculate the net flux crossing the surface S
The net flux crossing the surface S can be calculated using Gauss's Law.
Φ = Q / ε0
Since the permittivity of free space (ε0) is a constant, we can ignore it for the purpose of comparison.
So, the net flux crossing the surface S is given by:
Φ = Q
Step 3: Evaluate the integral to find the charge enclosed by the surface S
Now, we need to evaluate the integral to find the charge enclosed by the surface S.
Q = ∫(0 to 2π)∫(0 to 5)(12 sinϕ) * r * dr * dθ
Let's evaluate the integral step by step:
∫(0 to 2π)∫(0 to 5)(12 sinϕ) * r * dr * dθ
= 12 ∫(0 to 2π)∫(0 to 5)sinϕ * r * dr * dθ
= 12 ∫(0 to 2π)sinϕ * (∫(0 to 5)r * dr) * dθ
= 12 ∫(0 to 2π)sinϕ * (r^2 / 2) [from 0 to 5] * dθ
= 12 ∫(0 to 2π)(25 / 2)sinϕ * dθ
= 12 * (25 / 2) ∫(0 to 2π)sinϕ * dθ
=
Given:
Radius of the circular disc (r) = 5 m
Charge density (ρs) = 12 sinϕ μC/m^2
We need to find the net flux crossing the surface S.
Step 1: Calculate the total charge enclosed by the surface S
The total charge enclosed by the surface S can be found by integrating the charge density over the surface area of the circular disc.
Let's assume the charge enclosed by an infinitesimally small area element dA on the disc is dq. The charge dq can be calculated as:
dq = ρs * dA
The charge enclosed by the surface S is given by the integral of dq over the entire surface area of the circular disc.
Q = ∫dq
To perform the integration, we need to express the infinitesimal area element dA in terms of the polar coordinates (r, θ) on the disc.
dA = r * dr * dθ
The limits of integration for r are 0 to 5 m, and for θ are 0 to 2π (since it's a full circle).
Therefore, the charge enclosed by the surface S is given by:
Q = ∫(0 to 2π)∫(0 to 5)ρs * r * dr * dθ
= ∫(0 to 2π)∫(0 to 5)(12 sinϕ) * r * dr * dθ
Step 2: Calculate the net flux crossing the surface S
The net flux crossing the surface S can be calculated using Gauss's Law.
Φ = Q / ε0
Since the permittivity of free space (ε0) is a constant, we can ignore it for the purpose of comparison.
So, the net flux crossing the surface S is given by:
Φ = Q
Step 3: Evaluate the integral to find the charge enclosed by the surface S
Now, we need to evaluate the integral to find the charge enclosed by the surface S.
Q = ∫(0 to 2π)∫(0 to 5)(12 sinϕ) * r * dr * dθ
Let's evaluate the integral step by step:
∫(0 to 2π)∫(0 to 5)(12 sinϕ) * r * dr * dθ
= 12 ∫(0 to 2π)∫(0 to 5)sinϕ * r * dr * dθ
= 12 ∫(0 to 2π)sinϕ * (∫(0 to 5)r * dr) * dθ
= 12 ∫(0 to 2π)sinϕ * (r^2 / 2) [from 0 to 5] * dθ
= 12 ∫(0 to 2π)(25 / 2)sinϕ * dθ
= 12 * (25 / 2) ∫(0 to 2π)sinϕ * dθ
=
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A circular disc of radius 5 m with a charge density of ρs = 12 sinϕ μC/m2 is enclosed by a surface S. The net flux crossing the S is.a)1 μCb)2 μCc)0d)5 μCCorrect answer is option 'C'. Can you explain this answer? for GATE 2024 is part of GATE preparation. The Question and answers have been prepared according to the GATE exam syllabus. Information about A circular disc of radius 5 m with a charge density of ρs = 12 sinϕ μC/m2 is enclosed by a surface S. The net flux crossing the S is.a)1 μCb)2 μCc)0d)5 μCCorrect answer is option 'C'. Can you explain this answer? covers all topics & solutions for GATE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A circular disc of radius 5 m with a charge density of ρs = 12 sinϕ μC/m2 is enclosed by a surface S. The net flux crossing the S is.a)1 μCb)2 μCc)0d)5 μCCorrect answer is option 'C'. Can you explain this answer?.
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