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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
To find the electric flux density (D) at (R = 3m) in the arrangement of charged cylindrical sheets described, we use Gauss’s law in its integral form which relates the electric flux through a closed surface to the charge enclosed by that surface. For a cylindrical symmetry, the electric field (E) (and thus the electric flux density (D), since (D = epsilon_0 E) in vacuum or air where (epsilon_0) is the permittivity of free space) at a distance (R) from the axis is due to the enclosed charge per unit length ((lambda)) divided by (2piepsilon_0 R).
Given, we have three cylindrical sheets:
– The first with a surface charge density (sigma = 5 , text{C/m}^2) at (R = 2m),
– The second with (sigma = -2 , text{C/m}^2) at (R = 4m),
– The third with (sigma = -3 , text{C/m}^2) at (R = 5m).
Given charges are surface charge densities on cylindrical sheets; thus, the enclosed charge by a Gaussian cylinder of radius (R = 3m) only includes the charge from the first cylindrical sheet ((R = 2m)).
To calculate the total enclosed charge per unit length (