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To find the electric flux density (usually described by the symbol (D)) of a line charge using a cylindrical Gaussian surface, we need to integrate the charge density over the line charge’s length within the cylinder to find the total charge, but there’s a slight confusion in your question as it seems to combine concepts.
Given:
– The radius of the cylindrical Gaussian surface (r = 2) meters.
– The linear charge density (lambda = 3.14) units (since it’s stated as “charge density,” we’re interpreting it as linear despite the potential for different interpretations depending on context; units typically would be coulombs per meter ([C/m]) for a line charge).
The procedure involves using Gauss’s law, which in integral form states that the electric flux ((Phi_E)) through a closed surface is equal to the charge ((Q)) enclosed by the surface divided by the permittivity of free space ((varepsilon_0)):
[
Phi_E = frac{Q}{varepsilon_0}
]
For a line charge with linear charge density (lambda), the total charge (Q) enclosed by a cylindrical Gaussian surface of length (L) is:
[
Q = lambda L
]
However, the electric flux density (D) is related to the electric field (E) via the relationship:
[
D = varepsilon_0 E
d
Explanation: The electric field of a line charge is given by, E = λ/(2περ), where ρ is the
radius of cylinder, which is the Gaussian surface and λ is the charge density. The
density D = εE = λ/(2πρ) = 3.14/(2π X 2) = 1/4 = 0.25.