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To find the electric flux density ((D)) at (R = 4.5) m in the presence of three charged cylindrical sheets, we need to consider the charge densities ((sigma)) and their respective positions. The given cylindrical sheets have charge densities (sigma = 5), (sigma = -2), and (sigma = -3) C/m² at radii (R = 2) m, (R = 4) m, and (R = 5) m, respectively.
In cylindrical coordinates, given a uniformly charged infinite line or cylindrical sheet, the electric field ((E)) outside the cylinder, by Gauss’s law, relates to the enclosed charge. However, for non-conducting sheets, we deal directly with the surface charge and determine the electric displacement (flux density, (D)), which in free space and for linear materials is directly proportional to the electric field, with (D = epsilon_0 E), where (epsilon_0) is the vacuum permittivity ((8.854 times 10^{-12}) C²/N·m²).
However, because we are dealing with charges on cylindrical sheets and not points or infinite lines and considering the superposition principle (if there are multiple charges, the total electric field is the vector sum of the fields created by each charge alone), the approach is streamlined to focusing on the total enclosed charge by those cylinders within
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