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Three charged cylindrical sheets are present in three spaces with σ = 5 at R = 2m, σ = -2 at R = 4m and σ =-3 at R = 5m. Find the flux density at R = 4.5m.
- a)4/4.5
- b)3/4.5
- c)2/4.5
- d)1/4.5
Correct answer is option 'C'. Can you explain this answer?
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Three charged cylindrical sheets are present in three spaces with &sig...
Answer: c
Explanation: The Gaussian cylinder of R = 4.5m encloses sum of charges of two cylinders (R = 2m and R = 4m).
By Gauss law, ψ = Q
D(2πRL) = σ(2πRL), D(2π X 4.5) = Q1 + Q2 = σ1(2π X 2) + σ2(2π X 4), here σ1 = 5 and σ2 = -2. We get D = 2/4.5 units.
Explanation: The Gaussian cylinder of R = 4.5m encloses sum of charges of two cylinders (R = 2m and R = 4m).
By Gauss law, ψ = Q
D(2πRL) = σ(2πRL), D(2π X 4.5) = Q1 + Q2 = σ1(2π X 2) + σ2(2π X 4), here σ1 = 5 and σ2 = -2. We get D = 2/4.5 units.
Most Upvoted Answer
Three charged cylindrical sheets are present in three spaces with &sig...
A common center. The first sheet has a radius of r1 and a charge density of ρ1. The second sheet has a radius of r2 and a charge density of ρ2. The third sheet has a radius of r3 and a charge density of ρ3. The electric field at a point P in space, located a distance x from the common center, is given by E = E1 + E2 + E3, where E1, E2, and E3 are the electric fields due to the first, second, and third sheets, respectively.
The electric field due to a charged cylindrical sheet can be calculated using Gauss's law. Gauss's law states that the electric field through a closed surface is proportional to the total charge enclosed by that surface. For a cylindrical sheet, the electric field is perpendicular to the surface and has a constant magnitude.
To calculate the electric field due to the first sheet, we can consider a cylindrical Gaussian surface of radius r1 and length L, centered at the common center. The electric field through this surface is given by E1 = (ρ1/ε0) * (L/2πr1), where ε0 is the permittivity of free space.
Similarly, the electric field due to the second sheet can be calculated using a Gaussian surface of radius r2 and length L. The electric field through this surface is given by E2 = (ρ2/ε0) * (L/2πr2).
Finally, the electric field due to the third sheet can be calculated using a Gaussian surface of radius r3 and length L. The electric field through this surface is given by E3 = (ρ3/ε0) * (L/2πr3).
Summing up these three electric fields, we get the total electric field at point P as E = E1 + E2 + E3.
It is important to note that the direction of the electric field at point P depends on the direction of the individual electric fields due to the three sheets. If the charge densities on the sheets have the same sign, the electric fields due to the sheets will have the same direction, resulting in a stronger net electric field at point P. If the charge densities on the sheets have opposite signs, the electric fields due to the sheets will have opposite directions, resulting in a weaker net electric field at point P.
The electric field due to a charged cylindrical sheet can be calculated using Gauss's law. Gauss's law states that the electric field through a closed surface is proportional to the total charge enclosed by that surface. For a cylindrical sheet, the electric field is perpendicular to the surface and has a constant magnitude.
To calculate the electric field due to the first sheet, we can consider a cylindrical Gaussian surface of radius r1 and length L, centered at the common center. The electric field through this surface is given by E1 = (ρ1/ε0) * (L/2πr1), where ε0 is the permittivity of free space.
Similarly, the electric field due to the second sheet can be calculated using a Gaussian surface of radius r2 and length L. The electric field through this surface is given by E2 = (ρ2/ε0) * (L/2πr2).
Finally, the electric field due to the third sheet can be calculated using a Gaussian surface of radius r3 and length L. The electric field through this surface is given by E3 = (ρ3/ε0) * (L/2πr3).
Summing up these three electric fields, we get the total electric field at point P as E = E1 + E2 + E3.
It is important to note that the direction of the electric field at point P depends on the direction of the individual electric fields due to the three sheets. If the charge densities on the sheets have the same sign, the electric fields due to the sheets will have the same direction, resulting in a stronger net electric field at point P. If the charge densities on the sheets have opposite signs, the electric fields due to the sheets will have opposite directions, resulting in a weaker net electric field at point P.
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Three charged cylindrical sheets are present in three spaces with σ = 5 at R = 2m, σ = -2 at R = 4m and σ =-3 at R = 5m. Find the flux density at R = 4.5m.a)4/4.5b)3/4.5c)2/4.5d)1/4.5Correct answer is option 'C'. Can you explain this answer?
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Three charged cylindrical sheets are present in three spaces with σ = 5 at R = 2m, σ = -2 at R = 4m and σ =-3 at R = 5m. Find the flux density at R = 4.5m.a)4/4.5b)3/4.5c)2/4.5d)1/4.5Correct answer is option 'C'. Can you explain this answer? for Electrical Engineering (EE) 2024 is part of Electrical Engineering (EE) preparation. The Question and answers have been prepared according to the Electrical Engineering (EE) exam syllabus. Information about Three charged cylindrical sheets are present in three spaces with σ = 5 at R = 2m, σ = -2 at R = 4m and σ =-3 at R = 5m. Find the flux density at R = 4.5m.a)4/4.5b)3/4.5c)2/4.5d)1/4.5Correct answer is option 'C'. Can you explain this answer? covers all topics & solutions for Electrical Engineering (EE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Three charged cylindrical sheets are present in three spaces with σ = 5 at R = 2m, σ = -2 at R = 4m and σ =-3 at R = 5m. Find the flux density at R = 4.5m.a)4/4.5b)3/4.5c)2/4.5d)1/4.5Correct answer is option 'C'. Can you explain this answer?.
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