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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
To find the electric flux density (D) at (R = 6m) for a configuration involving multiple charged cylindrical sheets, we employ the concept of Gauss’s law. Gauss’s law in the context of electric flux density states that the net electric flux through a closed surface is equal to the charge enclosed by that surface.
Given:
1. (sigma = 5 , text{C/m}^2) at (R = 2m)
2. (sigma = -2 , text{C/m}^2) at (R = 4m)
3. (sigma = -3 , text{C/m}^2) at (R = 5m)
And we are asked to find the flux density (D) at (R = 6m).
To start, we calculate the charge per unit length (lambda) on each cylindrical sheet using the surface charge density (sigma) and the relation (lambda = sigma cdot 2pi R) because the charge per unit length for a cylindrical shell of radius (R) is the product of the surface charge density and the circumference of the cylinder.
1. For (R = 2m): (lambda_1 = 5 cdot 2pi (2) = 20pi , text{C/m})
2. For (R = 4m):