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prasanjit

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  1. Asked: August 27, 2024In: Education

    Gauss law can be evaluated in which coordinate system?

    prasanjit
    prasanjit Teacher
    Added an answer on September 21, 2024 at 9:35 pm

    Answer: d Explanation: The Gauss law exists for all materials. Depending on the Gaussian surface of the material, we take the coordinate systems accordingly. Suppose if the material is a coaxial cable, the Gaussian surface is in the form of cylinder. Thus we take Cylinder/Circular coordinate system.

    Answer: d
    Explanation: The Gauss law exists for all materials. Depending on the Gaussian surface of the material, we take the coordinate systems accordingly. Suppose if the material is a coaxial cable, the Gaussian surface is in the form of cylinder. Thus we take Cylinder/Circular coordinate system.

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  2. Asked: August 27, 2024In: Education

    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 = 6m.

    prasanjit
    prasanjit Teacher
    Added an answer on September 21, 2024 at 9:34 pm

    Answer: d Explanation: The radius R = 6m encloses all the three Gaussian cylinders. By Gauss law, ψ = Q D(2πRL) = σ(2πRL), D(2π X 6) = Q1 + Q2 + Q3 = σ1(2π X 2) + σ2(2π X 4) + σ3(2π X 5), here σ1 = 5, σ2 = -2 and σ3 = -3. We get D = -13/6 units.

    Answer: d
    Explanation: The radius R = 6m encloses all the three Gaussian cylinders.
    By Gauss law, ψ = Q

    D(2πRL) = σ(2πRL), D(2π X 6) = Q1 + Q2 + Q3 = σ1(2π X 2) + σ2(2π X 4) + σ3(2π X
    5), here σ1 = 5, σ2 = -2 and σ3 = -3. We get D = -13/6 units.

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  3. Asked: August 27, 2024In: Education

    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.

    prasanjit
    prasanjit Teacher
    Added an answer on September 21, 2024 at 9:33 pm

    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.

    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.

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  4. Asked: August 27, 2024In: Education

    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 = 3m.

    prasanjit
    prasanjit Teacher
    Added an answer on September 21, 2024 at 9:32 pm

    Answer: b Explanation: The radius is 3m, hence it will enclose one Gaussian cylinder of R = 2m. By Gauss law, ψ = Q D(2πRL) = σ(2πRL), D(2π X 3) = σ(2π X 2), Thus D = 10/3 units.

    Answer: b
    Explanation: The radius is 3m, hence it will enclose one Gaussian cylinder of R = 2m.
    By Gauss law, ψ = Q
    D(2πRL) = σ(2πRL), D(2π X 3) = σ(2π X 2), Thus D = 10/3 units.

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  5. Asked: August 27, 2024In: Education

    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 = 1m.

    prasanjit
    prasanjit Teacher
    Added an answer on September 21, 2024 at 9:31 pm

    Answer: a Explanation: Since 1m does not enclose any cylinder (three Gaussian surfaces of radius 2m, 4m, 5m exists), the charge density and charge becomes zero according to Gauss law. Thus flux density is also zero.

    Answer: a
    Explanation: Since 1m does not enclose any cylinder (three Gaussian surfaces of
    radius 2m, 4m, 5m exists), the charge density and charge becomes zero according
    to Gauss law. Thus flux density is also zero.

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  6. Asked: August 27, 2024In: Education

    Gauss law can be used to compute which of the following?

    prasanjit
    prasanjit Teacher
    Added an answer on September 21, 2024 at 8:59 pm

    Answer: c Explanation: Gauss law relates the electric flux density and the charge density. Thus it can be used to compute radius of the Gaussian surface. Permittivity and permeability are constants for a particular material.

    Answer: c
    Explanation: Gauss law relates the electric flux density and the charge density. Thus it
    can be used to compute radius of the Gaussian surface. Permittivity and permeability
    are constants for a particular material.

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  7. Asked: August 27, 2024In: Education

    Gauss law for magnetic fields is given by

    prasanjit
    prasanjit Teacher
    Added an answer on September 21, 2024 at 8:58 pm

    Answer: b Explanation: The divergence of magnetic flux density is always zero. This is called Gauss law for magnetic fields. It implies the non-existence of magnetic monopoles in any magnetic field.

    Answer: b
    Explanation: The divergence of magnetic flux density is always zero. This is called
    Gauss law for magnetic fields. It implies the non-existence of magnetic monopoles in
    any magnetic field.

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  8. Asked: August 27, 2024In: Education

    Gauss law cannot be used to find which of the following quantity?

    prasanjit
    prasanjit Teacher
    Added an answer on September 21, 2024 at 8:58 pm

    Answer: d Explanation: Permittivity is constant for a particular material(say permittivity of water is 1). It cannot be determined from Gauss law, whereas the remaining options can be computed from Gauss law.

    Answer: d
    Explanation: Permittivity is constant for a particular material(say permittivity of water is
    1). It cannot be determined from Gauss law, whereas the remaining options can be
    computed from Gauss law.

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      • 0
  9. Asked: August 27, 2024In: Education

    Find the potential due to a charged ring of density 2 units with radius 2m and the point at which potential is measured is at a distance of 1m from the ring.

    prasanjit
    prasanjit Teacher
    Added an answer on September 21, 2024 at 8:55 pm

    Answer: d Explanation: The potential due to a charged ring is given by λa/2εr, where a = 2m and r = 1m. We get V = 72π volts.

    Answer: d
    Explanation: The potential due to a charged ring is given by λa/2εr, where a = 2m and r = 1m. We get V = 72π volts.

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      • 0
  10. Asked: August 27, 2024In: Education

    The potential of a coaxial cylinder with charge density 1 unit , inner radius 1m and outer cylinder 2m is (in 109 )

    prasanjit
    prasanjit Teacher
    Added an answer on September 21, 2024 at 8:54 pm

    Answer: c Explanation: The potential of a coaxial cylinder will be ρl ln(b/a)/2πε, where ρl = 1, b =2m and a = 1m. We get V = 12.47 X 109 volts.

    Answer: c
    Explanation: The potential of a coaxial cylinder will be ρl ln(b/a)/2πε, where ρl = 1, b =2m and a = 1m. We get V = 12.47 X 109 volts.

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